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  华东师范大学学报(自然科学版)  2018 Issue (3): 30-37  DOI: 10.3969/j.issn.1000-5641.2018.03.004
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引用本文  

刘金梦, 宋卫东. 三参数射影平坦芬斯勒度量的构造[J]. 华东师范大学学报(自然科学版), 2018, (3): 30-37. DOI: 10.3969/j.issn.1000-5641.2018.03.004.
LIU Jin-meng, SONG Wei-dong. The explicit structure of projectively flat Finsler metrics with three parameters[J]. Journal of East China Normal University (Natural Science), 2018, (3): 30-37. DOI: 10.3969/j.issn.1000-5641.2018.03.004.

基金项目

国家自然科学基金(11371032);安徽省自然科学基金重点项目(KJ2017A795)

第一作者

刘金梦, 女, 硕士研究生, 研究方向为微分几何.E-mail:1813278900@qq.com

通信作者

宋卫东, 男, 教授, 硕士生导师, 研究方向为微分几何.E-mail:swd56@sina.com

文章历史

收稿日期:2017-04-24
三参数射影平坦芬斯勒度量的构造
刘金梦, 宋卫东     
安徽师范大学 数学计算机科学学院, 安徽 芜湖 241000
摘要:本文主要研究射影平坦芬斯勒度量,构造了一类含三参数的芬斯勒度量,并且得到了该度量是射影平坦的充要条件.另外,还给出了该度量有关旗曲率的表达式.
关键词芬斯勒度量    射影平坦    射影因子    
The explicit structure of projectively flat Finsler metrics with three parameters
LIU Jin-meng, SONG Wei-dong    
School of Mathematics & Computer Science, Anhui Normal University, Wuhu Anhui 241000, China
Abstract: In this paper, projectively flat Finsler metrics are considered. A class of projectively flat Finsler metrics with three parameters are formed. Moreover, the sufficient and necessary conditions for the measurement to be considered projectively flat was obtained. In particular, the flag curvature expression of projectively flat Finsler metrics are presented.
Key words: Finsler metrics    projectively flat    projectively factor    
0 引言

在芬斯勒几何中, 有不少具有标量旗曲率的芬斯勒度量是射影平坦的, 例如著名的Berwald度量[1]

$ F=\dfrac{\big(\sqrt{ \langle x, y\rangle^{2}+ |y|^{2}(1-|x|^{2})}+\langle x, y\rangle\big)^{2}}{(1-|x|^{2})^{2}\sqrt{ \langle x, y\rangle^{2}+ |y|^{2}(1-|x|^{2})}}. $

它是$B^{n}(r)\subset {\mathbb{R}}^{n}$上旗曲率$K=0$的射影平坦芬斯勒度量[2].

Berwald度量是由黎曼度量$\alpha :=\sqrt{a_{ij}(x)y^{i}y^{j}}$, 1-形式$\beta :=b_{i}y^{i}$$M^{n}$上的光滑正函数$\phi=\phi(s), s=\frac{\beta}{\alpha}$构成的$(\alpha, \beta)$度量[3], 由$(\alpha, \beta)$定义的芬斯勒度量可以表示成如下形式

$ F(x, y)=\alpha\phi(s), ~ s=\dfrac{\beta}{\alpha}. $ (1)

如果$\phi=\phi(s)$进一步满足

$ \phi(s) >0, ~\phi-s\phi_{s}+(b^{2}-s^{2})\phi_{ss}>0, ~|s|\leqslant b<b_{0}, $

其中$b=\|\beta\|_{\alpha}$, 则由式$(1)$定义的芬斯勒度量$F$称为$(\alpha, \beta)$度量.在$(\alpha, \beta)$度量基础上, 余昌涛等人[4-5]又提出了广义$(\alpha, \beta)$度量的概念, 定义为如下形式

$ F(x, y)=\alpha\phi(b^{2}, s), ~s=\dfrac{\beta}{\alpha}, $

其中$\alpha :=\sqrt{a_{ij}(x)y^{i}y^{j}}$为黎曼度量, $\beta :=b_{i}y^{i}$为1-形式.当$\phi(b^{2}, s)$只与$s$有关时, $(\alpha, \beta)$度量即为广义的$(\alpha, \beta)$度量.广义的$(\alpha, \beta)$度量是比$(\alpha, \beta)$度量应用更为广泛的一类芬斯勒度量.Berwald度量是广义$(\alpha, \beta)$度量的重要组成部分, 它可以表示成

$ F(x, y)=\alpha\phi(b^{2}, s)=\alpha(\sqrt{1+b^{2}}+s)^{2}, $

其中

$ \alpha=\dfrac{\sqrt{ \langle x, y\rangle^{2}+ |y|^{2}(1-|x|^{2})}}{1-|x|^{2}}, ~\beta=\dfrac{ \langle x, y\rangle}{(1-|x|^{2})^{\frac{3}{2}}}, $
$ s=\dfrac{\beta}{\alpha}, ~b^{2}=\dfrac{ |x|^{2}}{1-|x|^{2}}. $

通过对Berwald度量的研究, 文献[6]给出的一类芬斯勒度量同样可以表示为

$ F(x, y)=\alpha\phi(b^{2}, s)=\alpha(\sqrt{1+b^{2}}+s)^{2}, s=\dfrac {\beta}{\alpha}, $

其中

$ \alpha=\dfrac{\sqrt{|y|^{2}(1+\mu|x|^{2})-\mu\langle x, y\rangle^{2}}}{1+\mu|x|^{2}}, $
$ \beta=\dfrac{ \lambda\langle x, y\rangle+(1+\mu|x|^{2})\langle a, y\rangle-\mu\langle x, y\rangle\langle a, x\rangle}{(1+\mu|x|^{2})^{\frac{3}{2}}}, $
$ b^{2}=\dfrac{ \lambda^{2}|x|^{2}}{1+\mu|x|^{2}}+\dfrac{ 2\lambda\langle a, x\rangle}{1+\mu|x|^{2}}-\dfrac{\mu\langle a, x\rangle^{2}}{1+\mu|x|^{2}}+|a|^{2}. $

本文通过对以上芬斯勒度量的研究, 探讨${\mathbb{R}}^{n}$的开子集上一类含三参数的芬斯勒度量的解析构造[7].设$\xi$是任意常数, $\Omega=B^{n}(r)\subset {\mathbb{R}}^{n}$, 当$\xi<0$时, $r=\frac{1}{\sqrt{-\xi}}$; 当$\xi\geqslant0$时, $r=+\infty$, $|\cdot|$$\langle, \rangle$${\mathbb{R}}^{n}$上的标准范数和内积.设$F=\alpha\phi(b^{2}, s):~T\Omega\rightarrow[0, +\infty)$,

$ \alpha=\dfrac{\sqrt{\varepsilon|y|^{2}(1+\xi|x|^{2})+\kappa^{2}\langle x, y\rangle^{2}}}{1+\xi|x|^{2}}, $ (2)
$ \beta=\dfrac{ \kappa\langle x, y\rangle+\varepsilon(1+\xi|x|^{2})\langle a, y\rangle+\kappa^{2}\langle x, y\rangle\langle a, x\rangle}{(1+\xi|x|^{2})^{\frac{3}{2}}}, $ (3)

其中$\kappa$是任意常数, $\varepsilon$是任意正数, $a\in {\mathbb{R}}^{n}(|a|<1)$是常向量.

通过对式(2)和式(3)的研究, 构造了一类新的芬斯勒度量.

定理    设$F=\alpha\phi(b^{2}, s):~T\Omega\rightarrow[0, +\infty)$是由式(2)和式(3)给出的函数, 定义函数$\phi(b^{2}, s)= (\sqrt{1+b^{2}}+s)^{2}, s=\frac{\beta}{\alpha}$, 则以下结论成立.

(1) 若$\kappa^{2}+\varepsilon\xi=0$, 则

$ b^{2}=\varepsilon|a|^{2}+\dfrac{(1+\kappa\langle a, x\rangle)^{2}}{1+\xi|x|^{2}}-1. $

(2) $F$是射影平坦芬斯勒度量的充要条件是$\kappa^{2}+\varepsilon\xi=0.$

(3) 当$\kappa^{2}+\varepsilon\xi=0$时, $F$具有标量旗曲率, 且标量旗曲率为

$ K=\dfrac{1}{\sqrt{1+b^{2}}(\sqrt{1+b^{2}}+s)^{3}}\Big(\dfrac {\xi}{\varepsilon}+\dfrac{c^{2}}{1+b^{2}}\Big), $

其中

$ c^{2}=-\dfrac{\xi(1+\kappa\langle a, x\rangle)^{2}}{\varepsilon(1+\xi|x|^{2})}. $

    当$\xi=-1, \varepsilon=1, \kappa^{2}=1$时, $F$为广义的Berwald度量.

1 预备知识

定义1    设$F$$M^{n}$上的一个芬斯勒度量, 如果它可以表示为$F =\alpha\phi(b^{2}, s), s=\frac{\beta}{\alpha}$, 其中$\|\beta\|_{\alpha}\leqslant b_{0}$, $\phi=\phi(b^{2}, s)$是正光滑函数, 则$F$称为广义的$(\alpha, \beta)$度量.

引理1[6]    设$M$是一个$n$维流形, 如果对任意的黎曼度量$\alpha$和1-形式$\beta$, $b=\|\beta\|_{\alpha}<b_{0}$, $TM$上的函数$F =\alpha\phi(b^{2}, s), s=\frac{\beta}{\alpha}$, 是一个芬斯勒度量当且仅当$\phi=\phi(b^{2}, s)$是光滑正函数且满足

$ \phi>0, \phi-s\phi_{s}+(b^{2}-s^{2})\phi_{ss}>0, $ (4)

其中$s$$b$是满足$|s|\leqslant b<b_{0}$的任意实数.

引理2[8]    设$M$是一个$n$维流形, $F =\alpha\phi(b^{2}, s)$$M^{n}$上广义的$(\alpha, \beta)$度量, 则$F$是局部射影平坦芬斯勒度量, 如果$F$满足以下条件

(1) 函数$\phi(b^{2}, s)$满足下列方程

$ \phi_{ss}=2(\phi_{b}-s\phi_{bs}), $

其中$\phi_{b}$表示$\phi$$b^{2}$的一次偏导.

(2) $\alpha$是射影平坦的, $\beta$是闭的, 且$\beta$关于$\alpha$共形.

引理3[6]    设广义的$(\alpha, \beta)$度量$F =\alpha\phi(b^{2}, s), s=\frac{\beta}{\alpha}$, 是射影平坦芬斯勒度量, 则它的射影因子$P$可以表示为

$ P=\dfrac{2\alpha^{-1}(\phi-s\phi_{s})G_{\alpha}^{m}y_{m}+\phi_{s} (2b_{m}G_{\alpha}^{m}+r_{00})+2\alpha\phi_{b}(r_{0}+s_{0})}{2F}, $

其中$G_{\alpha}^{m}$是关于$\alpha$的测地系数, $r_{00}=r_{ij}y^{i}y^{j}, r_{0}=b^{j}r_{ij}y^{i}, s_{0}=b^{j}s_{ij}y^{i}, G_{\alpha}^{m}y_{m}=\frac{\alpha}{2}\alpha_{x^{k}}y^{k}$.

引理4[9]    设$F=F(x, y)$$U\subset {\mathbb{R}}^{n}$上的芬斯勒度量, 则$F=F(x, y)$是射影平坦芬斯勒的充要条件是$F$满足下面的方程

$ F_{x^{k}y^{l}}y^{k}-F_{x^{l}}=0. $

定义2[10]    在开集$U\in {\mathbb{R}}^{n}$上的芬斯勒流形$(M, F)$, 若$F$的所有测地线在$U$内部都是直线, 则称$F$是射影平坦的.而对于芬斯勒流形$(M, F)$, 若$M$上的任意一点, 均存在局部坐标系$\{x^{i}\}$, 使得$F$在坐标系下是射影平坦的, 则称$F$是局部射影平坦的.

这也等价于:一个芬斯勒度量是射影平坦的充要条件是它的测地系数$G^{i}$具有如下形式

$ G^{i}=Py^{i}, $

其中$P=P(x, y)$是芬斯勒度量$F(x, y)$的射影因子, 且$P(x, y)$具有一次正齐性, 即$P(x, \lambda y)=\lambda P(x, y), \forall\lambda>0$.设$(M, F)$是芬斯勒流形, $x\in M, ~y\in T_{x}M$, 对于包含$y$的截面$\Pi _{y}\in T_{x}M$, 令

$ K(\Pi_{y}, y):=\dfrac{g_{im}R^{i}_{k}u^{k}u^{m}} {F^{2}g_{ij}u^{i}u^{j}-[g_{ij}y^{i}u^{j}]^{2}}, $

其中$\Pi _{y}={\rm span}\{y, u\}$

$ R^{i}_{k}=2\dfrac{\partial G^{i}}{\partial x^{k}}-y^{j}\dfrac{\partial^{2} G^{i}}{\partial x^{j}\partial y^{k}}+2G^{j}\dfrac{\partial^{2} G^{i}}{\partial y^{j}\partial y^{k}}-\dfrac{\partial G^{i}}{\partial y^{j}}\cdot\dfrac{\partial G^{j}}{\partial y^{k}}. $

$K(\Pi_{y}, y)$称为$(M, F)$$x$点沿方向$y$的旗曲率.

$K(\Pi _{x}, y)=K(x, y)$仅是$(x, y)\in TM_{0}$的函数, 则称$(M, F)$具有标量旗曲率, $F=F(x, y)$称为具有标量旗曲率的度量.任何局部射影平坦芬斯勒度量必具有标量旗曲率[11], 且标量旗曲率满足以下等式

$ K=\dfrac{P^{2}-P_{x^{k}}y^{k}}{F^{2}}, $ (5)

其中$P$$F$的射影因子且

$ P=\dfrac{F_{x^{k}}y^{k}}{2F}. $
2 定理的证明

$ \omega=1+\xi|x|^{2}, \Delta=\varepsilon|y|^{2}(1+\xi|x|^{2})+ \kappa^{2}\langle x, y\rangle^{2}=\varepsilon|y|^{2}\omega+\kappa^{2}\langle x, y\rangle^{2}. $ (6)
$ \alpha^{2}=a_{ij}y^{i}y^{j}, \beta=b_{i}y^{i}. $ (7)

$ a_{ij}=\dfrac{\varepsilon\delta_{ij}}{\omega}+ \dfrac{\kappa^{2}x^{i}x^{j}}{\omega^{2}}, $ (8)
$ b_{i}=\dfrac{\kappa x^{i}+\varepsilon\omega a^{i}+\kappa^{2}\langle a, x\rangle x^{i}}{\omega^{\frac{3}{2}}}, $ (9)
$ b_{j}=\dfrac{\kappa x^{j}+\varepsilon\omega a^{j}+\kappa^{2}\langle a, x\rangle x^{j}}{\omega^{\frac{3}{2}}}. $ (10)

因为$\omega=1+\xi|x|^{2}>0$, $(a^{ij})=(a_{ij})^{-1}$, 所以

$ a^{ij}=\dfrac{\omega}{\varepsilon}\Big(\delta^{ij}-\dfrac{\kappa^ {2}x^{i}x^{j}}{\varepsilon+( \varepsilon\xi +\kappa^{2})|x|^{2}}\Big). $ (11)

结合式(6)–(11)得

$ \begin{align*} b^{2}&=\|\beta\|_{\alpha}^{2}= a^{ij}b_{i}b_{j}\\ & =\dfrac{\omega}{\varepsilon}\Big(\delta^{ij}-\dfrac{\kappa^{2}x^{i} x^{j}}{\varepsilon+( \varepsilon\xi +\kappa^{2})|x|^{2}}\Big)\Big (\dfrac{\kappa x^{i}+\varepsilon\omega a^{i}+\kappa^{2}\langle a, x\rangle x^{i}}{\omega^{\frac{3}{2}}}\Big)\Big(\dfrac{\kappa x^{j}+ \varepsilon\omega a^{j}+\kappa^{2}\langle a, x\rangle x^{j}}{\omega^ {\frac{3}{2}}}\Big)\\ &=\dfrac{1}{\varepsilon\omega^{2}}\Big\{[(\kappa+\kappa^{2} \langle a, x\rangle)^{2}|x|^{2}+2\varepsilon\omega\langle a, x\rangle(\kappa+\kappa^{2}\langle a, x\rangle)]\Big(1-\dfrac{\kappa^{2} |x|^{2}}{\varepsilon+(\varepsilon\xi+\kappa^{2})|x|^{2}}\Big)\\ &\quad +\varepsilon^{2}\omega^{2}\Big(|a|^{2}-\dfrac{\kappa^{2}\langle a, x\rangle^{2}}{\varepsilon+(\varepsilon\xi+\kappa^{2})|x|^{2}}\Big) \Big\}. \end{align*} $

由定理的条件$\kappa^{2}+\varepsilon\xi=0$$|a|$足够小, 利用$\omega=1+\xi|x|^{2}$可以得到

$ \begin{align*} b^{2}&=\dfrac{1}{\varepsilon\omega^{2}}\Big\{[(\kappa+\kappa^{2}\langle a, x\rangle)^{2}+2\varepsilon\omega \langle a, x\rangle(\kappa+\kappa^{2}\langle a, x\rangle)]\Big(1-\dfrac {\kappa^{2}|x|^{2}}{\varepsilon}\Big)+\varepsilon^{2}\omega^{2} \Big(|a|^{2}-\dfrac{\kappa^{2}\langle a, x\rangle^{2}} {\varepsilon}\Big) \Big\}\\ &=\dfrac{(\kappa+\kappa^{2}\langle a, x\rangle)^{2}|x|^{2}} {\varepsilon\omega}+2\langle a, x\rangle(\kappa+\kappa^{2} \langle a, x\rangle)+\varepsilon|a|^{2}-\kappa^{2}\langle a, x\rangle^{2}\\ &=\dfrac{(1+\kappa\langle a, x\rangle)^{2}}{\omega}(1-\omega)+ \varepsilon|a|^{2}+(1+\kappa\langle a, x\rangle)^{2}-1\\ &=\dfrac{(1+\kappa\langle a, x\rangle)^{2}}{\omega}+\varepsilon|a|^{2}-1. \end{align*} $

对任意的$\xi$, 都有$b^{2}<\frac{(1+\kappa\langle a, x\rangle)^{2}}{\omega}+\varepsilon|a|^{2}$, 所以$b$是有上界的.

根据定理中的函数$\phi(b^{2}, s)= (\sqrt{1+b^{2}}+s)^{2}, s=\frac{\beta}{\alpha}$, 可得

$ \phi>0, \phi_{s}=2(\sqrt{1+b^{2}}+s), \phi_{ss}=2. $ (12)

把式(12)代入式(4)可得

$ \phi-s\phi_{s}+(b^{2}-s^{2})\phi_{ss}=1+3(b^{2}-s^{2})>0. $

利用引理1知$F$是一个芬斯勒度量.

下面来求$\alpha$关于$x^{k}$的微分, 通过计算得

$ \alpha_{x^{k}}=\dfrac{1}{\omega^{2}}[\Delta^{-\frac{1}{2}}\omega (\varepsilon\xi|y|^{2}x^{k}+\kappa^{2}\langle x, y\rangle y^{k})-2\xi\Delta^{\frac{1}{2}}x^{k}]. $ (13)

由式(13)直接计算得到

$ \begin{align*} \alpha_{x^{k}y^{l}}&=\dfrac{1}{\omega^{2}}[-\Delta^{-\frac{3}{2}} \omega(\omega\varepsilon y^{l}+\kappa^{2}\langle x, y\rangle x^{l}) (\varepsilon\xi|y|^{2}x^{k}+\kappa^{2}\langle x, y\rangle y^{k})\\ &\quad \, +\Delta^{-\frac{1}{2}}\omega(\kappa^{2}x^{l}y^{k}+\kappa^{2} \langle x, y\rangle \delta^{kl})-2\Delta^{-\frac{1}{2}} \xi\kappa^{2} \langle x, y\rangle x^{l}x^{k}]\notag. \end{align*} $

$ \begin{align} \alpha_{x^{k}y^{l}}y^{k}&=\dfrac{1}{\omega^{2}}[-\Delta^{-\frac{3} {2}}\omega(\omega\varepsilon y^{l}+\kappa^{2}\langle x, y\rangle x^{l})(\varepsilon\xi+\kappa^{2})\langle x, y\rangle |y|^{2}) \notag\\ &\quad \, +\Delta^{-\frac{1}{2}}(\omega\kappa^{2}|y|^{2}x^{l}+\kappa^{2} \omega\langle x, y\rangle y^{l}-2\kappa^{2}\xi\langle x, y\rangle^{2}x^{l})]. \end{align} $ (14)

综合式(13)、(14)及$\Delta^{\frac{1}{2}}=\Delta^{-\frac{1}{2}} \cdot\Delta$得到

$ \alpha_{x^{k}y^{l}}y^{k}-\alpha_{x^{l}}=\dfrac{1} {\omega^{2}}\big\{[-\Delta^{-\frac{3}{2}}\omega(\omega \varepsilon y^{l}+\kappa^{2}\langle x, y\rangle x^{l})\langle x, y\rangle |y|^{2}+\Delta^{-\frac{1}{2}}\omega |y|^{2}x^{l}](\varepsilon\xi+\kappa^{2})\big\}. $

由引理4知$\alpha$是射影平坦当且仅当$\kappa^{2}+\varepsilon\xi=0$, 它的射影因子

$ \theta=\dfrac{\alpha_{x^{k}}y^{k}}{2\alpha}=-\dfrac{\xi\langle x, y\rangle}{\omega}. $ (15)

$\alpha$的截面曲率

$ ^{\alpha}K=\dfrac{\theta^{2}-\theta_{x^{m}}y^{m}}{\alpha^{2}}= \dfrac{\xi}{\varepsilon}. $ (16)

$b_{i|j}$表示$\beta$关于$\alpha$的协变导数的系数, 设

$ r_{ij}=\dfrac{1}{2}(b_{i|j}+b_{j|i}), ~s_{ij}=\dfrac{1}{2}(b_{i|j}-b_{j|i}), ~ r_{00}=r_{ij}y^{i}y^{j}, ~s^{i}_{j}=a^{ik}s_{kj}, $
$ s^{i}_{0}=s^{i}_{j}y^{j}, ~r_{i}=b^{j}r_{ij}, ~s_{i}=b^{j}s_{ij}, ~r_{0} =r_{i}y^{i}, ~b^{i}=a^{ij}b_{j}, $ (17)
$ s_{0}=s_{i}y^{i}, ~r^{i}=a^{ij}r_{j}, ~s^{i}=a^{ij}s_{j}, ~r=b^{i}r_{i}. $

容易看到$\beta$是闭的当且仅当$s_{ij}=0$.由于

$ b_{i|j}=\dfrac{\partial b_{i}}{\partial x^{j}}-b_{k}\Gamma^{k} _{ij}. $ (18)

由式(6)得$\frac{\partial \omega}{\partial x^{i}}=2\xi x^{i}$.利用式(8), 有

$ \dfrac{\partial a_{ij}}{\partial x^{l}}=\dfrac{\kappa^{2}}{\omega^{2}}(\delta_{il}x^{j}+ \delta_{jl}x^{i}+2\delta_{ij}x^{l})-\dfrac{4\kappa^{2}\xi x^{i}x^{j}x^{l}}{\omega^{3}}. $ (19)

根据式(11)、(19), $\kappa^{2}+\varepsilon\xi=0$$\omega=1+\xi|x|^{2}$, 可得Christoffel符号

$ \begin{align*} \Gamma^{k}_{ij}&=\dfrac{1}{2}a^{kl}\Big(\dfrac{\partial a_{il}} {\partial x^{j}}+\dfrac{\partial a_{jl}}{\partial x^{i}}- \dfrac{\partial a_{ij}}{\partial x^{l}}\Big)=\dfrac{\omega} {\varepsilon}(\delta^{kl}+\xi x^{k}x^{l})\Big[\dfrac{\kappa^{2}} {\omega^{2}}(\delta_{il}x^{j}+\delta_{jl}x^{i})- \dfrac{2\xi\kappa^{2}x^{i}x^{j}x^{l}}{\omega^{3}}\Big]\\ &=-\dfrac{\xi}{\omega}(\delta_{ik}x^{j}+\delta_{jk}x^{i})+ \dfrac{\xi\kappa^{2}}{\varepsilon\omega}(\delta_{il}x^{j}x^{k}x^{l}+ \delta_{jl}x^{i}x^{k}x^{l})\\ &\quad\, -\dfrac{2\xi\kappa^{2}}{\varepsilon\omega^{2}}\delta^{kl}x^{i} x^{j}x^{l}-\dfrac{2\kappa^{2}\xi^{2}}{\varepsilon\omega^{2}} |x^{2}|x^{i}x^{j}x^{k}\\ &=-\dfrac{\xi}{\omega}(\delta_{ik}x^{j}+\delta_{jk}x^{i}). \end{align*} $

利用$b_{k}=\frac{\kappa x^{k}+\varepsilon\omega a^{k}+ \kappa^{2}\langle a, x\rangle x^{k}}{\omega^{\frac{3}{2}}}$

$ b_{k}\Gamma^{k}_{ij}=-\dfrac{2\xi(\kappa+\kappa^{2}\langle a, x\rangle)x^{i}x^{j}}{\omega^{\frac{5}{2}}}- \dfrac{\xi\varepsilon}{\omega^{\frac{3}{2}}}(a_{i}x^{j} +a_{j}x^{i}), $ (20)
$ \dfrac{\partial b_{i}}{\partial x^{j}}=\dfrac{(\kappa+\kappa^{2}\langle a, x\rangle)\delta_{ij}+\kappa^{2}(a_{i}x^{j}+a_{j}x^{i})} {\omega^{\frac{3}{2}}}-\dfrac{3\xi(\kappa+\kappa^{2}\langle a, x\rangle)x^{i}x^{j}}{\omega^{\frac{5}{2}}}. $ (21)

将式(20)、(21)代入式(18)中, 经简单计算后得到

$ b_{i|j}=-\dfrac{\xi(\kappa+\kappa^{2}\langle a, x\rangle)x^{i}x^{j}}{\omega^{\frac{5}{2}}}+\dfrac{ (\kappa+\kappa^{2}\langle a, x\rangle)\delta^{ij}}{\omega^{\frac{3}{2}}}, $

$ s_{ij}=\dfrac{1}{2}(b_{i|j}-b_{j|i})=0, $
$ \begin{align*} r_{ij}&=\dfrac{1}{2}(b_{i|j}+b_{j|i})=-\dfrac{\xi(\kappa+ \kappa^{2}\langle a, x\rangle)x^{i}x^{j}}{\omega^{\frac{5}{2}}}+ \dfrac{ (\kappa+\kappa^{2}\langle a, x\rangle)\delta^{ij}} {\omega^{\frac{3}{2}}}\\ &=\dfrac{1}{\varepsilon}\cdot\dfrac{(\kappa+\kappa^{2}\langle a, x\rangle) }{\omega^{\frac{1}{2}}}\Big(-\dfrac{\xi\varepsilon x^{i}x^{j}}{\omega^{2}}+\dfrac{ \varepsilon\delta^{ij}} {\omega}\Big)\\ &= \dfrac{\kappa+\kappa^{2}\langle a, x\rangle }{\varepsilon\omega^{\frac{1}{2}}}a_{ij}. \end{align*} $

所以$\beta$是闭的, 且$\beta$关于$\alpha$是共形的, 共形因子为$c(x)=\frac{\kappa+\kappa^{2}\langle a, x\rangle }{\varepsilon\omega^{\frac{1}{2}}}$.

根据$\phi=(\sqrt{1+b^{2}}+s)^{2}$

$ \phi_{b}=(\sqrt{1+b^{2}}+s)(1+b^{2})^{-\frac{1}{2}}, ~ \phi_{bs}=(1+b^{2})^{-\frac{1}{2}}, $ (22)

由式(12)、(22)得

$ 2(\phi_{b}-s\phi_{bs})=2=\phi_{ss}. $

根据引理2知$F$是局部射影平坦的.利用式(17)和$F$是局部射影平坦的, 显然有

$ r_{00}=c\alpha^{2}, ~r_{0}=c\beta, ~r=cb^{2}, ~r^{i}=cb^{i}s^{i}_{0}, ~s_{0}=0, ~s^{i}=0. $ (23)

将式(15)、(23)代入引理3中, 得到

$ \begin{align*} P&=\dfrac{2\alpha^{-1}(\phi-s\phi_{s})G_{\alpha}^{m}y_{m}+\phi_{s} (2b_{m}G_{\alpha}^{m}+r_{00})+2\alpha\phi_{b}(r_{0}+s_{0})} {2F}\\ &=\dfrac{2\theta\alpha(\phi-s\phi_{s})+\phi_{s}(2\theta\beta+ c\alpha^{2})+2c\alpha\phi_{b}\beta}{2F}\\ &=\dfrac{2\theta\alpha\phi+c\alpha^{2}(2s\phi_{b}+\phi_{s})} {2F}\\ &=\theta+c\alpha(1+b^{2})^{-\frac{1}{2}}, \end{align*} $

其中

$ \theta=\dfrac{\alpha_{x^{k}}y^{k}}{2\alpha}= -\dfrac{\xi}{\omega}\langle x, y\rangle. $

利用式(15)和$\kappa^{2}+\varepsilon\xi=0$, 通过计算得

$ P_{x^{k}}y^{k}=\theta_{x^{k}}y^{k}+c_{x^{k}}y^{k} \alpha(1+b^{2})^{-\frac{1}{2}}+\alpha_{x^{k}}y^{k}c(1+b^{2})^ {-\frac{1}{2}}-\dfrac{1}{2}c\alpha[b^{2}]_{x^{k}}y^{k}(1+b^{2}) ^{-\frac{3}{2}}, $
$ P^{2}-P_{x^{k}}y^{k}=(\theta^{2}-\theta_{x^{k}}y^{k})+ \dfrac{(2\alpha\theta c-\alpha_{x^{k}}y^{k}c-c_{x^{k}}y^{k} \alpha )}{\sqrt{1+b^{2}}}+\dfrac{c^{2}\alpha^{2}}{1+b^{2}}+\dfrac {c\alpha[b^{2}]_{x^{k}}y^{k}}{2(1+b^{2})^{\frac{3}{2}}}, $
$ c_{x^{k}}y^{k}=\dfrac{\kappa^{2}\langle a, y\rangle}{\varepsilon\omega^{\frac{1}{2}}}-\dfrac{\xi( \kappa+\kappa^{2}\langle a, x\rangle)\langle x, y\rangle}{\varepsilon\omega^{\frac{3}{2}}}=-\dfrac{\xi} {\varepsilon}\beta, $ (24)
$ [b^{2}]_{x^{k}}y^{k}=\dfrac{2\kappa\langle a, y\rangle(1+\kappa\langle a, y\rangle)}{\omega}- \dfrac{2\xi\langle x, y\rangle(1+\kappa\langle a, y\rangle)^{2}}{\omega^{2}}=2c\beta. $ (25)

由式(5)、(15)、(16)、(24)、(25), 有

$ \begin{align*} K&=\dfrac{P^{2}-P_{x^{k}}y^{k}}{F^{2}}=\dfrac{1}{\alpha^{2} \phi^{2}}\Big[\alpha^{2}\dfrac{\xi}{\varepsilon}+\dfrac {\xi\alpha\beta}{\varepsilon\sqrt{1+b^{2}}}+(\sqrt{1+b^{2}}+s) \dfrac{c^{2}\alpha^{2}}{(1+b^{2})^{\frac{3}{2}}}\Big]\\ &=\dfrac{1}{\sqrt{1+b^{2}}(\sqrt{1+b^{2}}+s)^{3}}\Big(\dfrac{\xi} {\varepsilon}+\dfrac{c^{2}}{1+b^{2}}\Big), \end{align*} $

其中$c^{2}=-\frac{\xi(1+\kappa\langle a, x\rangle)^{2}}{\varepsilon\omega}$.

参考文献
[1] BERWALD L. Über die n-dimensionalen Geometrien Konstanter Krummung in denen die Geraen die Kurzesten sind[J]. Math Z, 1929, 30: 449-469. DOI:10.1007/BF01187782
[2] SHEN Z M. Projectively flat Finsler metrics of constant flat curvature[J]. Trans Amer Math Soc, 2003, 355: 1713-1728. DOI:10.1090/S0002-9947-02-03216-6
[3] MATSUMOTO M. The Berwald connection of Finsler space with an (α, β)-metric[J]. Tensor N S, 1991, 50: 18-21.
[4] YU C T, ZHU H M. On a new class of Finsler metric[J]. Differ Geom Appl, 2011, 29(2): 244-254. DOI:10.1016/j.difgeo.2010.12.009
[5] YU C T. On dually flat general (α, β)-metric[J]. Differ Geom Appl, 2015, 40: 111-122. DOI:10.1016/j.difgeo.2015.02.010
[6] SONG W D, WANG X S. A new class of Finsler metric with scalar flag curvature[J]. J Math Res Appl, 2012, 32(4): 485-492.
[7] YU C T, ZHU H M. Projectively flat general (α, β)-mertrics with constant flag curvature[J]. J Math Anal Appl, 2015, 429: 1222-1239. DOI:10.1016/j.jmaa.2015.04.072
[8] SHEN Z M, YILDIRIM G C. On a class of projectively flat metrics with constant flat curvature[J]. Canad J Math, 2008, 60(2): 443-456. DOI:10.4153/CJM-2008-021-1
[9] HAMAL G. Über die Geometrieen in denen die Geraden die Kvrzesten sind[J]. Math Ann, 1903, 57: 231-264. DOI:10.1007/BF01444348
[10] 莫小欢. 黎曼-芬斯勒几何基础[M]. 北京: 北京大学出版社, 2007.
[11] 沈一兵. 现代芬斯勒几何初步[M]. 北京: 高等教育出版社, 2012.