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
Berwald度量是由黎曼度量
$ F(x, y)=\alpha\phi(s), ~ s=\dfrac{\beta}{\alpha}. $ | (1) |
如果
$ \phi(s) >0, ~\phi-s\phi_{s}+(b^{2}-s^{2})\phi_{ss}>0, ~|s|\leqslant b<b_{0}, $ |
其中
$ F(x, y)=\alpha\phi(b^{2}, s), ~s=\dfrac{\beta}{\alpha}, $ |
其中
$ 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
$ \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
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
(1) 若
$ b^{2}=\varepsilon|a|^{2}+\dfrac{(1+\kappa\langle a, x\rangle)^{2}}{1+\xi|x|^{2}}-1. $ |
(2)
(3) 当
$ 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})}. $ |
注 当
定义1 设
引理1[6] 设
$ \phi>0, \phi-s\phi_{s}+(b^{2}-s^{2})\phi_{ss}>0, $ | (4) |
其中
引理2[8] 设
(1) 函数
$ \phi_{ss}=2(\phi_{b}-s\phi_{bs}), $ |
其中
(2)
引理3[6] 设广义的
$ 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
引理4[9] 设
$ F_{x^{k}y^{l}}y^{k}-F_{x^{l}}=0. $ |
定义2[10] 在开集
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}=Py^{i}, $ |
其中
$ 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
$ 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}}. $ |
若
$ K=\dfrac{P^{2}-P_{x^{k}}y^{k}}{F^{2}}, $ | (5) |
其中
$ P=\dfrac{F_{x^{k}}y^{k}}{2F}. $ |
令
$ \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) |
则
$ 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
$ 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
$ \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
根据定理中的函数
$ \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
下面来求
$ \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*} $ |
故
$ \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
$ \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
$ \theta=\dfrac{\alpha_{x^{k}}y^{k}}{2\alpha}=-\dfrac{\xi\langle x, y\rangle}{\omega}. $ | (15) |
$ ^{\alpha}K=\dfrac{\theta^{2}-\theta_{x^{m}}y^{m}}{\alpha^{2}}= \dfrac{\xi}{\varepsilon}. $ | (16) |
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
$ b_{i|j}=\dfrac{\partial b_{i}}{\partial x^{j}}-b_{k}\Gamma^{k} _{ij}. $ | (18) |
由式(6)得
$ \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),
$ \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}\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}}}, $ |
故
$ 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
根据
$ \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
$ 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
$ 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
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