活动标架下的双曲抛物面

The Hyperbolic Paraboloid under the Moving Frame

摘要: 双曲抛物面是微分几何中一常见研究对象，本文用活动标架理论得到一些双曲抛物面的活动标架基本量，进而去研究双曲抛物面。

Abstract: Hyperbolic paraboloid is a common object in differential geometry. In this paper, we use the theory of moving frame to get some basic quantities of Hyperbolic paraboloid, and then study Hyperbolic paraboloid.

活动标架法是微分几何中研究曲面的一个重要方法，能够很好地分析曲面的性质 [1]。双曲抛物面是微分几何的一个重要研究对象，在生活中有着广泛应用。参考文献 [2]，利用参数变换给双曲抛物面构造曲率线网参数表示；参考文献 [3] 分析通过双曲抛物面一条直母线的平面与其他直母线的位置关系，研究了双曲抛物面的直母线在特殊平面上的射影；参考文献 [4] 对双曲抛物面的直纹性进行探究，总结了双曲抛物面在建筑、电力工程、日常生活、宇宙学中的应用。下面根据活动标架理论，得到活动标架下双曲抛物面的一些基本量 [5]，以便进一步去分析双曲抛物面。

活动标架法首先要寻找一个活动标架，使得活动标架与双曲抛物面上的点一一对应起来。我们在曲面上选取正交坐标网，我们取双曲抛物面上的曲率线网 [2]。但是在计算中我们发现，利用双曲抛物面上的曲率线网计算过程很复杂。因此我们使用双曲抛物面上的常用参数 $r\left(u,v\right)=\left(u+v,u-v,-4uv\right)$，通过计算得到 $r_u=\left(1,1,-4v\right)$， $r_v=\left(1,-1,-4u\right)$，但是 $F=r_u\cdot r_v=\left(1,1,-4v\right)\cdot \left(1,-1,-4u\right)=16uv\ne 0$，不是我们想要的正交坐标网。为此在这里我们对 $r_u$ 和 $r_v$ 施行施密特正交化法，使其成为正交的单位向量 [6]。具体做法如下：

正交化： ${r^{\prime }}_u=r_u=\left(1,1,-4v\right)$，

${r^{\prime }}_v=r_v-\frac{\left(r_v,{r^{\prime }}_u\right)}{\left({r^{\prime }}_u,{r^{\prime }}_u\right)}{r^{\prime }}_u=\left(1-\frac{16uv}{2+16v^2},-1-\frac{16uv}{2+16v^2},-4u+\frac{64u{v}^2}{2+16v^2}\right)$。

单位化： ${r^{″}}_u=\frac{{r^{\prime }}_u}{\left|{r^{\prime }}_u\right|}=\frac{1}{\sqrt{2+16v^2}}\left(1,1,-4v\right)$，

${r^{″}}_v=\frac{{r^{\prime }}_v}{\left|{r^{\prime }}_v\right|}=\frac{2+16v^2}{\sqrt{8\left(1+8v^2\right)\left(8u^2+8v^2+1\right)}}\left(1,-1,-4u\right)+\frac{-16uv}{\sqrt{8\left(1+8v^2\right)\left(8u^2+8v^2+1\right)}}\left(1,1,-4v\right)$。

这样我们得到了一组正交的单位向量组，我们令：

$e_1={r^{″}}_u=\frac{1}{\sqrt{2+16v^2}}\left(1,1,-4v\right)$，

$e_2={r^{″}}_v=\frac{\sqrt{2+16v^2}}{2\sqrt{8u^2+8v^2+1}}\left(1,-1,-4u\right)+\frac{-8uv}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}\left(1,1,-4v\right)$，

$e_3=e_1\times e_2\text{=}\frac{1}{\sqrt{8u^2+8v^2+1}}\left(-2u-2v,2u-2v,-1\right)$，

这样我们就完成了活动标架法的第一步，即找到了一个双参数的活动标架。

定理1：双曲抛物面 $r\left(u,v\right)=\left(u+v,u-v,-4uv\right)$ 的相对分量为：

${\omega }_1^2=\frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v,{\omega }_1^3=\frac{4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v$，

${\omega }_2^1=\frac{-8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v,{\omega }_2^3=\frac{2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{-32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v$，

${\omega }_3^1=\frac{-4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v,{\omega }_3^2=\frac{-2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v$。

证明：对双参数的活动标架进行微分：

$\text{d}r=\left\{1,1,-4v\right\}\text{d}u+\left\{1,-1,-4u\right\}\text{d}v={\omega }^1{e}_1+{\omega }^2{e}_2$，

把 $e_1$ 和 $e_2$ 代入，得到：

$\begin{array}{l}\frac{{\omega }^1}{\sqrt{2+16v^2}}\left(1,1,-4v\right)+\frac{\sqrt{2+16v^2}{\omega }^2}{2\sqrt{8u^2+8v^2+1}}\left(1,-1,-4u\right)+\frac{-8uv{\omega }^2}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}\left(1,1,-4v\right)\\ =\left\{1,1,-4v\right\}\text{d}u+\left\{1,-1,-4u\right\}\text{d}v\end{array}$，

即 $\{\begin{array}{l}\frac{{\omega }^1}{\sqrt{2+16v^2}}+\frac{\sqrt{2+16v^2}{\omega }^2}{2\sqrt{8u^2+8v^2+1}}+\frac{-8uv{\omega }^2}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}=\text{d}u+\text{d}v\\ \frac{{\omega }^1}{\sqrt{2+16v^2}}-\frac{\sqrt{2+16v^2}{\omega }^2}{2\sqrt{8u^2+8v^2+1}}+\frac{-8uv{\omega }^2}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}=\text{d}u-\text{d}v\\ \frac{-4v{\omega }^1}{\sqrt{2+16v^2}}+\frac{-4u\sqrt{2+16v^2}{\omega }^2}{2\sqrt{8u^2+8v^2+1}}+\frac{32u{v}^2{\omega }^2}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}=-4v\text{d}u-4u\text{d}v\end{array}$，

解得 $\{\begin{array}{l}{\omega }^1=\sqrt{2+16v^2}\text{d}u+\frac{16uv}{\sqrt{2+16v^2}}\text{d}v\\ {\omega }^2=\frac{2\sqrt{8u^2+8v^2+1}}{\sqrt{2+16v^2}}\text{d}v\end{array}$。

$\text{d}{e}_1=\left(\frac{-16v\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}},\frac{-16v\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}},\frac{-4\text{d}v}{\sqrt{2+16v^2}}+\frac{64v^2\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}}\right)={\omega }_1^2{e}_2+{\omega }_1^3{e}_3$，

把 $e_2$ 和 $e_3$ 代入，得到：

$\begin{array}{l}{\omega }_1^2\frac{\sqrt{2+16v^2}}{2\sqrt{8u^2+8v^2+1}}\left(1,-1,-4u\right)+{\omega }_1^2\frac{-8uv}{\sqrt{\left(2+16v^2\right)\left(8u^2+8v^2+1\right)}}\left(1,1,-4v\right)\\ +{\omega }_1^3\frac{1}{\sqrt{8u^2+8v^2+1}}\left(-2u-2v,2u-2v,-1\right)\\ =\left(\frac{-16v\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}},\frac{-16v\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}},\frac{-4\text{d}v}{\sqrt{2+16v^2}}+\frac{64v^2\text{d}v}{{\left(2+16v^2\right)}^{\frac{3}{2}}}\right)\end{array}$，

解得： $\{\begin{array}{l}{\omega }_1^2=\frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\\ {\omega }_1^3=\frac{4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v\end{array}$。

$\begin{array}{c}\text{d}{e}_2^{\text{T}}=(\frac{-8u-8v-64u{v}^2-64v^3}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u\left[8u^2-8uv-64v^4-64u{v}^3+1\right]}{{\left(2+16v^2\right)}^{\frac{3}{2}}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v,\\ \frac{8u-8v+64u{v}^2-64v^3}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u\left[8u^2+8uv-64v^4+64u{v}^3+1\right]}{{\left(2+16v^2\right)}^{\frac{3}{2}}\sqrt{8u^2+8v^2+1}}\text{d}v,\\ \frac{-2\sqrt{2+16v^2}}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{128uv\left(4u^2+8v^2+1\right)}{{\left(2+16v^2\right)}^{\frac{3}{2}}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v)\\ ={\omega }_2^1{e}_1+{\omega }_2^3{e}_3\end{array}$，

把 $e_1$ 和 $e_3$ 代入，得到：

$\{\begin{array}{l}\frac{{\omega }_2^1}{\sqrt{2+16v^2}}\text{+}\frac{\left(-2u-2v\right){\omega }_2^3}{\sqrt{8u^2+8v^2+1}}=\frac{-8u-8v-64u{v}^2-64v^3}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u\left[\left(1+8v^2\right)\left(1-8v^2\right)+8u^2\right]}{{\left(2+16v^2\right)}^{\frac{3}{2}}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v\\ \frac{{\omega }_2^1}{\sqrt{2+16v^2}}\text{+}\frac{\left(2u-2v\right){\omega }_2^3}{\sqrt{8u^2+8v^2+1}}=\frac{8u-8v+64u^{2}v-64v^3}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u\left(8uv+8v^2+1\right)}{{\left(2+16v^2\right)}^{\frac{3}{2}}\sqrt{8u^2+8v^2+1}}\text{d}v\\ \frac{-4v{\omega }_2^1}{\sqrt{2+16v^2}}+\frac{-{\omega }_2^3}{\sqrt{8u^2+8v^2+1}}=\frac{-2\sqrt{2+16v^2}}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{128uv\left(4u^2+8v^2+1\right)}{{\left(2+16v^2\right)}^{\frac{3}{2}}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v\end{array}$，

解得 $\{\begin{array}{l}{\omega }_2^1=\frac{-8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\\ {\omega }_2^3=\frac{2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{-32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v\end{array}$。

$\begin{array}{c}\text{d}{e}_3=(\frac{-16v^2+16uv-2}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u^2+16uv-2}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v,\\ \frac{16v^2+16uv+2}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{-16u^2-16uv-2}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v,\\ \frac{8u}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u+\frac{8v}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}v)\\ ={\omega }_3^1{e}_1+{\omega }_3^2{e}_2\end{array}$

把 $e_1$ 和 $e_2$ 代入，解得 $\{\begin{array}{l}{\omega }_3^1=\frac{-4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v\\ {\omega }_3^2=\frac{-2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v\end{array}$。

通过上述计算，我们得到了相对分量，显然有 ${\omega }_1^2=-{\omega }_2^1,{\omega }_2^3=-{\omega }_3^2,{\omega }_1^3=-{\omega }_3^1$ 成立。

定理2：双曲抛物面的结构方程：

$\text{d}{\omega }^1={\omega }^2\wedge {\omega }_2^1=0$，

$\text{d}{\omega }^2={\omega }^1\wedge {\omega }_1^2=\frac{16u}{\sqrt{2+16v^2}\sqrt{8u^2+8v^2+1}}\text{d}u\wedge \text{d}v$，

$\text{d}{\omega }^3={\omega }^1\wedge {\omega }_1^3+{\omega }^2\wedge {\omega }_2^3=0$，

$\text{d}{\omega }_1^2={\omega }_1^3\wedge {\omega }_3^2=\frac{8}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v$，

$\text{d}{\omega }_2^3={\omega }_2^1\wedge {\omega }_1^3=0$，

$\text{d}{\omega }_3^1={\omega }_3^2\wedge {\omega }_2^1=\frac{32u}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v$。

证明：

$\text{d}{\omega }^1=\text{d}\left(\sqrt{2+16v^2}\text{d}u+\frac{16uv}{\sqrt{2+16v^2}}\text{d}v\right)=\frac{-16v}{\sqrt{2+16v^2}}\text{d}u\wedge \text{d}v+\frac{16v}{\sqrt{2+16v^2}}\text{d}u\wedge \text{d}v=0$，

${\omega }^2\wedge {\omega }_2^1=\frac{2\sqrt{8u^2+8v^2+1}}{\sqrt{2+16v^2}}\frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\wedge \text{d}v=0$，

故有 $\text{d}{\omega }^1={\omega }^2\wedge {\omega }_2^1$ 成立。

$\text{d}{\omega }^2=\text{d}\left(\frac{2\sqrt{8u^2+8v^2+1}}{\sqrt{2+16v^2}}\text{d}v\right)=\frac{2\frac{8u}{\sqrt{8u^2+8v^2+1}}}{\sqrt{2+16v^2}}\text{d}u\wedge \text{d}v=\frac{16u}{\sqrt{2+16v^2}\sqrt{8u^2+8v^2+1}}\text{d}u\wedge \text{d}v$，

$\begin{array}{c}{\omega }^1\wedge {\omega }_1^2=\left(\sqrt{2+16v^2}\text{d}u+\frac{16uv}{\sqrt{2+16v^2}}\text{d}v\right)\wedge \frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\\ =\frac{16u}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}u\wedge \text{d}v\end{array}$，

故有 $\text{d}{\omega }^2={\omega }^1\wedge {\omega }_1^2$ 成立。

而 $\text{d}{\omega }^3=0$，

${\omega }^1\wedge {\omega }_1^3+{\omega }^2\wedge {\omega }_2^3=\frac{4}{\sqrt{8u^2+8v^2+1}}\text{d}u\wedge \text{d}v+\frac{-4}{\sqrt{8u^2+8v^2+1}}\text{d}u\wedge \text{d}v=0$，

故有 $\text{d}{\omega }^3={\omega }^1\wedge {\omega }_1^3+{\omega }^2\wedge {\omega }_2^3$ 成立。

$\text{d}{\omega }_1^2=\text{d}\left(\frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\right)=\frac{8\left(1+8v^2\right)}{\left(1+8v^2\right){\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v=\frac{8}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v$，

${\omega }_1^3\wedge {\omega }_3^2=\frac{8}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v$，

故有 $\text{d}{\omega }_1^2={\omega }_1^3\wedge {\omega }_3^2$ 成立。

$\begin{array}{c}\text{d}{\omega }_2^3=\text{d}\left(\frac{2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{-32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v\right)\\ =\frac{-512u^{2}v-512v^3-64v+64v+512v^3+512u^{2}v}{{\left(8u^2+8v^2+1\right)}^2\sqrt{2+16v^2}}\text{d}u\wedge \text{d}v=0\end{array}$，

${\omega }_2^1\wedge {\omega }_1^3=\frac{8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\frac{4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v\wedge \text{d}v=0$，

故有 $\text{d}{\omega }_2^3={\omega }_2^1\wedge {\omega }_1^3$ 成立。

$\text{d}{\omega }_3^1=\text{d}\left(\frac{-4}{\sqrt{8u^2+8v^2+1}\sqrt{2+16v^2}}\text{d}v\right)=\frac{32u}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v$，

$\begin{array}{c}{\omega }_3^2\wedge {\omega }_2^1=\left(\frac{-2\sqrt{2+16v^2}}{8u^2+8v^2+1}\text{d}u+\frac{32uv}{\left(8u^2+8v^2+1\right)\sqrt{2+16v^2}}\text{d}v\right)\wedge \frac{-8u}{\sqrt{8u^2+8v^2+1}\left(1+8v^2\right)}\text{d}v\\ =\frac{32u}{\sqrt{2+16v^2}{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v\end{array}$，

故有 $\text{d}{\omega }_3^1={\omega }_3^2\wedge {\omega }_2^1$ 成立。

这样，我们得到了双曲抛物面结构方程。

应用：双曲抛物面是不可展曲面。

根据活动标架理论，高斯曲率 $K=\frac{-\text{d}{\omega }_1^2}{{\omega }^1\wedge {\omega }^2}$。

把双曲抛物面的活动标架基本量带入，得到双曲抛物面的高斯曲率：

$\begin{array}{c}K=\frac{-\text{d}{\omega }_1^2}{{\omega }^1\wedge {\omega }^2}=\frac{-\frac{8}{{\left(8u^2+8v^2+1\right)}^{\frac{3}{2}}}\text{d}u\wedge \text{d}v}{\left(\sqrt{2+16v^2}\text{d}u+\frac{16uv}{\sqrt{2+16v^2}}\text{d}v\right)\wedge \frac{2\sqrt{8u^2+8v^2+1}}{\sqrt{2+16v^2}}\text{d}v}\\ =\frac{-4}{{\left(8u^2+8v^2+1\right)}^2}\ne 0\end{array}$

因此得到双曲抛物面不可展。

本文得到了双曲抛物面的活动标架基本量，为今后利用活动标架法研究双曲抛物面奠定了基础。又利用双曲抛物面的活动标架基本量，验证了双曲抛物面是不可展曲面这一性质。

基金项目

内蒙古自治区青年科技英才支持计划(NJYT-19-A09)；内蒙古自然科学基金(2018MS01011)；国家自然科学基金(11661062)。

参考文献