"局部等距"的通常解释

hbghlyj

(图片来自PSTricks官网的示例)
以下证明来自这个讲义的61页
Example 101 The catenoid is the surface of revolution formed by rotating the curve $y=\cosh x$, known as a catenary, about the $x$-axis. So we can parameterise it as\[\mathbf r(x, θ) = (x, \cosh x \cos θ, \cosh x \sin θ), −π < θ < π, x ∈\mathbb R.\]The helicoid is formed in a “propeller-like” fashion by pushing the $x$-axis up the $z$-axis while spinning the $x$-axis at a constant angular velocity. So we can parameterise it as$$\mathbf s(X, Z) = (X\cos Z, X\sin Z, Z).$$[The need for differing notation – re $x$ and $X$ – will become apparent in due course.]
As some preliminary calculations we note that$$\mathbf{r}_{x} =(1, \sinh x \cos \theta, \sinh x \sin \theta),  \quad\mathbf{r}_{\theta} =(0,-\cosh x \sin \theta, \cosh x \cos \theta) $$$$\mathbf{r}_{x} \cdot \mathbf{r}_{x} =\cosh ^{2} x, \quad \mathbf{r}_{x} \cdot \mathbf{r}_{\theta}=0, \quad \mathbf{r}_{\theta} \cdot \mathbf{r}_{\theta} =\cosh ^{2} x $$And for the helicoid we have\begin{align*} \mathbf{s}_{X} &=(\cos Z, \sin Z, 0), \quad \mathbf{s}_{Z}=(-X \sin Z, X \cos Z, 1) \\ \mathbf{s}_{X} \cdot \mathbf{s}_{X} &=1, \quad\mathbf s_{X} \cdot\mathbf s_{Z}=0, \quad \mathbf{s}_{Z} \cdot \mathbf{s}_{Z}=1+X^{2} \end{align*}Now consider the map from the catenoid to the helicoid given by\[\mathbf r(x, θ) \mapsto\mathbf s(\sinh x, θ).\]We’ll show that for any curve $γ$ on the catenoid, its image, under the map to the helicoid, has the
same length.
If we consider the curve\[γ(t) =\mathbf r(x(t), θ(t))\quad a \le t \le b\]in the catenoid then $\gamma^{\prime}=x^{\prime} \mathbf{r}_{x}+\theta^{\prime} \mathbf{r}_{\theta}$ and so$$\left|\gamma^{\prime}\right|^{2}=\left(\mathbf{r}_{x} \cdot \mathbf{r}_{x}\right)\left(x^{\prime}\right)^{2}+2\left(\mathbf{r}_{x} \cdot \mathbf{r}_{\theta}\right) x^{\prime} \theta^{\prime}+\left(\mathbf{r}_{\theta} \cdot \mathbf{r}_{\theta}\right)\left(\theta^{\prime}\right)^{2}=\cosh ^{2} x\left(\left(x^{\prime}\right)^{2}+\left(\theta^{\prime}\right)^{2}\right)$$For the curve\[\Gamma(t) =\mathbf s(X(t), Z(t))\quad c\le t\le d\]in the helicoid we similarly have$$\left|\Gamma^{\prime}\right|^{2}=\left(X^{\prime}\right)^{2}+\left(1+X^{2}\right)\left(Z^{\prime}\right)^{2}$$Thus, a curve $\mathbf r(x(t), θ(t))$ where $a\le t\le b$ has length$$\int_{a}^{b}\left|\gamma^{\prime}\right| \mathrm{d} t=\int_{a}^{b} \sqrt{\cosh ^{2} x\left(\left(x^{\prime}\right)^{2}+\left(\theta^{\prime}\right)^{2}\right)} \mathrm{d} t=\int_{a}^{b} \cosh x \sqrt{\left(x^{\prime}\right)^{2}+\left(\theta^{\prime}\right)^{2}} \mathrm{~d} t$$For the image of the curve $\mathbf s(\sinh x, z)$ in the helicoid we have$$\int_{a}^{b}\left|\Gamma^{\prime}\right| \mathrm{d} t=\int_{a}^{b} \sqrt{\left(X^{\prime}\right)^{2}+\left(1+X^{2}\right)\left(Z^{\prime}\right)^{2}} \mathrm{d} t$$Now $X =\sinh x$ and $Z = θ$ so that the above equals\begin{align*} & \int_{a}^{b} \sqrt{\cosh ^{2} x\left(x^{\prime}\right)^{2}+\left(1+\sinh ^{2} x\right)\left(\theta^{\prime}\right)^{2}} \mathrm{~d} t \\=& \int_{a}^{b} \cosh x \sqrt{\left(x^{\prime}\right)^{2}+\left(\theta^{\prime}\right)^{2}} \mathrm{~d} t \\=& \int_{a}^{b}\left|\gamma^{\prime}\right| \mathrm{d} t \end{align*}So for any curve $γ$ in the catenoid, its image, under the map to the helicoid, has the same length.
This means that the map is an isometry between the surfaces (when distances are measured within the surface).

Czhang271828

非常经典的论题, 下依个人书写习惯抄题, 并给出"局部等距"的通常解释:

设 $X(u,v)=(\cosh u\cos v,\cosh u\sin v,u)$ 为悬链曲面 $S_1$ 之正则参数化表示, i.e.
$$
X:U(\subset\mathbb R^2)\to S_1(\mathbb R^3).
$$
同理, 记 $Y(\overline u,\overline v)=(\overline u\cos \overline v,\overline u\sin \overline v,\overline v)$​ 为螺旋曲面 $S_2$ 之正则参数表示. 此处, "正则"一词可理解为"处处存在局部逆映射".

当参数 $u$ 改变 $\Delta u$, $v$ 改变 $\Delta v$ 时, $X$ 改变 $X_u\cdot \Delta u+X_v\cdot \Delta v$​. 记 Riemann 度量
$$
g=(g_{ij})_{2\times 2}=\begin{pmatrix}X_u\cdot X_u&X_u\cdot X_v\\
X_v\cdot X_u&X_v\cdot X_v\end{pmatrix}.
$$
此处内积即通常内积. 在建立曲面上的度量后, 单位长度 $\mathrm ds=\sqrt{\sum_{i,j=1}^2  g_{ij}\mathrm du^i\mathrm du^j}$, 如 $S$ 上则 $\mathrm ds=\sqrt{g_{11}\mathrm (\mathrm du)^2+2g_{12}\mathrm du\mathrm dv+g_{22}(\mathrm dv)^2}$. 显然, 曲面 $S_1$ 在 $p(\in S_1)$ 处与 $S_2$ 某点局部等距若且仅若存在局部双射 $\varphi:U_p(\subset S_1)\to U_{\varphi(p)}(\subset S_2)$ 使得 $p$ 在 $S_1$ 上的 Riemann 度量与 $\varphi(p)$ 在 $S_2$ 上的 Riemann 度量相等.

以 $S_1$ 与 $S_2$ 为例, 度量 $g$ 与 $\overline g$​ 分别为
$$
g=\begin{pmatrix}\cosh^2u&0\\0&\cosh^2 u\end{pmatrix},\quad \overline g=\begin{pmatrix}1&0\\0&\overline u^2+1\end{pmatrix}.
$$
记 $(u,v)=(u(\overline u,\overline v),v(\overline u,\overline v))$​. 若存在局部等距映射, 据链式法则, 有
$$
\left(\dfrac{\partial(u,v)}{\partial(\overline u,\overline v)}\right)'\cdot g\cdot \left(\dfrac{\partial(u,v)}{\partial(\overline u,\overline v)}\right)=\overline g.
$$
代入得
$$
\begin{pmatrix}\partial_{\overline u} u&\partial_{\overline u}v\\\partial_{\overline v} u&\partial_{\overline v}v\end{pmatrix}\cdot \begin{pmatrix}\cosh^2u&0\\0&\cosh^2 u\end{pmatrix}\cdot \begin{pmatrix}\partial_{\overline u} u&\partial_{\overline v}u\\\partial_{\overline u} v&\partial_{\overline v}v\end{pmatrix}=\begin{pmatrix}1&0\\0&\overline u^2+1\end{pmatrix}.
$$
此处仅需证明存在等距映射, 因此观察出某个解 $(u,v)=(u(\overline u,\overline v),v(\overline u,\overline v))$ 即可. 视 $u=u(\overline u)$, $v=v(\overline v)$, 解常微分方程
$$
\left\{\begin{align*}
\dfrac{\mathrm du}{\mathrm d\overline u}=&\dfrac{1}{\cosh u},\\
\dfrac{\mathrm dv}{\mathrm d\overline v}=&\dfrac{\sqrt{\overline u^2+1}}{\cosh u}.\\
\end{align*}\right.
$$
取 $\overline u=\sinh u$, $v=\overline v$ . 因此 $S_1$ 到 $S_2$ 的局部等距由映射
$$
\varphi:S_1\to S_2,X^{-1}(u,v)\to Y(\sinh u,v)
$$
给出. 之所以不存在全局的等距映射, 是因为 $S_1$ 与 $S_2$ 的拓扑结构有殊. 悬链曲面与无限长圆柱之侧面具有相同的拓扑结构(同胚), 螺旋曲面与上半平面同胚, 而无限长圆柱之侧面与上半平面不同胚, 因为前者表面上的某些闭曲线无法连续变化至单点,  但上半平面上的一切闭曲线均可连续变换至单点, 因此两者不具有相同的拓扑结构. 是故 $S_1$ 与 $S_2$ 间无法建立全局的等距映射.

局部(全局)等距 $\Leftrightarrow$ 存在参数化使得度量局部(全局)相等.

hbghlyj

Czhang271828 发表于 2022-1-15 08:39
螺旋曲面与上半平面同胚

如何证明呢？

hbghlyj

Czhang271828 发表于 2022-1-15 08:39
螺旋曲面与上半平面同胚

螺旋曲面与整个平面同胚吧

数学帮助论坛上有一个证明：mathhelpforum.com/t/helicoid.148224/

Informally : the slit plane $\mathbb{C}\setminus\mathbb{R}_{\ge0}$ is easily seen to be homeomorphic to a strip without its boundary, (any branch of the logarithm doing the job). Now lift the cut to the helicoid, and map each slice of the helicoid to a strip, identifying the appropriate boundaries of two strips if the corresponding boundaries of the slices are identified on the helicoid. In this way, the helicoid is mapped homeomorphically on the plane.

首先 $\rm Log$ 是从 slit plane $\mathbb{C}\setminus\mathbb{R}_{\ge0}$ 到 strip $\mathbb{R}\times(0,2\pi)$ 的同胚，逆映射为 $\exp$
于是就能把 $\mathbb{R}^2$ 分成许多条 strip，再通过 $\exp$ 分别把每一个 strip$$\mathbb{R}\times(n\cdot2\pi,n\cdot2\pi+2\pi)$$映射到 catenoid 夹在平面$z=n\cdot2\pi$与$z=(n+1)2\pi$之间的部分$$\cases{x=r\cos\theta\\y=r\sin\theta\\z=\theta}\qquad n\cdot2\pi<\theta<n\cdot2\pi+2\pi$$如何验证相邻两片的接缝$\theta=n\cdot2\pi$处映射是连续的

hbghlyj

Czhang271828 发表于 2022-1-15 08:39
此处仅需证明存在等距映射, 因此观察出某个解 $(u,v)=(u(\overline u,\overline v),v(\overline u,\overline v))$ 即可.

1#的动画中，是连续变化的，如何写出连续变化中间某时刻的等距映射

hbghlyj

Czhang271828 发表于 2022-1-15 08:39
非常经典的论题

如何证明这两个曲面都是minimal surfaces (that is, surfaces whose mean curvature is 0)？