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Let $s$ denote the surface of revolution $$(x,y,z)=(\cos \theta \cosh v, \sin \theta \cosh v,v)$$

where $0 < \theta < 2 \pi$ and $-\infty < v < \infty$

Let $s'$ denote the surface $$(x',y',z')=(u \cos \theta, u \sin \phi, \phi)$$ where $0 < \phi < 2 \pi$ and $ -\infty < u < \infty$

Let $f$ be the mapping which takes the point $(x,y,z)$ on $s$ to the point $(x',y',z')$ on $s'$ where $\theta=\phi$ and $u=\sinh v$.

How do I show that $f$ is an isometry form $s$ to $s'$.

I would love to see a proof.

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  • $\begingroup$ In the first formula, replace $vv$ by $v,v$ $\endgroup$ – Jean Marie Apr 30 '16 at 18:45
  • $\begingroup$ @jeanmarie would you be able to please do an explicit proof? I really want to know it for the exam $\endgroup$ – Al jabra Apr 30 '16 at 19:33
  • $\begingroup$ Why haven't you changed $vv$ int $v,v$ ? $\endgroup$ – Jean Marie Apr 30 '16 at 20:36
  • $\begingroup$ @jeanmarie just changed it. I thought you were talking about starting of the proof haha Sorry about that $\endgroup$ – Al jabra Apr 30 '16 at 20:54
  • $\begingroup$ An hint: this is a classical isometric mapping between a catenoid and a helicoid; with this keywords you should have no difficulty to find references beginning for example by mathworld.wolfram.com/Catenoid.html $\endgroup$ – Jean Marie Apr 30 '16 at 21:00
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Let $$X(\theta,v)=(\cos\theta\cosh v,\sin\theta\cosh v,v)\qquad 0<\theta <2\pi,\quad -\infty<v<\infty$$ be a parametrization of $S$. Consider

$X_{\theta}:=\frac{\partial}{\partial \theta}X(\theta,v)=(-\sin\theta \cosh v, \cos\theta \cosh v, 0),$ $X_{v}:=\frac{\partial}{\partial v}X(\theta,v)=(\cos\theta \sinh v, \sin\theta \sinh v, 1).$

Now, let $p\in S$ and $w \in T_{p}S$ and let $I\subseteq\mathbb{R}$ be a open interval with $0\in I$, consider a curve $\alpha:I\rightarrow\mathbb{R}^{2}$ where $\alpha(t)=(\theta(t),v(t))$ such that $X(\alpha(0))=p$ and $\left.\frac{d}{dt}\right|_{t=0}X(\alpha(t))=w$. (this curve always exists)

Note that $\left\{X_{\theta},X_{v}\right\}$ ($\theta$ and $v$ evaluated in $t=0$) is a base of $T_{p}S $. Then, note that $$w=\left.\frac{d}{dt}\right|_{t=0}X(\alpha(t))=\theta'(0)X_{\theta}+v'(0)X_{v}.$$

Therefore

$$\begin{array}{rcl} \left\langle w,w\right\rangle_{p} &=& \left\langle \theta'(0)X_{\theta}+v'(0)X_{v},\theta'(0)X_{\theta}+v'(0)X_{v}\right\rangle_{p} \\ &=& \theta'(0)^{2}\left\langle X_{\theta},X_{\theta}\right\rangle_{p}+v'(0)\theta'(0)\left\langle X_{v},X_{\theta}\right\rangle_{p}+\theta'(0)v'(0)\left\langle X_{\theta},X_{v}\right\rangle_{p}+v'(0)^{2}\left\langle X_{v},X_{v}\right\rangle_{p} \\ &=& \theta'(0)^{2}\left\langle X_{\theta},X_{\theta}\right\rangle_{p}+2v'(0)\theta'(0)\left\langle X_{v},X_{\theta}\right\rangle_{p}+v'(0)^{2}\left\langle X_{v},X_{v}\right\rangle_{p} \\ &=& \theta'(0)^{2}\cosh^{2}v+2v'(0)\theta'(0) 0+v'(0)^{2}\cosh^{2}v\\ &=& \theta'(0)^{2}\cosh^{2}v+v'(0)^{2}\cosh^{2}v \end{array}$$

For other hand, note that $$X'(\phi,u)=(u\cos\phi,u\sin\phi,\phi)\qquad 0<\phi<2\pi,\quad -\infty<u<\infty$$ is the original parameterization of $S'$. Let us make the following change of parameters: $$\phi=\theta, \qquad u=\sinh v, \qquad 0<\theta<2\pi,\quad -\infty<v<\infty.$$ which is possible since the map is clearly one-to-one, and the Jacobian $$\frac{\partial(\phi,u)}{\partial(\theta,v)}=\cosh v$$ is nonzero everywhere. Thus, a new parametrization of the helicoid is $$\overline{X}'(\theta,v)=\left(\sinh v \cos\theta,\sinh v \sin \theta,\theta\right).$$ So, consider

$\overline{X}'_{\theta}:=\frac{\partial}{\partial \theta}\overline{X}'(\theta,v)=(-\sin\theta \sinh v, \cos\theta \sinh v, 1),$ $\overline{X}'_{v}:=\frac{\partial}{\partial v}\overline{X}'(\theta,v)=(\cos\theta \cosh v, \sin\theta \cosh v, 0).$
Note that $f$ is: $$\begin{array}{rcl} f:S&\longrightarrow & S' \\ p&\longrightarrow& f(p):=\overline{X}' \circ X^{-1}(p) \end{array}$$

Furthermore, $\left\{\overline{X}'_{\theta},\overline{X}'_{v}\right\}$ ($\theta$ and $v$ evaluated in $t=0$) is a base of $T_{f(p)}S $. Then, we have $$df_{p}(w)=\left.\frac{d}{dt}\right|_{t=0}f(X(\alpha(t)))=\left.\frac{d}{dt}\right|_{t=0}\overline{X}' \circ X^{-1}(X(\alpha(t)))=\theta'(0)\overline{X}'_{\theta}+v'(0)\overline{X}'_{v}.$$

Therefore

$$\begin{array}{rcl} \left\langle df_{p}(w),df_{p}(w)\right\rangle_{f(p)} &=& \left\langle \theta'(0)\overline{X}'_{\theta}+v'(0)\overline{X}'_{v},\theta'(0)\overline{X}'_{\theta}+v'(0)\overline{X}'_{v}\right\rangle_{f(p)} \\ &=& \theta'(0)^{2}\left\langle \overline{X}'_{\theta},\overline{X}'_{\theta}\right\rangle_{f(p)}+v'(0)\theta'(0)\left\langle \overline{X}'_{v},\overline{X}'_{\theta}\right\rangle_{f(p)}+\theta'(0)v'(0)\left\langle \overline{X}'_{\theta},\overline{X}'_{v}\right\rangle_{f(p)}+v'(0)^{2}\left\langle \overline{X}'_{v},\overline{X}'_{v}\right\rangle_{f(p)} \\ &=& \theta'(0)^{2}\left\langle \overline{X}'_{\theta},\overline{X}'_{\theta}\right\rangle_{f(p)}+2v'(0)\theta'(0)\left\langle \overline{X}'_{v},\overline{X}'_{\theta}\right\rangle_{f(p)}+v'(0)^{2}\left\langle \overline{X}'_{v},\overline{X}'_{v}\right\rangle_{f(p)} \\ &=& \theta'(0)^{2}\cosh^{2}v+2v'(0)\theta'(0) 0+v'(0)^{2}\cosh^{2}v\\ &=& \theta'(0)^{2}\cosh^{2}v+v'(0)^{2}\cosh^{2}v \end{array}$$ Therefore, $\left\langle w,w\right\rangle_{p}=\left\langle df_{p}(w),df_{p}(w)\right\rangle_{f(p)}$ for all $w\in T_{p}S$.

So, for all $w_{1},w_{2}\in T_{p}S$, we have: $$\begin{array}{rcl} \left\langle w_{1},w_{2}\right\rangle_{p} &=& \frac{1}{2}\left\langle w_{1}+w_{2},w_{1}+w_{2}\right\rangle_{p}- \frac{1}{2}\left\langle w_{1},w_{1}\right\rangle_{p}-\frac{1}{2}\left\langle w_{2},w_{2}\right\rangle_{p}\\ &=& \frac{1}{2}\left\langle df_{p}(w_{1}+w_{2}),df_{p}(w_{1}+w_{2})\right\rangle_{f(p)}-\frac{1}{2}\left\langle df_{p}(w_{1}),df_{p}(w_{1})\right\rangle_{f(p)}-\frac{1}{2}\left\langle df_{p}(w_{2}),df_{p}(w_{2})\right\rangle_{f(p)}\\ &=& \left\langle df_{p}(w_{1}),df_{p}(w_{2})\right\rangle_{f(p)} \end{array}.$$

Therefore, $f$ is is an isometry from $S$ to $S'$.

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Let $\mathbf{x}: [0, 2 \pi] \rightarrow \mathbb{R}^n $ be a regular injective patch and let $\mathbf{y} : [0, 2 \pi] \rightarrow \mathbb{R}^n$ your second patch. Now: \begin{equation} ds^{2}_\mathbf{x}=g_{x,11} du²+g_{x,12} du dv + g_{x,22} dv² \end{equation} And \begin{equation} ds^{2}_\mathbf{y}=g_{y,11} du²+g_{y,12} du dv + g_{y,22} dv² \end{equation} denote the induced Riemannian metrics on \mathbf{x} and \mathbf{y}. Then the map: \begin{equation} \Phi=y \circ x^{-1} : \mathbf{x}[0,2 \pi] \rightarrow \mathbf{y}[0,2 \pi] \end{equation} is a local isometry if and only if \begin{equation} ds^{2}_\mathbf{x}=ds^{2}_\mathbf{y} \end{equation} In your specific case the first fundamental form of the parametrization (x,y,z) is: \begin{equation} ds^{2}_\mathbf{x}=cosh^{2}(v) (d\theta^{2}+dv^{2}) \end{equation} while for the parametrization (x',y',z') we have: \begin{equation} ds^{2}_\mathbf{y}= du^{2}+(1+u^{2}) d\phi^{2} \end{equation} if $u=sinh(v)$, $du = cosh(v) dv$, and $\theta=\phi$ we have: \begin{equation} ds^{2}_\mathbf{y}= cosh(v)^{2} dv^{2}+cosh(v)^{2} d\theta^{2}=ds^{2}_\mathbf{x} \end{equation}

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