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.

  • $\begingroup$ 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

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 $$w=\left.\frac{d}{dt}\right|_{t=0}X(\alpha(t))=\theta'(0)X_{\theta}+v'(0)X_{v}.$$


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

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 $$df_{p}(w)=\left.\frac{d}{dt}\right|_{t=0}f(X(\alpha(t)))=\left.\frac{d}{dt}\right|_{t=0}\overline{X}' \circ X^{-1}(X(\alpha(t)))=\theta'(0)\overline{X}'_{\theta}+v'(0)\overline{X}'_{v}.$$


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


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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.