Exercise :

Consider the surfaces : $$\Phi_1 : f_1(\theta, v) = \left( \cos \theta \cosh v, \sin \theta \cosh v, v\right), \; (\theta, v) \in (0,2 \pi) \times \mathbb R$$ $$\Phi_2 : f_2(\phi, u) = \left( u\cos \phi , u\sin \phi, \phi\right), \; (\theta, v) \in (0,2 \pi) \times \mathbb R$$ Check if the following mapping is an isometry between them : $$f: \Phi_1 \to \Phi_2 \;; f_1(\theta, v) \mapsto f_2(\theta, \sinh v)$$

Thoughts-Question :

To start off, this is a Differential Geometry related question which I am not that experienced, thus if it feels trivial, excuse me.

From my continuous experience, interest and studying of a whole differnt subject (Functional Analysis - Operator Theory), I know very well that a Linear Isometry is essentialy achieved if $\|Av\|_Y = \|v\|_X$ where $A:X \to Y$ is a linear operator. This means that they are distance preserving. It is a global isometry if it also is surjective.

Now, a similar correspondance can be found in Differential Geometry. Specifically, if we have $2$ surfaces, $\Phi_1$ and $\Phi_2$, then the function $f: \Phi_1 \to \Phi_2$ is an isometry if and only if $f:\Phi_1 \to \Phi_2$ is a differentiable mapping which is an inective and surjective local isometry.

Now, I am having a hard time proving the following statements. First of all, I start by constructing my function as stated by the exercise body :

$$f(f_1(\theta,v)) = f_2(\theta, \sinh v)$$ $$\implies$$ $$f(\cos\theta\cosh v, \sin \theta\cosh v, v) = (\sinh v \cos \theta, \sinh v\sin \theta, \theta)$$

So, checking the statements needed, first of all, that $f$ is differentiable.

Now, how does one show that this $f$ is injective and surjective ?

Also, what about the local isometry ? I know that we can check if it is a local isometry or not, since the fundamental quantities of the fundamental form must oblige the following relations : $$E_p = E_{f(p)}, \; F_p = F_{f(p)}, \; G_p = G_{f(p)}$$

I am kind of confused on the calculations of the fundamental quantities though. In a solved (but poorly elaborated) example I've seen, one must first calculate the inverse of $f$ and then correlate the argument of $f$ with what it's mapped to.

I would really appreciate any thorough elaboration which can help me how to handle showing the injectivity, surjectivity but most importantly on how to find the fundamental quantities stated.


Let's try to make the question a bit more precise. Define

$$ S_1 = \{ f_1(\theta, v) \, | \, (\theta, v) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{x,y,z}, \\ S_2 = \{ f_2(\phi, u) \, | \, (\phi, u) \in (0,2\pi) \times \mathbb{R} \} \subseteq \mathbb{R}^3_{a,b,c}. $$

Then $S_1,S_2$ are both parametric surfaces in $\mathbb{R}^3$. In order to make things less confusing, it is comfortable to think of each $S_i$ as living in a different copy of $\mathbb{R}^3$. To emphasize this point I can give different names to the coordinates of $\mathbb{R}^3$ in which each $S_i$ lives (this what my non-standard notation $\mathbb{R}^3_{x,y,z},\mathbb{R}^3_{a,b,c}$ mean).

The functions $f_i \colon (0,2\pi) \times \mathbb{R} \rightarrow S_1$ given by the formulas above are global parametrizations for the surfaces $S_i$. Let's write $$ f_1(\theta,v) = (x(\theta,v), y(\theta,v),z(\theta,v)). $$ The function $f_1$ is one-to-one and onto $S_1$ so given a point $p = (x_0,y_0,z_0) \in S_1$, we have a unique point $$f^{-1}(p) = f^{-1}(x_0,y_0,z_0) = (\theta(p), v(p)) = (\theta(x_0,y_0,z_0), v(x_0,y_0,z_0))$$ such that $$ f_1(\theta(p),v(p)) = f_1(\theta(x_0,y_0,z_0),v(x_0,y_0,z_0)) = p $$ and similarly for $f_2$.

Now, let's define a map $F \colon S_1 \rightarrow S_2$ by the formula $$ F(p) = f_2(\theta(p), \sinh v(p)). $$ This is related to your definition for if we write $p = f_1(\theta,v)$ then $$ F(f_1(\theta,v)) = f_2(\theta, \sinh v). $$ The local representation of the map $F$ between the surfaces is the map $$\tilde{F} = f_2^{-1} \circ F \circ f_1 \colon (0,2\pi) \times \mathbb{R} \rightarrow (0,2\pi) \times \mathbb{R}$$ and is given by $$ \tilde{F}(\theta,v) = (\theta, \sinh v). $$

You are asked whether $F$ is an isometry between $S_1$ and $S_2$. For $F$ to be an isometry, it needs to satisfy two conditions:

  1. The map $F$ needs to be one-to-one and onto.
  2. For all $p \in S_1$ and $v,w \in T_p(S_1)$, we should have $\left< v, w \right> = \left< dF|_p(v), dF|_p(w) \right>$. That is, $F$ should infinitesimally preserve the length of tangent vectors. Here, $dF \colon T_p(S_1) \rightarrow T_{F(p)} S_2$ is the differential of the map $F$.

Now, it turns out that instead of checking this directly from the definitions for $F$, you can deduce everything from the representation $\tilde{F}$ of the map $F$. Namely, $F$ will be an isometry if and only if:

  1. The map $\tilde{F}$ needs to be one-to-one and onto.
  2. The map $\tilde{F}$ needs to preserve the (local representations of the) first fundamental form. For each $(\theta,v) \in (0,2\pi) \times \mathbb{R}$, we should have $$ E_1(\theta,v) = E_2(\tilde{F}(\theta,v)), F_1(\theta,v) = F_2(\tilde{F}(\theta,v)), G_1(\theta,v) = G_2(\tilde{F}(\theta,v)) $$ where $E_i,F_i,G_i$ are the coefficients of the first fundamental form of $S_i$ with respect to the parametrization $f_i$.

How do we calculate the $E_i,F_i,G_i$? By the formulas

$$ E_1(\theta,v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial \theta} \right>, \,\, F_1(\theta, v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial v} \right>, \,\, G_1(\theta, v) = \left< \frac{\partial f_1}{\partial v}, \frac{\partial f_1}{\partial v} \right> $$ and similarly $$ E_2(\phi,u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial \phi} \right>, \,\, F_2(\phi, u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial u} \right>, \,\, G_2(\phi, u) = \left< \frac{\partial f_2}{\partial u}, \frac{\partial f_2}{\partial u} \right>. $$

For example,

$$ E_1(\theta,v) = \left< \frac{\partial f_1}{\partial \theta}, \frac{\partial f_1}{\partial \theta} \right> = \| \left( -\sin \theta \cosh v, \cos \theta \cosh v, 0 \right) \|^2 = \cosh^2(v), \\ E_2(\phi, u) = \left< \frac{\partial f_2}{\partial \phi}, \frac{\partial f_2}{\partial \phi} \right> = \| \left( -\sin \phi u, \cos \phi u, 1 \right) \| = 1 + u^2$$

and we indeed see that $$\cosh^2(v) = E_1(\theta,v)) = E_2(\tilde{F}(\theta,v)) = E_2(\theta, \sinh v) = 1 + \sinh^2 v. $$

I'll leave the rest of the calculations for you.

  • $\begingroup$ Thanks a lot for the very thorough, elaborative and constructive answer. I had an intuition that the fundamental quantities between the two formulas should be checked, but I was hesitant, as in another example regarding the proof of a conformal map, the calculations of them were very weird (?) as they were calculated via an inverse function. $\endgroup$ – Rebellos Jun 15 at 16:44
  • 1
    $\begingroup$ @Rebellos: Sure. The point is that in your question (and many other questions), instead of giving you explicitly the map $F$ between the surfaces $S_i$, they actually are giving you the map $\tilde{F}$ which is the local parametrization of $F$ (that is, they are describing $F$ in terms of coordinates both in the domain and range). Hence, there is no need to invert or anything, just work with $\tilde{F}$. $\endgroup$ – levap Jun 15 at 16:46
  • $\begingroup$ I found that out, after checking around the internet. Or actually, I thought that it may be it, because I found a paper elaborating on how you could re-parametrize an equation to show that a helicoid and catenoid are isometric (or something like that). In the conformal case here for example, how would one proceed though ? Do I need to find again that $\bar{F}$ function which does the trick you elaborated so instructively ? $\endgroup$ – Rebellos Jun 15 at 16:48
  • $\begingroup$ @Rebellos: You can definitely do that and it won't involve much work. But you can also work directly from the definition with $f$ and not with the local representation $\tilde{f}$. Unlike in this question, in your question about conformal maps, the $f$ is given as the restriction of a global map on $\mathbb{R}^3$ so it is easy to calculate the differential of $f$ directly. $\endgroup$ – levap Jun 15 at 16:54
  • $\begingroup$ @Rebellos: Another disadvantage of working with $\tilde{f}$ and not $f$ in the other questions is that unlike here, in the case of the sphere there is no global parametrization so you will need to construct $\tilde{f}$ for more than one parametrization. $\endgroup$ – levap Jun 15 at 16:57

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