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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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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