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Let $G(s,t)=(\cosh s \cdot \cos t, \sinh s \cdot \sin t)$, and let $A=\{(s,t)\in\mathbb{R^2}|s>0, 0<t<2\pi\}$. Prove that $G_{|A}$ has global inverse and give a graphical representation of $G(A)$.

After verifying that $G_{|A}\in C^1$ and $\det(JG_{|A})\neq 0\ \ \forall (s,t)\in A$, I want to prove that $G_{|A}$ is injective. I start with $\cosh \tilde s \cdot \cos \tilde t = \cosh s \cdot \cos t$ and $\sinh \tilde s \cdot \sin \tilde t = \sinh s \cdot \sin t$, but I don't know how to prove that $\tilde s = s$ and $\tilde t = t$.

$\cosh s$ and $\sinh s$ are injective in $A$, but I don't know if this helps.

And what's the idea behind the graph I am asked to draw? Thanks in advance!

$\textbf{Attempt}$

Let $u=\cosh s \cdot \cos t,\ t=\arccos(\frac{u}{\cosh s})$

$v = \sinh s \cdot \sin t=\sinh s\cdot \sin(\arccos(\frac{u}{\cosh s}))=\sinh s\cdot\sqrt{1-(\frac{u}{\cosh s})^2}$

Given $u$, we can prove $v=g_u(s)$ is injective by verifying if its first derivative is strictily monotonic,

$g_u(s)'=\cosh s\cdot \sqrt{1-(\frac{u}{\cosh s})^2}+\sinh s\cdot\frac{1}{2\sqrt{1-(\frac{u}{\cosh s})^2}}\cdot2\frac{u}{\cosh s}\cdot\frac{u}{\cosh^2s}\cdot \sinh s=$

$=\cosh s\cdot \sqrt{1-(\frac{u}{\cosh s})^2}+\frac{u^2\sinh^2 s}{\cosh^3\sqrt{1-(\frac{u}{\cosh s})^2}}>0$, because $\cosh s>1,\ \sinh s>0\ \forall s>0,$

and $1-(\frac{u}{\cosh s})^2=1-\frac{\cosh^2s\cdot\cos^2 t}{\cosh^2s}=1-\cos^2t=\sin^2t\in[0,1]\ \forall t\in(0,2\pi)$

But $\sin t=0$ if $t=\pi$, so the proof might fail here.

Anyway, we can repeat the same prodecure to prove that $u$ is injective. From $v$, we have that: $t=\arcsin(\frac{v}{\sinh s})\to u=\cosh s\cdot \cos(\arcsin(\frac{v}{\sinh s}))=\cosh s\cdot\sqrt{1-(\frac{v}{\sinh s})^2}$

Given $v$, we prove $u=h_v(s)$ is injective:

$h_v(s)'= \sinh s\cdot\sqrt{1-(\frac{v}{\sinh s})^2}+\cosh s\cdot\frac{1}{2\sqrt{1-(\frac{v}{\sinh s})^2}}\cdot2\frac{v}{\sinh s}\cdot\frac{v}{\sinh^2s}\cdot \cosh s=$

$=\sinh s\cdot \sqrt{1-(\frac{v}{\sinh s})^2}+\frac{v^2\cosh^2 s}{\sinh^3\sqrt{1-(\frac{v}{\sinh s})^2}}>0$, because $\cosh s>1,\ \sinh s>0\ \forall s>0,$

and $1-(\frac{v}{\sinh s})^2=1-\frac{\sinh^2s\cdot\sin^2 t}{\sinh^2s}=1-\sin^2t=\cos^2t\in[0,1]\ \forall t\in(0,2\pi)$

And we have a similar problem: $\cos t=0$ if $t=\pi/2$.

What's wrong in my attempt? Am I making wrong assumptions?

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  • $\begingroup$ Isn't it enough for the Jacobian to be invertible everywhere? Or are you just particularly interested in proving it directly? $\endgroup$ – preferred_anon May 2 '19 at 21:19
  • $\begingroup$ If $G_{|A}\in C^1$ and $\det(JG_{|A})\neq 0\ \ \forall (s,t)\in A$ then, by the Inverse Function Theorem, $\forall(x,y)\in A$, $\exists\ \underline{\text{local}}$ diff inverse of $G$ around the point $(x,y)$. To prove that exists a $\underline{\text{global}}$ inverse of $G_{|A}$, I must prove that the function is injective. Am I right or am I making bad assumptions? $\endgroup$ – Gibbs May 2 '19 at 21:36
  • $\begingroup$ Ah no I think you're right, my bad memory. I had a go at the algebra but couldn't get it out $\endgroup$ – preferred_anon May 2 '19 at 21:39
  • $\begingroup$ Are you sure the second component of $G$ isn't $\cosh s \sin t$? That would (1) make the problem a lot less technically unpleasant, and (2) relate $G$ to a familiar parameterization of a well-known surface. $\endgroup$ – Travis Willse May 2 '19 at 21:50
  • $\begingroup$ $G$ is correct as it is written in the question. Maybe my textbook is wrong, but I don't think so... $\endgroup$ – Gibbs May 2 '19 at 22:06
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Hint Under the usual identification $\Bbb R^2 \leftrightarrow \Bbb C$, $(u, v) \leftrightarrow u + iv$, the transformation $G$ is $$G(s + it) = \cosh s \cos t + i \sinh s \sin t ,$$ but writing $z := s + it$ this is precisely $$G(z) = \cosh z .$$

We need to show that $G\vert_A$ admits an inverse---which would necessarily be a branch of the inverse hyperbolic cosine---where $A$ is the open half-strip $\{\operatorname{Re} z > 0, 0 < \operatorname{Im} z < 2 \pi\}$. But by definition $G\vert_A^{-1}$ is exactly the principal branch of the inverse hyperbolic cosine, namely $$\operatorname{arcosh} w = \log(w + \sqrt{w - 1} \sqrt{w + 1}) .$$ Here, the branches of $\log$ and $\sqrt{\cdot}$ are the principal ones.

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As for the graphic representation note that for $u=\cosh s\cdot\cos t$, $v=\sinh s\cdot\sin t$: $$ \frac{u^2}{\cosh^2 s}+\frac{v^2}{\sinh^2 s} = 1$$ $$ \frac{u^2}{\cos^2 t}-\frac{v^2}{\sin^2 t} = 1$$ Which means that the lines of constant $s$ are ellipses, and the lines of constant $t$ are hyperbolae. You can also check that they will always intersect at right angles. I believe that for the graphical representation of $G$ you're supposed to sketch the image of a grid under the action of $G$.

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  • $\begingroup$ This is nicely put! The fact that the lines of constant parameter values $s, t$ is, by the way, an immediate consequence of the fact that we can view $(s, t) \to G(s, t)$ as an analytic function on $\Bbb C$ (see my answer). In particular it is conformal and so preserves angles everywhere that $G'(z) \neq 0$, which holds everywhere in $A$. $\endgroup$ – Travis Willse May 6 '19 at 22:20
  • $\begingroup$ That should be "...that the lines of constant parameter values $s, t$ are orthogonal...". $\endgroup$ – Travis Willse May 8 '19 at 17:48

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