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!


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?

  • $\begingroup$ 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

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.

| cite | improve this answer | |

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 $G$.

| cite | improve this answer | |
  • $\begingroup$ 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

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.