# Proving $G(s,t)=(\cosh s \cdot \cos t, \sinh s \cdot \sin t)$ has global inverse in a certain region

Let $$G(s,t)=(\cosh s \cdot \cos t, \sinh s \cdot \sin t)$$, and let $$A=\{(s,t)\in\mathbb{R^2}|s>0, 0. Prove that $$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?

• Isn't it enough for the Jacobian to be invertible everywhere? Or are you just particularly interested in proving it directly? – preferred_anon May 2 '19 at 21:19
• 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? – Gibbs May 2 '19 at 21:36
• Ah no I think you're right, my bad memory. I had a go at the algebra but couldn't get it out – preferred_anon May 2 '19 at 21:39
• 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. – Travis Willse May 2 '19 at 21:50
• $G$ is correct as it is written in the question. Maybe my textbook is wrong, but I don't think so... – 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.
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 $$G$$.
• 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 $A$. – Travis Willse May 6 '19 at 22:20
• That should be "...that the lines of constant parameter values $s, t$ are orthogonal...". – Travis Willse May 8 '19 at 17:48