Edit: I was able to show that the two maps are isometries. Only need help with the latter part of the problem, namely, that radial straight lines are geodesics relative to the hyperbolic metric.
The problem statement below (see link to image) is taken from Elementary topics in differential geometry by John A. Thorpe, chapter 24 Riemannian metrics.
The hyperbolic metric is defined in the book by
$$g(v, w) = \frac{4 v \bullet w}{(1 – ||p||^{2})^{2}},$$
where $p$ is a point in the unit disc $U = \{ (x_{1}, x_{2}) \in \mathbb{R}^{2} : x_{1}^{2} + x_{2}^{2} < 1 \}$. Now, the book defines an isometry as a bijective local isometry; where a local isometry is defined as preserving dot products from tangent spaces such that
$$d \psi(v) \bullet d\psi(w) = v \bullet w.$$
Where $d \psi(v)$ is the differential which can be computed by $d \psi(v) = (\nabla_{v} \psi_{1}, \ldots , \nabla_{v} \psi_{n}) = (\nabla \psi_{1} \bullet v, \ldots , \nabla \psi_{n} \bullet v)$ (at least on surfaces $S \subseteq R^{n}$).
Now, when trying to show that the maps in (i) and (ii) (see problem statement) are isometries, I am confused to whether I have to show that
$$d \psi(v) \bullet d\psi(w) = v \bullet w \quad \text{ or } \quad g(d \psi(v), d\psi(w)) = g(v, w)?$$
If it is the latter, then when computing the differential do we use the usual dot product or again the hyperbolic metric?
As another idea that I have but unsure whether it is valid: When I took functional analysis, we defined an isometry of two metric spaces as a bijective map $\phi$ from one metric space into the other such that $d(\phi(x), \phi(y)) = d(x, y)$. Would this work here as well if we treat the hyperbolic metric as an inner product and then induce a norm from it and from the norm we induce a distance?
Edit: I was able to solve (i). Namely, one has to show that $g(d \psi(v), d\psi(w)) = g(v, w)$ and using $d \psi(v) = (\nabla_{v} \psi_{1}, \ldots , \nabla_{v} \psi_{n}) = (\nabla \psi_{1} \bullet v, \ldots , \nabla \psi_{n} \bullet v)$ with the usual dot product from $\mathbb{R}^{n}$. Will try the same approach with (ii). Any ideas on the last question of the problem statement?
Best Answer
To show that straight lines from the origin are geodeics, you can make use of the fact that isometries preserve geodesics. More precisely, if $\psi:M\to N$ is an isometry between Riemannian manifolds ($M$ and $N$ may be the same) and $\gamma:(a,b)\to M$ is a path in $M$, then $\gamma$ is a geodesic in $M$ if and only if $\psi\circ\gamma$ is a geodesic in $N$. A proof of this will depend on what definition of a geodesic you're using.
Additionally, we know from ODE theorems on existence and uniqueness of solutions that if $\gamma(0)$ and $\dot{\gamma}(0)$ are specified, there is a unique maximal geodesic $\gamma:(a,b)\to M$ satisfying these initial conditions.
Combining these two, it's straightforward to show that if $\psi:M\to M$ is an isometry, $\gamma:(a,b)\to M$ is a maximal geodesic, $\psi(\gamma(0))=\gamma(0)$, and $d\psi(\dot{\gamma}(0))=\dot{\gamma(0)}$, then $\psi\circ\gamma=\gamma$. That is, if the position and velocity of a geodesic is invariant under an isometry at one point, then the entire geodesic is invariant under that isometry.
Using the above claim, we can make a symmetry argument. In the Hyperbolic metric on the $2$-ball we can choose a geodesic $\gamma$ with $\gamma(0)=(0,0)$ and $\dot{\gamma}(0)=(1,0)$. Since these are invariant under the isometry $\psi(x,y)=(x,-y)$ we conclude that $\psi\circ\gamma=\gamma$, and thus $\gamma$ must lie entirely on the $x$-axis. This determines $\gamma$ up to parameterization. Since rotations about the origin are also isometries, all of the rotations of $\gamma$ are also geodesics.
As for the point about distance functions, it's important to remember that the Riemannian metric is an inner product on tangent spaces, not on the manifold itself. The construction you describe would give a distance function on a tangent space, but not on the underlying manifold. That said, it is possible to derive a distance function on a connected Riemannian manifold, namely $$ d(p,q)=\inf\left\{\text{length}(\gamma):\gamma\text{ a piecewise differentiable curve joining $p$ and $q$}\right\} $$ With this distance function, a connected Riemannian manifold is a metric space, and the two different notions of isometry always agree. Establishing this requires a bit more work than on vector spaces, though, and it's not particularly useful for this problem.