Isometries on the Hyperbolic Plane

Question

This paper states in definition 1.12 that a function $\phi:\mathbb{H}\to\mathbb{H}$ is an isometry of the hyperbolic plane if for all $z\in\mathbb{H}$ and $v,w\in T_z\mathbb{H}$, $$\langle v,w\rangle_z=\langle D_{\phi_z}(v),D_{\phi_z}(w)\rangle_{\phi(z)}$$Note that all of the above notation is formally defined in the first 3 pages of the cited paper. It is then stated that this definition of isometry can be found to be equivalent to the standard "distance preserving function" definition through some "simple calculation", but I am failing to see this calculation. Why are the two definitions equivalent?

Answer 1

This is only a partial answer, as I believe the remark in the cited paper is a bit too optimistic (I mean, the result is true, but so not obvious as claimed).

Let $(M,\langle,\rangle)$ be a Riemannian manifold.

By definition, the distance between $x$ and $y$ is the infimum of length $l(\gamma)$ of differentiable paths $\gamma:[0,1]\to M$, where $$l(\gamma)=\int_0^1 \sqrt{\langle \dot{\gamma}(t),\dot{\gamma}(t) \rangle_{\gamma(t)}}dt.$$

Assume $\phi:M\to M$ satisfies $\langle v,w \rangle_x=\langle D\phi_x.v, D\phi_x. w \rangle_\phi(x)$. Then the distance between $\rho(\phi(x),\phi(y))$ is smaller than the distance $\rho(x,y)$ between $x$ and $y$. To see this, let $\gamma$ be a path between $x$ and $y$. Then $\phi\circ \gamma$ is a path between $\phi(x)$ and $\phi(y)$, of length $$l(\phi\circ \gamma)=\int_0^1 \sqrt{\langle D\phi_{\gamma(t)}.\dot{\gamma}(t),D\phi_{\gamma(t)}.\dot{\gamma}(t) \rangle_{\phi(\gamma(t))}}dt =l(\gamma).$$ Minimizing over all possible $\gamma$ gives $$\rho(\phi(x),\phi(y))\leq \rho(x,y).$$

If $\phi$ is a diffeomorphism, then the same thing applied to $\phi^{-1}$ gives the reverse inequality. So we have proved that a differentiable map preserving locally the scalar product is distance-decreasing, and by the inverse function theorem, a local isometry. My belief is that the "simple calculation" refers to this.

Note that it does not seem completly trivial (at least to me) to prove that local isometries of $\mathbb{H}\to \mathbb{H}$ are indeed diffeomorphism (both surjective and injective are not completly obvious): this certainly has to do with $\mathbb{H}$ being simply connected (or, in this case, some complex analysis may be sufficient).

For the converse, one should prove that a isometry is a differentiable function; once this holds, the differential has to preserve the scalar product. But again, this is not completely immediate.