Isometries of the hyperbolic plane

Question

$\color{brown}{\text{I was reading about}}$ the Mobius transformations $\mathcal{M}$ of the hyperbolic plane $\mathbb{H} := \{a + bi: b > 0\} \subset \mathbb{C}:= \mathbb{R}^2$ and its isometries in general. Let us assume we use the convention $a + bi := (a, b)$. To recall, the hyperbolic plane $\mathbb{H}$ is a smooth manifold as it is an open subset of $\mathbb{R}^2$. So, it inherits the already existing smooth structure. We can indeed consider it as a Riemannian manifold by defining the Riemannian metric

$$\left<\xi_1, \xi_2\right>_z := \mathrm{Re}\left(\dfrac{\xi_1\overline{\xi_2}}{\mathrm{Im}(z)^2}\right).$$

Furthermore, if $\xi \in T_z\mathbb{H}$ as $T_z\mathbb{H} \cong \mathbb{C} = \mathbb{R}^2$, one can always go back and forth between $\xi$ being a tangent vector or just a complex number. Finally, a map $T:\mathbb{H}\rightarrow \mathbb{H}$ is called an isometry if for all $z \in \mathbb{H}$ and $\xi_1, \xi_2 \in T_z\mathbb{H} \cong \mathbb{C}$ provided that $\left<DT(z)\xi_1, DT(z)\xi_2\right>_{T(z)} = \left<\xi_1, \xi_2\right>_z$.

$\color{brown}{\text{As an example of isometries that don't belong to}\ \mathcal{M}}$, I saw $z\mapsto -\bar z$ and $z\mapsto \dfrac{-1}{z}$. However these maps are not complex differentiable. Shouldn't an isometry be a biholomorphism? Maybe what people expect from isometries are just diffeomorphisms? How can one show that these maps are isometries?

Update: After discussing with @coiso and @didier it seems that I had more or less the answer to showing that the inner product is preserved for the second map but had a few computational mistake.

Answer 1

The map $f: \overset{\mathbb{H}}{\overbrace{\mathbb{R}\times \mathbb{R}^+}} \rightarrow \overset{\mathbb{H}}{\overbrace{\mathbb{R}\times \mathbb{R}^+}}$ given by $z \mapsto -\dfrac{1}{\bar z}$ is indeed given by

$$f(a, b):= -{\left(\dfrac{1}{a + bi}\right)} = - {\left(\dfrac{a - bi}{a^2 + b^2}\right)} = \dfrac{-a + bi}{a^2 + b^2} = \left(\dfrac{-a}{a^2 + b^2}, \dfrac{b}{a^2 + b^2}\right).$$ As $b \neq 0$ for any $(a, b) \in \mathbb{H}$, we can be sure that $f$ is differntiable in the calculus sense. The matrix of the Frechet derivative of $f$ at a point $z:= (a, b)$ is equal to

$$\left[\begin{array}{cc} \dfrac{a^2 - b^2}{(a^2 + b^2)^2} & \dfrac{-2ab}{(a^2 + b^2)^2} \\ \dfrac{2ab}{(a^2 + b^2)^2} & \dfrac{a^2 - b^2}{(a^2 + b^2)^2} \end{array}\right] = \dfrac{1}{(a^2 + b^2)^2} \left[\begin{array}{cc} a^2 - b^2 & -2ab \\ 2ab & a^2 - b^2 \end{array}\right].$$

We are now ready to check whether $f$ is an isometry or not. First we note that $$\boxed{\left<(x_1, y_1), (x_2, y_2)\right>_{(a, b)} = \dfrac{x_1x_2 + y_1y_2}{b^2}} $$ We now expand $$\left<Df(a, b)(x_1, y_1), Df(a, b)(x_2, y_2)\right>_{f(a, b)}$$ to get $$ \dfrac{1}{(a^2 + b^2)^4} \left<( \color{blue}{(a^2 - b^2)x_1 - 2aby_1}, \color{green}{2abx_1 + (a^2 - b^2)y_1}) , (\color{blue}{(a^2 - b^2)x_2 - 2aby_2}, \color{green}{2abx_2 + (a^2 - b^2)y_2}) \right>_{\dfrac{(-a, b)}{(a^2 + b^2)}} $$ which is $$ \dfrac{\dfrac{1}{(a^2 + b^2)^4}\left( \begin{aligned} \color{blue}{(a^2 - b^2)^2x_1x_2 + (2ab)^2y_1y_2 + 2ab(a^2 - b^2) (x_1y_2 + x_2y_1)} \\ + \color{green}{(a^2- b^2)^2y_1y_2 + (2ab)^2x_1x_2 - 2ab(a^2 - b^2) (x_1y_2 + x_2y_1)} \end{aligned} \right)}{\left(\dfrac{b}{a^2 + b^2}\right)^2} $$ which simplifies to $$ \dfrac{ \dfrac{1}{(a^2 + b^2)^2}\Big((a^2 - b^2)^2 + (2ab)^2\Big)(x_1x_2 + y_1y_2)}{b^2}. $$

This last expression also simplifies to

$$ \boxed{ \dfrac{ (x_1x_2 + y_1y_2)}{b^2} } $$

which is exactly the expression we had in the first box.