Isometry in Hyperbolic space

Let $\mathbb{H}^2=\{ (x,y)\in\mathbb{R}|\ y>0 \}$ the hyperbolic space with the metric $g=(dx^2+dy^2)/y^2$. Let $$\psi(x,y)=(\psi_1(x,y),\psi_2(x,y))=\left(\mathrm{Re}\frac{az+b}{cz+d},\mathrm{Im}\frac{az+b}{cz+d}\right),$$ where $z=x+iy$.
The numbers are chosen such that $ad-bc=1$.

I want to show that $\psi$ is an isometry, that is
\begin{equation}
\psi^*g=g\quad (1)
\end{equation}.
In coordinates the above equation is
\begin{equation}
g_{\mu\nu}(x,y)=g_{\alpha\beta}(\psi(x,y))\frac{\partial \psi_\alpha(x,y)}{\partial x_\mu}\frac{\partial \psi_\beta(x,y)}{\partial x_\nu},\quad (2)
\end{equation}
where $x_1=x$ and $x_2=y$.

I have trouble with the calculation. I’m sure that I can use (1) to show the isometry property. Unfortunately I failed. So I used the coordinate definition.

First I noted, that
\begin{align*}
g_{\alpha\beta}(\psi(x,y))&=\begin{pmatrix}
\frac{1}{\psi_2}&0\\0&\frac{1}{\psi_2}
\end{pmatrix}\\
\frac{\partial \psi_\alpha(x,y)}{\partial x_\mu}&=\begin{pmatrix}
\mathrm{Re}\frac{1}{(cz+d)^2}&\mathrm{Re}\frac{i}{(cz+d)^2}\\
\mathrm{Im}\frac{1}{(cz+d)^2}&\mathrm{Im}\frac{i}{(cz+d)^2}
\end{pmatrix}
\end{align*}

By doing the matrix calculation for the right hand side of the equation (2) I don’t get the final result $g_{\mu\nu}(x,y)$.

Hence the following questions arise:

  1. How can I use the coordinate free equation to prove the claim?
  2. Where is the mistake in the above calculations or in other words: Is my way absolutely nonsense?

Solutions Collecting From Web of "Isometry in Hyperbolic space"

As noted in the comment, it is easier to work with complex coordinate. It will be even easier if we break down the general transformation to simpler ones: Note that when $a, c\neq 0$,
$$\begin{split}
\frac{az + b}{cz+d} &= \frac ac \frac{cz + \frac{cb}{a}}{cz+ d}\\
& = \frac ac – \frac{d – \frac{cb}{a}}{cz + d} \\
&= \frac ac – \frac 1a \frac{ad-cb}{cz+ d} \\
&= \frac ac – \frac 1a \frac{1}{cz+ d}.
\end{split} $$
If $a=0$,
$$ \frac{b}{cz+d} = b\frac{1}{cz+d},$$
and when $c=0$,
$$\frac{az+ b}{d} = \frac{1}{d} (az+b).$$
So the mapping are composition of translation (by a real number), scaling (by a positive constant) and inversion $I(z) = -\frac{1}{z}$ . As composition of isometries is still an isometry, it suffices to consider only translation, scaling and inversion. The metric is given by
$$ g =\frac{1}{y^2} (dx^2 + dy^2),$$
so it is obvious that translations (by a real number) are isometries. For a scaling, let $A >0$ and $L_A (z) = Az$. Then
$$ L_A^* g = \frac{1}{(L_A(y))^2} \big( d (L_A x)^2 + d(L_A y)^2\big) = \frac{1}{A^2 y^2} (A^2 dx^2 + A^2 dy^2) = g. $$
Lastly, using complex coordinate $z =x+ iy$, then
$$ dx = \frac 12 (dz + d\bar z), \ \ dy = \frac 1{2i}(dz – d\bar z)\Rightarrow g = \frac{1}{(\text{Im}z)^2} dzd\bar z. $$
Using
$$ -\frac 1z = -\frac{\bar z}{|z|^2} \ \ \ \Rightarrow \text{Im}\big( I(z)\big) = \frac{1}{|z|^2} \text{Im} z, $$

\begin{split} \Rightarrow I^* g &= \frac{1}{(\text{Im} I(z)\big)^2} d(I(z)) d\overline{(I(z))} \\
&= \frac{|z|^4}{(\text{Im} z)^2} \bigg( \frac{1}{z^2} dz\bigg)\bigg( \frac{1}{\bar z^2} d\bar z\bigg) \\
& = \frac{1}{(\text{Im}z)^2} dzd\bar z = g
\end{split}
Thus $I$ is also an isometry.