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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.

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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:

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

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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.

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