Unique metric for the Hyperbolic Half Plane Model? I was reading today that there is a unique metric (up to multiplicative constant) that preserves distances wrt to linear fractional transformations:
$$z \mapsto \frac{az + b}{cz + d}$$
of the upper half complex plan to itself ($a, b, c, d \in \mathbb{R}$).  This metric is hyperbolic distance.
I'd like to read the proof of the result.  Whereabouts can I find it? Or I would appreciate a sketch of why it's true.  I am wondering if it's the sort of thing where you "guess" the hyperbolic metric and then prove it's unique; or if you somehow construct it from the set of transformations.
Thanks.
 A: First, note that the transformation rule defines an action of $GL_+(2, \Bbb R)$ (the group of $2 \times 2$ real matrices with positive determinant) on the upper half-plane, namely via
$$\pmatrix{a & b\\c & d} \cdot z := \frac{a z + b}{c z + d}.$$
For any such transformation, if we scale each of $a, b, c, d$ by $\lambda \in \Bbb R - \{0\}$, the transformation is unchanged, and the quantity $a d - b c$ scales by a factor of $\lambda^2$; so, we don't lose anything by restricting our attention to the transformations for which $a d - b c = 1$, that is, by considering only the restriction of our action to $SL(2, \Bbb R)$.*
So, we're now looking for a metric invariant under the given $SL(2, \Bbb R)$-action on $\Bbb H$. In particular, it must be invariant under the isotropy action at any point. Direct computation shows that the isotropy subgroup at the convenient point $i \in \Bbb H$ is
$$\left\{\pmatrix{a & -b \\ b & a} : a^2 + b^2 = 1 \right\} \cong SO(2);$$ the isotropy action thus preserves a unique inner product (up to scale) on the tangent space $T_i \Bbb H$, and some computation shows that
$$dx\vert_i^2 + dy\vert_i^2 = dz\vert_i \,d\bar{z}\vert_i$$
is such an inner product (here, $x, y$ are the usual real coordinates on $\Bbb C$, namely the ones characterized by $z = x + i y$). Pulling this metric back by arbitrary elements $g \in SL(2, \Bbb R)$ gives that the metric
$$\color{#bf0000}{\frac{dx^2 + dy^2}{y^2} = \frac{dz \,d\bar{z}}{(\Im z)^2}},$$
the usual hyperbolic metric, is invariant under $SL(2, \Bbb R)$.
Remark *Given such a transformation with coefficients $a, b, c, d$, the transformation with $-a, -b, -c, -d$ defines the same transformation, so the $SL(2, \Bbb R)$-action in fact descends to an action of $PSL(2, \Bbb R) = SL(2, \Bbb R) / \{\pm I\}$, and this latter action is faithful. It's often more convenient, though, to work with $SL(2, \Bbb R)$, whose elements are bona fide matrices, at the cost of a little redundancy.
A: If $g$ is the hyperbolic metric of curvature $-1$ on $\mathcal{H}\colon = \{ \mathcal{Im} z > 0\}$, $g = \frac{d x^2+ d y^2}{y^2}$, then it's easy to show that $g$ is invariant under all the transformations $z \mapsto \frac{az + b}{c z +d}$, where $a,b,c,d$ real and $ad - b c >0$. We want uniqueness of invariant metrics. Let $g'$ be any other Riemannian metric on $\mathcal{H}$. Then there exists a unique function $\phi>0$ so that $g' = \phi\cdot g$. Let $T$ be any diffeomorphism of $\mathcal{H}$. Then $T^{*} g'= T^{*} (\phi \cdot g) = T^{*}\phi \cdot T^{*}g $. Now assume that $T$ is a linear fractional transformation as above. Then $T^{*} g = g$ ( $g$ is invariant). Therefore,  to have $g'$ invariant under all the linear fractional transforms we must have $\phi$ invariant. Since the group $GL(2,\mathbb{R})_{+}$ acts transitively on $\mathcal{H}$, this implies $\phi$ constant.
$\bf{Added:}$ Let's show that the metric
$$\frac{d x^2 + d y^2}{y^2}$$
is invariant.
It's best to use complex coordinates: $$\frac{d x^2 + d y^2}{y^2}= \frac{d z \,d \bar z}{(\mathcal{Im} z)^2}$$
Let $z'= T z = \frac{az + b}{cz + d}$, where $a,b,c,d \in \mathbb{R}$.  We have
$$z' - \bar z' = \frac{az + b}{cz + d}- \frac{a\bar z + b}{c\bar z + d}= \frac{(ad - b c) ( z - \bar z)}{|c z + d|^2} $$
so
$$\mathcal{Im} z' = \frac{(ad - b c)\mathcal{Im} z}{|c z + d|^2}$$
while 
\begin{eqnarray}
d z' = \frac{(a d - b c) d z}{(c z + d)^2} \\
d \bar z' = \frac{(a d - b c) d \bar z}{(c\bar z + d)^2}
\end{eqnarray}
Putting all together we conclude that $\frac{d z \,d \bar z}{(\mathcal{Im} z)^2}$ is invariant under the transformation $T$.
Added: a useful formula to check that the curvature is $-1$: the curvature of the metric $\rho(dx^2 + d y^2)$ is $$-\frac{\Delta(\log \rho)}{2\rho}$$
