# Length and area in hyperbolic geometry

I am reading a book about modern geometries by Michael Henle. He gives formulas for length of a curve and area of a region (in upper half plane model: $l(\gamma)=\int _a^b \frac{|z'(t)|}{y(t)}dt, A(R)=\iint_{R} \frac{dxdy}{y^{2}})$ as definitions. I am pretty sure these aren't Ad Hoc definitions though, but derived from something else. Am I right?

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– Will Jagy Sep 18 '12 at 18:44
So in the wikipedia article they start with a metric. Can one start, as the book did, with transformations, instead of with a metric (define the geometry to be the upper half plane where two sets are congruent if one set can be obtained by applying tranforms of the form (az+b)/(cz+d) on the other) and derive the metric from that? (Sound improbable.) – tsho Sep 18 '12 at 19:03
I don't particularly know. Give it a try. – Will Jagy Sep 18 '12 at 19:05

## 1 Answer

The invariance under transformations of the form $(az+b)/(cz+d)$ determines the metric up to a constant scalar factor. Think of a metric as a prescription of length of each tangent vector to the plane (the length may depend on the point at which the vector is attached to the plane). To normalize the metric, let's say that the vector $v_0=\langle 1,0\rangle$ attached at the point $i=(0,1)$ has unit length. The transformation $f(z)=(az+b)/(cz+d)$ (where we may assume $ad-bc=1$ for simplicity) has derivative $f'(z)=1/(cz+d)^2$. Thus, it maps $v_0$ to the vector $1/(ci+d)^2$ attached to the plane at $w=f(i)=(az+b)/(cz+d)$. It follows that this vector also has invariant length $1$. Since the Euclidean length of this vector is $1/(c^2+d^2)$, the invariant metric scales it by $c^2+d^2$.

Next observe that $$\mathrm{Im} w= \frac{(ai+b)(-ci+d)}{c^2+d^2}=\frac{1}{c^2+d^2}$$ Hence, the scaling factor at $w$ is $1/\mathrm{Im}\, w$. Note that this factor applies to vectors in all directions, because by varying $a,b,c,d$ we can map $v_0$ into a vector of any direction at any point.

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