Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 2 / 3G (Geometry of Group Actions)

Table of contents
- Solution 3G

Solution

Use the curvature-minus-one Poincare disc model, whose Riemannian metric is

d s^{2} = \frac{4 ( d x ^{2} + d y ^{2} )}{( 1 - x ^{2} - y ^{2} ) ^{2}} .

Its geodesics are Euclidean diameters and arcs of circles meeting the boundary circle orthogonally. The hyperbolic area of a measurable region

D

is

Area (D) = \int_{D} \frac{4}{( 1 - x ^{2} - y ^{2} ) ^{2}} d x d y .

An isometry can move the center of a circle to the origin. Its hyperbolic radius

r

and Euclidean radius

ρ

then satisfy

r = \int_{0}^{ρ} 2 d u / (1 - u^{2}) = 2 artanh ρ

, so

ρ = tanh (r /2)

. Integrating the area element of a surface in polar coordinates yields

A (r) = \int_{0}^{2 π} \int_{0}^{ρ} \frac{4 u}{( 1 - u ^{2} ) ^{2}} d u d θ = \frac{4 π ρ ^{2}}{1 - ρ ^{2}} = 2 π (cosh r - 1) .

Integrating the line element around the circle gives

C (r) = \frac{4 π ρ}{1 - ρ ^{2}} = 2 π sinh r, A^{'} (r) = 2 π sinh r = C (r) .

A geometric definition of Pi valid in this geometry is

π = r ↓ 0 lim \frac{C ( r )}{2 r} = r ↓ 0 lim \frac{A ( r )}{r ^{2}} .

Indeed

sinh r = r + O (r^{3})

and

cosh r = 1 + r^{2} /2 + O (r^{4})

. The circumference-to-diameter ratio at a fixed positive radius is not constant, so the limiting qualification is necessary.

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