DIFF GEOM: The Poincare Half-Plane

The fundamental form can also be used to study surfaces which cannot be constructed in ordinary space. In particular, we can define a new geometry on the plane by giving it a non-Euclidean fundamental form. This is exactly what Henri' Poincare' did in order to study hyperbolic geometry, which is geometry in which there are infinitely many parallel lines to a line l through a point not on l itself.

Specifically, Poincare modeled hyperbolic geometry by assigning the fundamental form

ds2 =  du2+dv2 v2

(3)

to the upper half of the uv-plane. The result is called the Poincare half-plane and has a unique geometry that differs appreciably from the usual half-plane.

That is, distances between ( u,v) points are now defined by (3), so that the length of a curve á u(t) ,v( t) ñ , t in [ a,b] , is given by

l =   ó õ b a  1 v2   æ è du dt ö ø 2    +  1 v2    æ è dv dt ö ø 2     dt

or equivalently for functions v = f( u) , by the integral

l =   ó õ b a  1 v2    +  1 v2    æ è dv du ö ø 2     du

 

For example, the shortest path between the points ( -1,1) and ( 1,1) is not a straight line under the Poincare metric (3). Because distances become shorter as v increases, the shortest path between ( -1,1) and (1,1) is a curve with an intercept greater than v = 1.           

EXAMPLE 6    Compute the distances from ( -1,1) to ( 1,1) in the Poincare half-plane along the line segment v = 1.  Then compute it along the upper semi-circle v = ( 2-u² )^1/2.  Which path is longer with respect to the Poincare metric? Solution: The distance from ( -1,1) to (1,1) along the line segment v = 1 is given by

l =   ó õ 1 -1  1 1    +  1 1  (0) 2    du =  ó õ 1 -1  du = 2

If v = ( 2-u²) ^1/2, then v' = u(2-u²) ^-1/2. Thus, the distance from ( -1,1) to ( 1,1) along the upper semicircle v = ( 2-u²)^1/2 is given by

l = ó õ 1 -1   1 [ ( 2-u2) 1/2] 2 +  1 [ ( 2-u2) 1/2] 2 ( u(2-u2) -1/2) 2   du

This simplifies to

l = ó õ 1 -1   1 ( 2-u2) 1/2 1+  u2 2-u2   du = ó õ 1 -1   1 ( 2-u2) 1/2  2-u2 2-u2 +  u2 2-u2   du = ó õ 1 -1   1 ( 2-u2) 1/2  2 2-u2   du = 2 ó õ 1 0   Ö2 2-u2 du

since the integrand is even. Partial fractions then leads to

l =  2 2 ó õ 1 0  æ è  1 Ö2 - u +  1 Ö2 + u ö ø du = ( -ln( Ö2 - u) +ln( Ö2 + u))  1 0 =   ln( Ö2 -1 ) -ln( Ö2 + 1 )

Since ln( Ö2 - 1) -ln( Ö2 + 1) = 1.762, the distance from ( -1,1) to ( 1,1) is shorter along the circle v = ( 2-u²) ^1/2 than along the straight line.

       

In fact, it is shown in the accompanying worksheet that the geodesics of the hyperbolic plane are vertical lines and semi-circles centered on the x-axis. That is, semi-circles centered on the x-axis and vertical lines are the ''straight lines'' in the Poincare half-plane.  Given a semi-circle C and a point P not on the semi-circle, there are infinitely many other semi-circles centered on the x-axis that pass through P. This means that in the Poincare half plane, there are infinitely many ''parallel lines'' to a given ''line'' l through a point P not on l.

We say that the geometry of the Poincare half-plane is non-Euclidean because it does not satisfy the parallel postulate that is foundational to Euclidean Geometry.