StereographicProjection.ps
From the perfect symmetry of the sphere, planes n and p' create equal angles with the straight line NP'. The line of intersection of n and p' is perpendicular to the line NP. Since n is parallel to the image plane, the image plane forms the same angle that p' does with the projecting ray PP'. It also intersects p' in a straight line perpendicular to PP'.
From this, we can draw the conclusion that this mapping is angle-preserving! Here's how:
If r' is tangent to the sphere at P' and r is the image of r', then r and r' form equal angles with PP'. This is because r is created by the intersection of the image plane with the plane containing r' and NP'.
BUT, if two straight lines r and r' are respective intersections of a plane e with planes p and p' - e containing straight line PP' and planes p and p' form equal angles with PP' while intersecting in a straight line perpendicular to PP' - then r and r' also form equal angles with PP'.
With the perfect symmetry of the sphere, we can now extend this property. Take s' as another tangent to the sphere at P'. With s as its image, the angle formed by r and s is equal to the angle formed by r' and s'.
Therefore, "Stereographic Projection reproduces the angles on the sphere without distortion", and so the mapping is angle-preserving.
We can also draw the conclusion that this mapping is also circle-preserving!
Take k' to be an arbitrary circle on the sphere and NOT passing through N. Then each point on k' has a plane tangent to the sphere and all of these planes create a circular cone with vertex S. Since k' does not pass through N, NS is not tangent to the sphere at N, hence it is not parallel to the image plane. Take M to be the point where NS meets the image plane.
Let P' be an arbitrary point on k' and P its image. Now P'S is tangent to the sphere at P' and PM is the image of P'S. The angle PP'S is equal to the angle P'PM. Next, we draw a line passing through S and parallel to PM. We put the point P" where this new line intersects line NP. So P" is either the same point as P' or the triangle P'P"S has equal angles at P' and P", making it an isosceles triangle with SP' equal to SP". We now have similar triangles with equal ratios: PM/P'S = PM/P"S = MN/SN, and by solving this we get: PM = P'S*MN/SN. Since S has the same distance from all points of k', we discover that P'S is constant and from the formula previously constructed, it follows that PM is also a constant. If PM is a constant, then k is a circle with M as its centre.
We have now shown that Stereographic Projection of a sphere onto an image plane maps circles, not passing through N, onto circles in the plane. By reversing the previous argument we can show that every circle on the image plane is the image of a circle on the sphere. If a circle is able to move on the sphere, it can approach a circle that is passing through N, making NS approach a tangent to the sphere at N and causing M to approach infinity. It then follows that the circles passing through N have images that are straight lines on the image plane. Hence, the set all circles on the sphere corresponds to the set of all circles and straight lines in the plane. So, "Stereographic Projection is circle-preserving".
Now we will look at any mapping, a', of the sphere onto itself. a' maps all of the circles on the sphere into circles (still on the sphere). ie. a' could be a rotation of the sphere about some diameter. With Stereographic Projection, a' creates a mapping a of the image plane into itself. a will map the set all circles and straight lines into itself. A map such as a of the plane into itself is a "circle-preserving transformation".
We are now in a position where we can finally construct the hyperbolic plane.
First, let the Hyperbolic Plane be represented by the interior of a circle m lying in a horizontal plane. At the centre of m, place a sphere having the same radius as m, touching the plane at that centre. Now project, by Vertical Parallel Projection, the circumference and interior of m onto the lower hemisphere bounded by circle l, congruent to m. This hemisphere has now become a new model for the Hyperbolic Plane. Every chord g of m is projected into a semicircle v of the sphere meeting l at right angles. These semicircles are regarded as the images of the hyperbolic straight lines.
Now, using Stereographic Projection, map the hemisphere back into the plane. The image of the hemisphere covers the interior of a circle k and is the new model for the Hyperbolic Plane.
With the angle- and circle- preserving nature of Stereographic Projection, the semicircles v have now become arcs n , perpendicular to k . The diameters of k will be included as the limiting cases in this class of circular arcs.
This final model of the Hyperbolic Plane is due to Poincaré.
There is a one-to-one correspondence between the set of all circular arcs perpendicular to k , and the set of all the chords of m . Therefore, any two points A and B in the interior of k can be joined by one and only one circular arc perpendicular to k . Let R and S be the two points where this arc meets k .
The hyperbolic distance between A and B can be calculated from the formula:
s = c/2 * |log AR*BS / BR*AS|
If A', B', R', and S' are the points of the original model that give rise to A, B, R, and S in the Poincaré model, then this relation holds:
AR*BS / BR*AS = [A'R'*B'S' / B'R'*A'S']^0.5
And so the formula for the hyperbolic distance in Poincaré's model is:
s = c * |log AR*BS / BR*AS|
Now, the Euclidean angles in Poincaré's model are equal to the hyperbolic angles multiplied by a fixed proportionality factor, x . But, since the angle 360 degrees of a full rotation is reproduced in the Hyperbolic Plane without any change, x must be 1. Thus, "Poincaré's model preserves angles".
Keeping in mind that the axioms of congruence are valid in the Hyperbolic Plane, we can draw a triangle A'B'M congruent to ABC, where the centre M of k corresponds to the point C. Since every circle perpendicular to k that passes through M will degenerate into a diameter of k and M is exterior to all other circles perpendicular to k, then in the figure above, the hyperbolic straight lines A'M and B'M can represented by Euclidean straight lines while the hyperbolic straight line A'B' is represented by a circular arc. Now, the Euclidean angles at A' and B' are smaller in the triangle A'B'M formed by two straight lines and a circular arc than the angles in the rectilinear triangle A'B'M, formed by three straight lines. We can then conclude that the sum of the interior angles of triangle A'B'M must be less than 180 degrees.
Because Poincaré's model preserves angles, the same is true for the sum of Hyperbolic angles in the Hyperbolic triangle A'B'M and also its congruent partner, triangle ABC.
Jennifer Montgomery
Cohn-Vossen S, Hilbert D. Geometry and the imagination. Nemenyi P, = translator; New York: Chelsea Publishing Company; 1952. 357 p. Translation of: Anshauliche geometrie.