StereographicProjection.ps

StereographicProperties(i).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:

StereographicProperties(ii).ps

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'.

StereographicProperties(iii).ps

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.

StereographicProperties(iv).ps

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.

HyperbolicPlane.ps

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* .

Hyperbolic(i).ps

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"**.

Hyperbolic__ii_.ps

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.