We let
be a sphere in Euclidean three space. We want to obtain
a picture of the sphere on a flat piece of paper or a plane.
Whenever one projects a higher dimensional object onto a lower dimensional object, some type of distortion must occur. There are a number of different ways to project and each projection preserves some things and distorts others.
Later we will explain why we choose stereographic
projection, but first we describe it.
We shall map the sphere
onto the plane containing its equator.
Connect a typical point
on the surface of the sphere to the north pole
by a straight line in three space. This line will intersect the equatorial plane at some point
. We call
the projection of
.
Using this recipe
every point of the sphere except the North pole projects to some point on the equatorial plane. Since we want to include the North pole in our picture, we add an extra point
, called
the point at infinity, to the equatorial plane
and we view
as the image of
under stereographic projection.
to be the unit sphere,
so that
plane is the equatorial plane. The typical point
on the sphere
has coordinates
. The typical point
in the equatorial plane, whose coordinates are
, will be called
.
.
.
.
For this to be true where do we have to think of
as lying: interior to
or exterior to it?
-axis?
What projects onto the
?
Call the set of points that project onto
the prime meridian.
-axis?
onto any vertical line, not necessarily the
axis?
Stereographic projection preserves circles and angles. That is, the image of a circle on the sphere is a circle in the plane and the angle between two lines on the sphere is the same as the angle between their images in the plane. A projection that preserves angles is called a conformal projection.
We will outline two proofs of the fact that stereographic projection preserves circles, one algebraic and one geometric. They appear below.
Before you do either proof, you may want to clarify in your own mind what a circle on the surface of a sphere is. A circle lying on the sphere is the intersection of a plane in three space with the sphere. This can be described algebraically. For example, the sphere of radius 1 with center at the origin is given by
An arbitrary plane in three-space is given by
for some arbitrary choice of the constants
,
,
, and
.
Thus a circle on the unit sphere is any set of points whose coordinates
simultaneously satisfy equations 2 and 3.
The fact that the points
,
and
all lie on one line can be expressed by the fact that
for some non-zero real number
. (Here
.)
The idea of the proof is that one can
use equations 2 and 4 to write
as a function of
and
,
as a function of
and
, and
as a function of
and
to simplify equation 3 to an equation in
and
.
Since the equation in
and
so obtained is clearly the equation of a circle in the
plane, the projection of the intersection of 2 and 3 is a circle.
To be more precise:
Equation 4 says that
.
Set
and verify that

If

lies on the plane,

Thus

Or

Whence,

Or

Recalling that

, we see
Since the coefficients of the

and the

terms are the same, this is the equation of a circle in the plane.
The geometric proofs sketched below use the following principle:
It doesn't really make much difference if instead of projecting onto the equatorial plane, we project onto another horizontal plane (not through N), for example the plane that touches the sphere at the South pole, S. Just what difference does this make?
, we
project onto the horizontal plane
through
. Then by symmetry
the tangent planes
and
at
and
make the same angle
with
, as also does
, by properties of parallelism (see figure # 1 at the end of this handout).
So
and
are images of each other in the (``mirror'') plane
through
and perpendicular to
.
For a point
on the sphere near
, the line
is nearly parallel to
, so that for points near
, stereographic projection is approximately
the reflection in
.
is where some cone touches the sphere, say the cone
of tangent lines to the sphere from a point
.
Now project onto the horizontal plane
through
.
In figure # 2 which NEED NOT be a vertical plane,
the four angles
are equal, for the same reasons as before, so that
. The image of
is therefore the horizontal circle of the
same radius centered at
.
in the
circle of radius
centered at
is the unique point
on the
ray
for which
.