How are Stereographic Projections Derived in Differential Geometry?

[niall14]

In cartesian coordinates (x.y,z) on the and (X,Y) on the plane, the projection and its inverse are given by the following formulae:

(X,Y)=(x/1-z,y/1-z)

(x,y,z)=(2X/1+X^2+Y^2, 2Y/1+X^2+Y^2, -1+X^2+Y^2/1+X^2 +Y^2)

This relates to the field of differntial geo.Anybody have a proof to where thes equations are derived?

Thanks very much

 

[Kreizhn]

The derivation is straightforward if not somewhat cumbersome. Do you know how stereographic projection is defined? You take the straightline through a pole and a point on the sphere and project it onto the plane. Consider a fixed point on the sphere (x,y,z) and find the line through the North pole and (x,y,z). Then find where it intersects the plane z=0.

 

[niall14]

thanks for your reply.not really to sure what u mean by 'project'.if you could give me the first few lines of the proof I am sure i could figure it out.

 

[Kreizhn]

I'll give you one direction, the other follows similarly but is definitely the harder direction. Consider the north pole $(0,0,1)$ and let $(x_0,y_0,z_0)$ be a point on the sphere. Then the line between the north pole and this point is given by
$$x = x_0 t, \qquad y = y_0 t, \qquad z = 1 + (z_0 - 1)t$$
which is easily calculated via basic geometric results. Notice that when $t=0$ we get the North pole, and when $t = 1$ we get the point on the sphere. Now we want to calculate where this line hits the z=0 plane, so in particular
$$z= 0 = 1+(z_0 - 1)t \quad \Rightarrow t = \frac1{1-z_0}$$
Substituting this value into the equation of the line we get the intersection of the line with the plane
$$\left( \frac{x_0}{1-z_0}, \frac{y_0}{1-z_0}, 0 \right)$$
and we drop the third coordinate since we want to identify the sphere with the plane.

Again, the other direction is a bit harder and more technical, but follows precisely the same idea. Give it a try yourself and see if you can get the right answer.

 

[niall14]

thanks ill give it a go