# Urgent Stereographic projection question

1. May 22, 2007

### Mathman23

Urgent Stereographic projection question....

1. The problem statement, all variables and given/known data

Given the unit sphere $$S^2= \{x^2 + y^2 + (z-1)^2 = 1\}$$

Where N is the Northpole = (0.0.2) the stereographic projection

$$\pi: S^2 \sim \{N\} \rightarrow \mathbb{R}^2$$ carries a point p of the sphere minus the north pole N onto the intersection of the xy plane with the straight line which N to p. Then if $$(u,v) = \pi(x,y,z)$$ where $$(x,y,z) \in S^2 \sim \{N\}$$ and $$(u,v) \in xy$$ plane.

(1)Then I need to show the following:

$$\pi^{-1}: \mathbb{R}^2 \rightarrow S^2$$

$$\pi^{-1} \Biggr( \begin{array}{ccc}x = \frac{4u}{u^2+v^2+4} \\ y = \frac{4v}{u^2+v^2+4} \\ z = \frac{2(u^2+v^2)}{u^2+v^2+4} \end{array}$$

(2)Show that this can be covered by two coordinant neighbourhoods using stereographic projection.

2. Relevant equations

Could somebody please give me a small hint maybe or how to proceed here??

3. The attempt at a solution

(1) Here I need to find the unique line from N to xy-plane, which I get to

x = ut, y = vt and z = 2-2t. For t = 0; (x,y,z) = (0,0,2) and for t = 1; (x,y,z) = (u,v,0)

This unique line intersects with the sphere at:

$$(ut)^2 + (vt)^2 + (1-2t)^2 = 1 \Rightarrow (4+u^2 + v^4) \cdot t^2 - 4t =0$$

where we interested at the solution $$t = \frac{4}{4+u^2+v^2}$$

Thus the point of intersection with the sphere is

$$x = \frac{4u}{u^2+v^2+4}$$ and $$y = \frac{4v}{u^2+v^2+4}$$ and $$z = \frac{2(u^2+v^2)}{u^2+v^2+4}$$

$$\pi_{1}(u,v) = \frac{4u}{u^2+v^2+4},-\frac{4v}{u^2+v^2+4}, \frac{2(u^2+v^2)}{u^2+u^2+4}$$

(2)

If lets say a parameterzation around the North Pole by sending (u,v) to (u,-v,0), the resulting map

$$\pi_{2}(s,t) = \frac{4s}{s^2+t^2+4},-\frac{4t}{s^2+t^2+4}, \frac{2(s^2+t^2)}{s^2+t^2+4}$$

it sends the unit circle to the equator and the unit disk to the Northern hemisphere.

It covers the sphere except for the Southpole which is covered by the parameterization in (1). Ergo these two maps or neighbourhood cover the entire sphere.

By simple Algebra

$$\pi_{2}(u,v) = \pi_{1}(\frac{4s}{s^2 + t^2}, \frac{-4t}{s^2 +t^2})$$

$$\pi_{1}(s,t) = \pi_{2}(\frac{4u}{u^2 + v^2}, \frac{-4v}{u^2 +v^2})$$

Q.e.d.

Is this correct?? Or am I on the wrong track?? I hope somebody would answer cause you guys are my only hope :|
Best Regards
Fred

Last edited: May 23, 2007