# A steographic problem(did I get it right?)

1. Mar 10, 2009

### Cauchy1789

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

A given sphere $$S^2$$ is given by $$x^2 + y^2 + (z-1)^2 = 1$$
where stereographical projection $$\pi:\pi: S^2 \thilde \{N\} \rightarrow \mathbb{R}^2$$
which carries a point p = (x,y,z) of the sphere minus Northpole N = (0,0,2) onto the intersection of the xy plane which a straight line which connects N to p.

(u,v)= pi(x,y,z)

3. The attempt at a solution

Show that $$\pi^{-1}: \mathbb{R}^2 \rightarrow S^2$$ is given by

$$\pi^{-1}(x,y,z) = (\frac{4u}{u^2 + v^2 + 4}, \frac{4v}{u^2 + v^2 + 4}, \frac{2(u^2+v^2)}{u^2 + v^2 + 4})$$

We can construct the line from N to p. such that x = us, y = vs and z = 2-2s

Thus the intersection of the line L from N to p.
$$(us)^2 + (vs)^2 + (1-2s)^2 = 1$$

which gives us an $$s = \frac{4}{u^2 + v^2 + 4}$$

thus L(s) = $$(\frac{4u}{u^2+v^2 +4}),(\frac{4v}{u^2+v^2 +4}), (\frac{2(u^2 +v^2}{u^2+v^2 +4})$$

Show that the sphere can the coverd by stereographic projection can be covered by the two coordinant neighbourhoods.

Let a mapping of of parameterization around the Northpole of the the sphere be the same to project the point (u,v) to (u,-v,0) which gives

$$\pi_2(s,t) = (\frac{4s}{s^2+t^2 +4}),-(\frac{4t}{s^2+t^2 +4}), (\frac{2(s^2 +t^2}{s^2+t^2 +4})$$

which coveres the sphere from the equator to the North pole, while

$$\pi_1(u,v) = (\frac{4u}{u^2+v^2 +4}),-(\frac{4v}{u^2+v^2 +4}), (\frac{2(u^2 +v^2}{u^2+v^2 +4})$$

thus they are covered by two neighbourhoods...

How is this??