Stereographic Projection

[Mathman23]

Hi there,

Look at the topic from my textbook "Stereographic Projection". Please inform me if I have understood it correctly :)

Lets take specific example from my textbook $$S' = \{(x,y,z)|x^2 + y^2 + (z-1)^2 = 1\}$$ is a sphere where $$N = (0,0,2)$$ and $$P = (x,y,z)$$ can be viewed as steographic projection $$g: \sim\{N \} \rightarrow \mathbb{R}^2$$ which carries a point of the sphere S' minus the North Pole N onto the intersection of line from N to p with the xy-plane. Then $$g(x,y,z) = (u,v)$$ and $$(x,y,z) \in S' \sim \{N\}$$ and finally $$(u,v) \in \mathbb{R}^2$$.

(1) show that $$g^{-1}: \mathbb{R}^2 \rightarrow S'$$ is $$\left( \begin{array}{ccc}x\\y\\z\end{array} \right) = \left( \begin{array}{ccc}\frac{2(u^2 + v^2)}{u^2 + v^2 +4} \\\\ \frac{2u}{u^2 + v^2 +4} \\\\ \frac{2v}{u^2 + v^2 +4} \end{array} \right)$$


Solution (1):

If I write S' in terms of u and v I get

$$u^2 + v^2 = \frac{z}{2-z}$$

then $$z = \frac{2(u^2 + v^2)}{u^2 + v^2 +1}$$

where its possible to see $$2-z = \frac{2}{u^2 + v^2 +1}$$

Therefore $$x = \frac{2u}{u^2 + v^2 +1}$$

and

$$y = \frac{2v}{u^2 + v^2 +1}$$

p.s. I treat $$g(x,y,z) = \left\{\begin{array}{ccc}\frac{x + iy}{2-z} \ \ \mathrm{if \ z \neq 2} \\ \infty \ \ \mathrm{if \ z=2} \end{array}$$

and that $$g: S' \rightarrow \mathbb{C}^*$$ is a unique plane which containes (0,0,1) and (0,0,2) and (x,y,z). Futhermore $$g(x,y,z) = u + iv$$

This is true is $$g^{-1}(u+iv) = (\frac{2u}{u^2 + v^2 +1})^2 + (\frac{2v}{u^2 + v^2 +1})^2 + ((\frac{2(u^2 + v^2)}{u^2 + v^2 +1})-1)^2=1$$

How can this solution be true, since the textbook claims that

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

$$\mathbb{C}^* = \mathbb{C} \cup \{\infty\}$$ is a Riemann Sphere. Then the stereographic projection from S' to $$\mathbb{C}^{*}$$ is defined by setting g(x,y,z) equal to the point where the line from (x,y,z) to (0,0,2) intersects with the plan z=0. This is defined for all points on S' except (0,0,2).

2) I am then told to show that sphere S' can be covered by two coordinant neighborhoods. I have an idear on howto show this: If their exist a Coordinant neighborhood around the Northpole and Southpole of S', then if their respective domains intersect, then the sphere is covered by these two coordinant neighborhoods? Aren't they?

Best Regards

Fred

p.s. Could somebody here please inform me what I am doing wrong here in 1?

 

[lippyka]

Solution (2): Yes, you are correct that the sphere S' can be covered by two coordinate neighborhoods. If a coordinate neighborhood is defined around the North pole and South pole of S', then the intersection of their respective domains will cover the entire sphere. However, in order to prove this, you will need to show that the intersection of the two coordinate neighborhoods covers all points on the sphere S'. You can do this by showing that all points on the sphere can be expressed in terms of the two coordinate neighborhoods.