3

This question is from my tutorial problem set:

One way to define a system of coordinates for the sphere $S^2$ given by $x^2+y^2+(z-1)^2=1$ is to consider the stereographic projection $\pi:S^2-\{N\} \to \mathbb{R}^2$ which carries a point $(x,y,z)$ of $S^2$ minus north pole $N =(0,0,2)$ onto intersection of $xy$ plane with the straight line connecting $N$ to $p.$ Let $(u,v)=\pi(x,y,z).$

(I) Show that $\pi^{-1}:\mathbb{R}^2 \to S^2$ is given by $\pi^{-1}(u,v) =\dfrac{1}{u^2+v^2+4}\left(4u,4v,2u^2+2v^2\right)$

(II) Let $U$ be $uv$ plane. Using stereographic projection, define a coordinate function $\textbf{x}:U \to S^2 -\{N\}.$

For (I), it clear that $\pi^{-1}:\mathbb{R}^2 \to S^2$ isn't bijective since $(0,0,2) \not\in $ range of $\pi^{-1}.$ It seems natural to define $\textbf{x}(u,v) =\dfrac{1}{u^2+v^2+4}\left(4u,4v,2u^2+2v^2\right).$ So, what is the difference between (I) and (II) ?

(III): Find a coordinate function $\textbf{y}: U \to S^2-\{(0,0,0)\}.$

Isn't this exactly the same as (I) and (II) ?

Alexy Vincenzo
  • 5,400
  • 1
    Not quite. Stereographic projection from the south pole is different from stereographic projection from the north pole, although presumably you'll figure it out just fine. :) By the way, ordinarily $S^2$ denotes the unit sphere centered at the origin, so this notation is confusing to lots of us. – Ted Shifrin Feb 02 '14 at 03:55
  • Thank you. So (I) and (II) are essentially the same? Could you advise on finding the stereographic projection from south pole? – Alexy Vincenzo Feb 02 '14 at 03:57
  • 1
    Just redo the algebra, doing lines through $(0,0,0)$ instead of lines through $(0,0,2)$. But, clearly, you'll need a different plane, one that doesn't pass through the origin! (Ordinarily we do this with the $xy$-plane and the unit sphere centered at the origin.) – Ted Shifrin Feb 02 '14 at 04:08

0 Answers0