# Stereographic projection problem

Can anyone help me find the explicit representation of the inverse northern pole of the stereographic parametrizations of a unit sphere denoted by $SP^{-1}_{-}$ denoting the bottom half below the equator of the sphere. For instance I think I know what the upper half of the projection looks like that is $SP_+$ and $SP^{-1}_{+}$ that is (Note the domain is from the complex numbers $\mathbb{C}$ not the real plane):

$SP_{+}=(\frac{2Re(z)}{|z|^2+1},\frac{2Im(z)}{|z|^2+1}, \frac{|z|^2-1}{|z|^2+1})$ and $SP^{-1}_{+} = \frac{x_1+x_2i}{1-x_3}$.

Further note: $SP_{+}:\mathbb{C} \rightarrow \mathbb{S^2}-\{(0,0,1)\}$, $SP_{-}:\mathbb{C} \rightarrow \mathbb{S^2}-\{(0,0,-1)\}$. I can not however get the bottom half of the stereographic parametrisation and its inverse that is $SP^{-1}_{-}$ and $SP_{-}$ . Can anyone help me please?

• Just to clarify what you mean, you're looking for the corresponding stereographic projection formulae, where instead of projecting from $(0, 0, 1)$, you instead project from $(0, 0, -1)$? – Theo Bendit Feb 15 '18 at 4:46
• Yes that is correct. Thank You. – Aurora Borealis Feb 15 '18 at 5:01

You're reflecting the problem about the $x$-$y$ plane. The same point on the complex plane will stereographically project to the point $(x, y, -z)$ instead of $(x, y, z)$. So,

$$z \in \mathbb{C} \mapsto \left(\frac{2\Re(z)}{|z|^2 + 1}, \frac{2\Im(z)}{|z|^2 + 1}, -\frac{|z|^2-1}{|z|^2+1}\right) \in \mathbb{R}^3.$$

In other words, if $\sigma$ is the projection from $(0, 0, 1)$, and $R$ is the reflection about the $x$-$y$ plane, then we are looking at $R \circ \sigma$. The inverse will therefore be $\sigma^{-1} \circ R^{-1} = \sigma^{-1} \circ R$, which produces the map

$$(x_1, x_2, x_3) \in \mathbb{R}^3 \mapsto \frac{x_1 + i x_2}{1 + x_3} \in \mathbb{C}.$$

I hope that helps!

• Ok, that was very helpful thank you. Now I have another question, for instance if we consider some function (complex in nature) such that $f(z)= \frac{az+b}{cz+d}$ such that $a,b,c,d \in \mathbb{C}-\{0\}$ and $ad\ne bc$. If we define some map $\Psi:\mathbb{S^2}\rightarrow \mathbb{S^2}$ such that $\Psi(p)=SP_+ (a/c)$ if $p = (0,0,1)$ , and $(0,0,1)$ if $p = SP_+ (-d/c)$, and finally $SP_+(f(SP^{-1}_+(p)))$ otherwise. Question: how can we show that the tangent map $d\Psi$ is indeed invertible? – Aurora Borealis Feb 15 '18 at 5:34
• Look, I think this is the topic for a new question. I'm not sure I follow your question, and I think it'd benefit from not being stuffed into a comment. :-) – Theo Bendit Feb 15 '18 at 5:41
• Ok, anyways your help was much appreciated. – Aurora Borealis Feb 15 '18 at 6:00