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?
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!