Stereographic projection when the "North/South Pole" is not given by $(0,...,\pm 1)$?

Question

Straight forward enough... what if My point is arbitrary, how can I get a new stereographic projection?

Answer 1

Let $p=(p_1,\dots,p_n)$ be the "Pole", i.e., center of projection, lying on the unit sphere. The role of the $x_n$-coordinate is now taken by the linear functional $\varphi$ defined by $x\mapsto x\cdot p$. Note that $\varphi(p)=1$.

Given a point $x$ on the sphere, rescale the vector $x-p$ so that the $\varphi$ value of the scaled vector is $-1$. Then add $p$ to get a point in the kernel of $\varphi$. In a formula, $$x\mapsto p-\frac{x-p}{\varphi(x-p)} = p+\frac{x-p}{1-x\cdot p}$$

Inverse map: given $x$ in the kernel of $\varphi$, look for scalar $t$ such that $(1-t)p+tx$ has unit norm. Since $x\cdot p$, this amounts to asking $(1-t)^2+ t^2|x|^2 =1$. Simplify to $t(1+|x|^2)=2$. The result is $$x\mapsto p + \frac{2}{1+|x|^2}(x-p) = \frac{|x|^2-1}{|x|^2+1} p+\frac{2 }{ |x|^2+1}x$$