Ngô Quốc Anh

April 20, 2016

Stereographic projection, 6

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:08

I want to propose an alternative way to calculate the Jacobian of the stereographic projection $\mathcal S$. In Cartesian coordinates $\xi=(\xi_1, \xi_2,...,\xi_{n+1})$ on the sphere $\mathbb S^n$ and $x=(x_1,x_2,...,x_n)$ on the plane, the projection and its inverse are given by the formulas $\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}$

and $\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n$.

It is well-known that the Jacobian of the stereographic projection $\mathcal S: \xi \mapsto x$ is $\displaystyle \frac{\partial \xi}{\partial x} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}.$

The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points $x, y \in \mathbb R^n$ and denote $\xi = \mathcal S(x)$ and $\eta = \mathcal S(y)$.

The Euclidean distance between $\xi$ and $\eta$ is $\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.$

Note that $\displaystyle \sum_{i=1}^n |\xi_i - \eta_i|^2=\sum_{i=1}^n 4\left( \frac{x_i}{1+|x|^2}-\frac{y_i}{1+|y|^2}\right)^2=\frac{4|x|^2}{(1+|x|^2)^2} + \frac{4|y|^2}{(1+|y|^2)^2} - \frac{8\langle x,y \rangle}{(1+|x|^2)(1+|y|^2)}=\frac 4{(1+|x|^2)(1+|y|^2)} \left[ |x|^2 \frac{1+|y|^2}{1+|x|^2} + |y|^2 \frac{1+|x|^2}{1+|y|^2} - 2\langle x,y \rangle\right]$

and $\displaystyle |\xi_{n+1} - \eta_{n+1}|^2 = 4 \left( \frac 1{1+|x|^2} - \frac 1{1+|y|^2} \right)^2=\frac 4{(1+|x|^2)(1+|y|^2)}\frac {(|x|^2-|y|^2)^2}{(1+|x|^2)(1+|y|^2)}.$

Therefore, $\displaystyle |\xi -\eta|^2 =\frac 4{(1+|x|^2)(1+|y|^2)}|x-y|^2.$

In other words, we have just shown that the Euclidean distance has been scaled by $2/((1+|x|^2)(1+|y|^2))^{1/2}$. Therefore, the standard metric on $\mathbb S^n$ induced from $\mathbb R^{n+1}$ fulfills $\displaystyle g_{ij}(x) = \left( \frac 2{1+|x|^2} \right)^2 \delta_{ij}.$

Hence the standard volume element on $\mathbb S^n$ satisfies $\displaystyle dg(x) = \left( \frac 2{1+|x|^2} \right)^n dx.$

Thus the Jacobian of the stereographic projection $\mathcal S: \xi \mapsto x$ at the point $x$ is the coefficient $\big(2/(1+|x|^2)\big)^n$.