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.

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