# Ngô Quốc Anh

## July 11, 2010

### Stereographic projection

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 0:47

In geometry, the stereographic projection, usually denoted by $\pi$, is a particular mapping (function) that projects a sphere onto a plane. The projection is defined on the entire sphere, except at one point – the projection point. Where it is defined, the mapping is smooth and bijective. It is conformal, meaning that it preserves angles. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures.

Intuitively, then, the stereographic projection is a way of picturing the sphere as the plane, with some inevitable compromises. Because the sphere and the plane appear in many areas of mathematics and its applications, so does the stereographic projection; it finds use in diverse fields including complex analysis, cartography, geology, and photography. In practice, the projection is carried out by computer or by hand using a special kind of graph paper called a stereonet or Wulff net.

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 $\pi : \xi \mapsto x$ and its inverse $\pi^{-1}: x \mapsto \xi$ 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

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

Let us show you an example making use of the projection. We assume $v(x)$ verifies the following PDE

$\displaystyle -\Delta v = \frac{n(n-2)}{4}v^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n$.

Then the transformed function $u(\xi)$, to be exact $u(\pi^{-1}(x))$, given by

$\displaystyle v(x)=u(\pi^{-1}(x))\left( \frac{2}{1+|x|^2}\right)^\frac{n-2}{2}$

satisfies the following PDE

$\displaystyle -\Delta_g u + \frac{n(n-2)}{4}u = \frac{n(n-2)}{4}u^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n$.

where $\Delta_g$ denotes the Laplace-Beltrami operator with respect to the standard metric $g$ on $\mathbb S^n$.

Similarly, if function $u(\xi)$ verifying the PDE

$\displaystyle -\Delta_g u + \frac{n(n-2)}{4}u = \frac{n(n-2)}{4}u^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n$

then a new function $v(x)$ given by

$\displaystyle v(x)=u(\pi(x))\left( \frac{2}{1+|x|^2}\right)^\frac{n-2}{2}$

will satisfy the following PDE

$\displaystyle -\Delta v = \frac{n(n-2)}{4}v^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n$.

More general, PDE

$\displaystyle -\Delta_g u(\xi) + \frac{n(n-2)}{4}u(\xi) = K(\xi)u(\xi)^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n$

becomes

$\displaystyle -\Delta v(x) = K(\pi^{-1}(x))v(x)^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n$.

In conclusion, using the stereographic projection we can transfer some geometric problems on sphere $\mathbb S^n$ to ones in the whole space $\mathbb R^n$.