# Ngô Quốc Anh

## October 3, 2017

### Stereographic projection not at the North pole and an example of conformal transformation on S^n

Filed under: Riemannian geometry — Tags: , — Ngô Quốc Anh @ 12:09

Denote $\pi_P : \mathbb S^n \to \mathbb R^n$ the stereographic projection performed with $P$ as the north pole to the equatorial plane of $\mathbb S^n$. Clearly when $P$ is the north pole $N$, i.e. $N = (0,...,0,1)$, then $\pi_N$ is the usual stereographic projection.

Clearly, for arbitrary $x \in \mathbb S^n$, the image of $x$ is

$\displaystyle \pi_P : x \mapsto y = P+\frac{x-P}{1-x \cdot P}.$

For the inverse map, it is not hard to see that

$\displaystyle \pi_P^{-1} : y \mapsto x =\frac{|y|^2-1}{|y|^2+1}P+\frac 2{|y|^2+1}y.$

Derivation of $\pi_P$ and $\pi_P^{-1}$ are easy, for interested reader, I refer to an answer in . Let us now define the usual conformal transformation $\varphi_{P,t} : \mathbb S^n \to \mathbb S^n$ given by

$\displaystyle \varphi_{P,t} : x \mapsto \pi_P^{-1} \big( t \pi_P ( x) \big)$

## 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

$\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.$

## August 9, 2011

### Stereographic projection, 5

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 14:39

I put here several common formulas that are needed when we work on stereographic projection. First we try to calculate

$\displaystyle\Delta \left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right).$

We do step by step.

Step 1. A direct computation leads us to

$\displaystyle\begin{gathered} {\partial _i}\left( {{{\left( {\frac{2}{{1 + |x{|^2}}}} \right)}^{n - 2}}} \right) = (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}{\partial _i}\left( {\frac{2}{{1 + |x{|^2}}}} \right) \hfill \\ \qquad= (n - 2){\left( {\frac{2}{{1 + |x{|^2}}}} \right)^{n - 3}}\frac{{ - 4{x_i}}}{{{{(1 + |x{|^2})}^2}}}. \hfill \\ \end{gathered}$

Therefore,

## July 15, 2011

### Stereographic projection, 4

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 19:31

It turns out that, via the stereographic projection, equation

$\displaystyle - {\Delta _{{\mathbb S^n}}}u = \lambda u + {u^{\frac{{n + 2}}{{n - 2}}}}$

on $\mathbb S^n$ with $u>0$ becomes

$\displaystyle - {\Delta _{{\mathbb R^n}}}u = V(x) u + {u^{\frac{{n + 2}}{{n - 2}}}}, \quad x \in \mathbb R^n$

with the following property

$u(x) \to 0, \quad |x|\to +\infty$

where

$\displaystyle V(x) = \frac{{n(n - 2) + 4\lambda }}{{{{(1 + |x|^2)}^2}}}.$

A very simple consequence is that for the prescribing scalar curvature equation, the term $V$ disappears as we already notice that

$\displaystyle\lambda = -\frac{n(n-2)}{4}.$

## February 14, 2011

### Stereographic projection, 3

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 18:07

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 not hard to see that

$\displaystyle\frac{{\partial ({\xi _1},...,\xi _n,\xi_{n+1} )}}{{\partial ({x_1},...,x_n,\xi_{n+1})}} = \frac{2^n}{{{{\left( {1 + {{\left| x \right|}^2}} \right)}^{2n}}}}\det \left( {\begin{array}{*{20}{c}} {1 + {{\left| x \right|}^2} - 2x_1^2}&{ - 2{x_1}{x_2}}& \cdots &{ - 2{x_1}{x_n}} & 0 \\ { - 2{x_2}{x_1}}&{1 + {{\left| x \right|}^2} - 2x_2^2}& \cdots &{ - 2{x_2}{x_n}} & 0 \\ \vdots & \vdots & \ddots & \vdots \\ { - 2{x_n}{x_1}}&{ - 2{x_n}{x_2}}& \cdots &{1 + {{\left| x \right|}^2} - 2x_n^2} & 0\\ * & * & \cdots & * & 1 \end{array}} \right).$

## January 31, 2011

### Stereographic projection, 2

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 19:15

This is a sequel to this topic where we have recalled several properties of the stereographic projection $\pi : \mathbb S^n \to \mathbb R^n$. Recall that by the following transformation

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

we know that

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

and

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

are equivalent in the weak sense.

The way to see it comes from the following identities

$\displaystyle \int_{{\mathbb{S}^n}} {|\nabla u(\xi ){|^2} + \frac{{n(n - 2)}}{4}u{{(\xi )}^2} = \int_{{\mathbb{R}^n}} {|\nabla v(x){|^2}} }$

and

$\displaystyle \int_{{\mathbb{S}^n}} {|u(\xi ){|^{\frac{{2n}}{{n - 2}}}} = \int_{{\mathbb{R}^n}} {|v(x){|^{\frac{{2n}}{{n - 2}}}}} }$

where $u \in H^1(\mathbb S^n)$.

## 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 $\mathbb R^n$.