Contents

# Contents

## Idea

Stereographic projection is the name for a specific homeomorphism (for any $n \in \mathbb{N}$) form the n-sphere $S^n$ with one point $p \in S^n$ removed to the Euclidean space $\mathbb{R}^n$

$S^n \backslash \{p\} \overset{\simeq}{\longrightarrow} \mathbb{R}^n \,.$

One thinks of both the $n$-sphere as well as the Euclidean space $\mathbb{R}^n$ as topological subspaces of $\mathbb{R}^{n+1}$ in the standard way, such that they intersect in the equator of the $n$-sphere. For $p \in S^n$ one of the corresponding poles, the stereorgraphic projection is the map which sends a point $x \in S^{n}\backslash \{p\}$ along the line connecting it with $p$ to the equatorial plane.

If one applies stereographic projection to both possible poles $p_+, p_- \in S^n$ of the sphere given a fixed equatorial plane, then one obtains two different homeomorphisms

$\left\{ \mathbb{R}^n \underoverset{\simeq}{\phi_i}{\longrightarrow} S^n\backslash \{p_i\} \right\}_{i \in \{+,-\}} \,.$

The set of these two projections constitutes an atlas that exhibits the n-sphere as a topological manifold, in fact a differentiable manifold and in fact as a smooth manifold.

For $n = 2$ and with $\mathbb{R}^2 \simeq \mathbb{C}$ regarded as the complex plane, then this atlas realizes the 2-sphere as a complex manifold: the Riemann sphere.

## Definition

We consider the ambient canonical coordinates of $\mathbb{R}^{n+1}$, in terms of which the $n$-sphere is the topological subspace whose underlying subset is presented as follows:

$S^n \;\simeq\; \left\{ x = (x_1, \cdots, x_{n+1}) \in \mathbb{R}^{n+1} \,\vert\, \underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1 \right\} \;\subset\; \mathbb{R}^{n+1} \,.$

Without restriction, we may identify the given pole point in the coordinates with

$p = (1,0,\ldots,0)$

hence the corresponding equatorial hyperplane with

$\mathbb{R}^n \;\simeq\; \left\{ x = (x_1, x_2, \cdots, x_{n+1}) \in \mathbb{R}^{n+1} \,\vert\, x_1 = 0 \right\} \;\subset\; \mathbb{R}^{n+1} \,.$
###### Proposition

(standard stereographic projection)

The function which sends a point $x \in S^{n} \backslash p \subset \mathbb{R}^{n+1}$ to the intersection of the line through $p$ and $x$ with the equatorial hyperplane is a homeomorphism which is given in terms of ambient coordinates by

$\array{ \mathbb{R}^{n+1} \supset \;\;\; & S^n \backslash (1,0, \cdots, 0) &\overset{\phantom{AA} \phantom{AA}}{\longrightarrow}& \mathbb{R}^{n} & \;\;\; \subset \mathbb{R}^{n+1} \\ & (x_1, x_2, \cdots, x_{n+1}) &\overset{\phantom{AAAA}}{\mapsto}& \frac{1}{1 - x_1} \left( 0 , x_2, \cdots, x_{n+1} \right) } \,.$
###### Proof

First consider more generally the stereographic projection

$\sigma \;\colon\; \mathbb{R}^{n+1} \backslash (1,0,\cdots, 0) \longrightarrow \mathbb{R}^n = \{x \in \mathbb{R}^{n.1} \,\vert\, x_1 = 0 \}$

of the entire ambient space minus the point $p$ onto the equatorial plane, still given by mapping a point $x$ to the unique point $y$ on the equatorial hyperplane such that the points $p$, $x$ any $y$ sit on the same straight line.

This condition means that there exists $d \in \mathbb{R}$ such that

$p + d(x-p) = y \,.$

Since the only condition on $y$ is that $y_1 = 0$ this implies that

$p_1 + d(x_1-p_1) = 0 \,.$

This equation has a unique solution for $d$ given by

$d = \frac{1}{1 - x_1}$

$\sigma(x_1, x_2, \cdots, x_{n+1}) = \frac{1}{1-x_1}(0,x_2, \cdots, x_n) \,$

Since rational functions are continuous, this function $\sigma$ is continuous and since the topology on $S^n\backslash p$ is the subspace topology under the canonical embedding $S^n \backslash p \subset \mathbb{R}^{n+1} \backslash p$ it follows that the restriction

$\sigma\vert_{S^n \backslash p} \;\colon\; S^n\backslash p \longrightarrow \mathbb{R}^n$

is itself a continuous function (because its pre-images are the restrictions of the pre-images of $\sigma$ to $S^n\backslash p$).

To see that $\sigma \vert_{S^n \backslash p}$ is a bijection of the underlying sets we need to show that for every

$(0, y_2, \cdots, y_{n+1})$

there is a unique $(x_1, \cdots , x_{n+1})$ satisfying

1. $(x_1, \cdots, x_{n+1}) \in S^{n} \backslash \{p\}$, hence

1. $x_1 \lt 1$;

2. $\underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1$;

2. $\underset{i \in \{2, \cdots, n+1\}}{\forall} \left(y_i = \frac{x_i}{1-x_1} \right)$.

The last condition uniquely fixes the $x_{i \geq 2}$ in terms of the given $y_{i \geq 2}$ and the remaining $x_1$, as

$x_{i \geq 2} = y_i \cdot (1-x_1) \,.$

With this, the second condition says that

$(x_1)^2 + (1-x_1)^2 \underset{r^2}{\underbrace{\underoverset{i = 2}{n+1}{\sum}(y_i)^2}} = 1$

hence equivalently that

$(r^2 + 1) (x_1)^2 - (2 r^2) x_1 + (r^2 - 1) = 0 \,.$

By the quadratic formula the solutions of this equation are

\begin{aligned} x_1 & = \frac { 2 r^2 \pm \sqrt{ 4 r^4 - 4 (r^4 - 1) } } { 2 (r^2 + 1) } \\ & = \frac { 2 r^2 \pm 2 } { 2 r^2 + 2 } \end{aligned} \,.

The solution $\frac{ 2 r^2 + 2 }{ 2 r^2 + 2 } = 1$ violates the first condition above, while the solution $\frac{ 2 r^2 - 2 }{ 2 r^2 + 2 } \lt 1$ satisfies it.

Therefore we have a unique solution, given by

$\left( \sigma\vert_{S^n \backslash \{p\}} \right)^{-1}(0,y_2, \cdots, y_{n+1}) \;=\; \left( \frac{2 r^2 - 2}{2 r^2 +2} , \left( 1- \frac{2 r^2 - 2}{2 r^2 +2} \right) y_2 , \cdots , \left( 1- \frac{2 r^2 - 2}{2 r^2 +2} \right) y_{n+1} \right)$

In particular therefore also an inverse function to the stereographic projection exists and is a rational function, hence continuous. So we have exhibited a homeomorphism as required.

## Generalizations

Of course the construction in prop. does not really depend on the specific coordinates chosen there.

More generally, for any point $p\in S^n$, considered as a vector in $\mathbb{R}^{n+1}$, let $W\subset \mathbb{R}^{n+1}$ be the linear subspace that is the orthogonal complement to the linear span $span\{v\}$. Then there is a homeomorphism

$\sigma \;\colon\; S^n\setminus \{v\} \stackrel{\simeq}{\longrightarrow} W$

given by sending a point $p\in S^n\setminus \{v\}$ to the point in $W$ that is the intersection of the line through $v$ and $p$ with $W$.

More generally, one may take for $W$ any affine subspace of $\mathbb{R}^{n+1}$ normal to $span\{v\}$ and not containing $v$. For instance one option is to take $W$ to be the tangent space to $S^n$ at $-v$, embedded as a subspace of $\mathbb{R}^{n+1}$.

(see e.g. Cook-Crabb 93, section 1)

## Properties

### One-point compactification

The inverse map $\sigma^{-1}$ exhibits $S^n$ as the one-point compactification of $W$.

### Extra geometric structure

In most cases of interest one is doing geometry using stereographic projection, so the sphere and the subspace $W$ are equipped with extra structure. Taking the explicit case above ($v=(1,0,\ldots,0)$), as the formula is so simple, some structures are automatically preserved, for instance:

• conformal structure

• orthogonalgroup action fixing $v,-v$ pointwise (note that these two points are sent to “$\infty$” and $0$, respectively, under $\sigma$)

## Over other fields

Stereographic projection is a general geometric technique which can be applied to other commutative rings $k$ besides $k = \mathbb{R}$. We give a brief taste of this for the case of fields. For simplicity, we focus on conic sections, i.e., solutions sets in the projective plane $\mathbb{P}^2(k)$ to homogeneous polynomials of degree $2$.

Via stereographic projection, all pointed nonsingular conic sections $C \subset \mathbb{P}^2(k)$ are isomorphic and can be identified explicitly with a projective line $\mathbb{P}^1(k)$ by means of a stereographic projection.

Geometrically, if $p$ is the chosen basepoint of $C$ and $L \subset \mathbb{P}^2(k)$ is a line not incident to $p$, then for any other point $q$ of $C$ the unique line $L(p, q)$ incident to $p$ and $q$ intersects $L$ in exactly one point, denoted $\phi(q)$. (Here $\phi(p)$ is taken to be the intersection of the tangent to $p$ at $C$ with $L$; this can be considered the basepoint of $L$.) In the opposite direction, to each point $x$ of $L$, the line $L(p, x)$ intersects $C$ in $p$ and (since a quadratic with one root will also have another root) another point $q$ (which might be the same as $p$; this happens precisely when $L(p, x)$ is the tangent to $C$ at $p$); this gives the inverse $\phi^{-1}(x) = q$. In this way we obtain an isomorphism $\phi: C \to L$ of subvarieties.

###### Example

For the case $k = \mathbb{Q}$ and the rational conic $C = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: x^2 + y^2 = 1\}$, stereographic projection from the point $(-1, 0)$ on $C$ to the rational line $x = 0$ defines an isomorphism of algebraic varieties $C \to \mathbb{P}^1(\mathbb{Q})$. As the calculations above show, its inverse takes $(0, t)$ (for $t \in \mathbb{Q}$) to $(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2})$.

A Pythagorean triple is a triple of integers $(a, b, c)$ such that $a^2 + b^2 = c^2$, so that $(\frac{a}{c}, \frac{b}{c})$ lies on $C$. The isomorphism indicates that for each such point on $C$ (written as a pair of fractions in reduced form), there is a rational number $t = \frac{p}{q}$ (again in reduced form) so that

$\frac{a}{c} = \frac{1 - t^2}{1 + t^2} = \frac{q^2 - p^2}{q^2 + p^2}, \qquad \frac{b}{c} = \frac{2 t}{1 + t^2} = \frac{2 p q}{q^2 + p^2}.$

If we assume $a, b, c$ are coprime, and further (re)arrange the triple so that $a$ is odd and $b$ is even, one may quickly conclude from the first of this pair of equations that $p, q$ must have opposite parity. A little further argumentation then shows $q^2 - p^2, 2 p q, q^2 + p^2$ must be mutually coprime, so that in fact $a = q^2 - p^2, b = 2 p q$, and $c = q^2 + p^2$. Thus we arrive at the classical parametrization of Pythagorean triples, essentially on the basis of the geometry of stereographic projection!