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

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

One thinks of both the nn-sphere as well as the Euclidean space ℝ n\mathbb{R}^n as topological subspaces of ℝ n+1\mathbb{R}^{n+1} in the standard way, such that they intersect in the equator of the nn-sphere.

For p∈S np \in S^n one of the corresponding poles, the stereographic projection is the map which sends a point x∈S n\{p}x \in S^{n}\backslash \{p\} along the line connecting it with pp to the equatorial plane.

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

{ℝ n⟶≃ϕ iS n\{p i}} i∈{+,−}. \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=2n = 2 and with ℝ 2≃ℂ\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 ℝ n+1\mathbb{R}^{n+1}, in terms of which the nn-sphere is the topological subspace whose underlying subset is presented as follows:

S n≃{x=(x 1,⋯,x n+1)∈ℝ n+1|∑i=1n+1(x i) 2=1}⊂ℝ n+1. 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,…,0) p = (1,0,\ldots,0)

hence the corresponding equatorial hyperplane with

ℝ n≃{x=(x 1,x 2,⋯,x n+1)∈ℝ n+1|x 1=0}⊂ℝ n+1. \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∈S n\p⊂ℝ n+1x \in S^{n} \backslash p \subset \mathbb{R}^{n+1} to the intersection of the line through pp and xx with the equatorial hyperplane is a homeomorphism which is given in terms of ambient coordinates by

ℝ n+1⊃ S n\(1,0,⋯,0) ⟶AAAA ℝ n ⊂ℝ n+1 (x 1,x 2,⋯,x n+1) ↦AAAA 11−x 1(0,x 2,⋯,x n+1). \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

σ:ℝ n+1\(1,0,⋯,0)⟶ℝ n={x∈ℝ n.1|x 1=0} \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 pp onto the equatorial plane, still given by mapping a point xx to the unique point yy on the equatorial hyperplane such that the points pp, xx any yy sit on the same straight line.

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

p+d(x−p)=y. p + d(x-p) = y \,.

Since the only condition on yy is that y 1=0y_1 = 0 this implies that

p 1+d(x 1−p 1)=0. p_1 + d(x_1-p_1) = 0 \,.

This equation has a unique solution for dd given by

d=11−x 1 d = \frac{1}{1 - x_1}

and hence it follow that

σ(x 1,x 2,⋯,x n+1)=11−x 1(0,x 2,⋯,x n) \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 σ\sigma is continuous and since the topology on S n\pS^n\backslash p is the subspace topology under the canonical embedding S n\p⊂ℝ n+1\pS^n \backslash p \subset \mathbb{R}^{n+1} \backslash p it follows that the restriction

σ| S n\p:S n\p⟶ℝ n \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 σ\sigma to S n\pS^n\backslash p).

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

(0,y 2,⋯,y n+1) (0, y_2, \cdots, y_{n+1})

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

(x 1,⋯,x n+1)∈S n\{p}(x_1, \cdots, x_{n+1}) \in S^{n} \backslash \{p\}, hence
1. x 1<1x_1 \lt 1;
2. ∑i=1n+1(x i) 2=1\underoverset{i = 1}{n+1}{\sum} (x_i)^2 = 1;
∀i∈{2,⋯,n+1}(y i=x i1−x 1)\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≥2x_{i \geq 2} in terms of the given y i≥2y_{i \geq 2} and the remaining x 1x_1, as

x i≥2=y i⋅(1−x 1). x_{i \geq 2} = y_i \cdot (1-x_1) \,.

With this, the second condition says that

(x 1) 2+(1−x 1) 2∑i=2n+1(y i) 2⏟r 2=1 (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−(2r 2)x 1+(r 2−1)=0. (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

x 1 =2r 2±4r 4−4(r 4−1)2(r 2+1) =2r 2±22r 2+2. \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 2r 2+22r 2+2=1\frac{ 2 r^2 + 2 }{ 2 r^2 + 2 } = 1 violates the first condition above, while the solution 2r 2−22r 2+2<1\frac{ 2 r^2 - 2 }{ 2 r^2 + 2 } \lt 1 satisfies it.

Therefore we have a unique solution, given by

(σ| S n\{p}) −1(0,y 2,⋯,y n+1)=(2r 2−22r 2+2,(1−2r 2−22r 2+2)y 2,⋯,(1−2r 2−22r 2+2)y n+1) \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∈S np\in S^n, considered as a vector in ℝ n+1\mathbb{R}^{n+1}, let W⊂ℝ n+1W\subset \mathbb{R}^{n+1} be the linear subspace that is the orthogonal complement to the linear span span{v}span\{v\}. Then there is a homeomorphism

σ:S n∖{v}⟶≃W \sigma \;\colon\; S^n\setminus \{v\} \stackrel{\simeq}{\longrightarrow} W

given by sending a point p∈S n∖{v}p\in S^n\setminus \{v\} to the point in WW that is the intersection of the line through vv and pp with WW.

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

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

Properties

One-point compactification

The inverse map σ −1\sigma^{-1} exhibits S nS^n as the one-point compactification of WW.

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

conformal structure
orthogonal group action fixing v,−vv,-v pointwise (note that these two points are sent to “∞\infty” and 00, respectively, under σ\sigma)

Over other fields

Stereographic projection is a general geometric technique which can be applied to other commutative rings kk besides k=ℝk = \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 ℙ 2(k)\mathbb{P}^2(k) to homogeneous polynomials of degree 22.

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

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

Example

For the case k=ℚk = \mathbb{Q} and the rational conic C={(x,y)∈ℚ×ℚ:x 2+y 2=1}C = \{(x, y) \in \mathbb{Q} \times \mathbb{Q}: x^2 + y^2 = 1\}, stereographic projection from the point (−1,0)(-1, 0) on CC to the rational line x=0x = 0 defines an isomorphism of algebraic varieties C→ℙ 1(ℚ)C \to \mathbb{P}^1(\mathbb{Q}). As the calculations above show, its inverse takes (0,t)(0, t) (for t∈ℚt \in \mathbb{Q}) to (1−t 21+t 2,2t1+t 2)(\frac{1 - t^2}{1 + t^2}, \frac{2 t}{1 + t^2}).

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

ac=1−t 21+t 2=q 2−p 2q 2+p 2,bc=2t1+t 2=2pqq 2+p 2.\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,ca, b, c are coprime, and further (re)arrange the triple so that aa is odd and bb is even, one may quickly conclude from the first of this pair of equations that p,qp, q must have opposite parity. A little further argumentation then shows q 2−p 2,2pq,q 2+p 2q^2 - p^2, 2 p q, q^2 + p^2 must be mutually coprime, so that in fact a=q 2−p 2,b=2pqa = q^2 - p^2, b = 2 p q, and c=q 2+p 2c = q^2 + p^2. Thus we arrive at the classical parametrization of Pythagorean triples, essentially on the basis of the geometry of stereographic projection!

References

Textbook accounts:

Tom Apostol, Fig. 1.3 in: Mathematical Analysis 1973 (pdf)