Mathematics.

geometric function theory

The Riemann Sphere

Complex Analysis30 minDifficulty6 out of 10

You should know: mobius transformations

Overview

The Riemann sphere is the extended complex plane ℂ ∪ {∞} realized as an honest compact surface — a sphere — via stereographic projection. Sit the unit sphere in ℝ³ with its south pole touching the origin of the complex plane (or, in the standard convention, centered at the origin with the plane through the equator), and project each point of the sphere from the north pole through to where the line hits the plane. Every point of the sphere except the north pole itself maps to a unique finite complex number, and as points on the sphere approach the north pole, their images run off to infinity in every direction on the plane — so declaring the north pole to correspond to a single point ∞ compactifies the plane into a sphere with no boundary and no distinguished points. This turns ℂ ∪ {∞} into a genuine compact Riemann surface on which Möbius transformations act as the full group of holomorphic automorphisms, and on which every rational function becomes a well-defined map of the sphere to itself, poles included.

Intuition

Imagine a lamp at the north pole of a glass sphere resting on a sheet of paper (the complex plane, touching at the south pole). Every point on the sphere casts a shadow on the paper, except the north pole itself, whose 'shadow' would have to be a light ray parallel to the paper — it lands nowhere finite. Points near the north pole cast shadows farther and farther out on the paper in every direction, which is exactly the behavior we want to capture by a single point ∞: no matter which direction you recede to infinity in the plane, you're approaching the same point on the sphere. This is why the Riemann sphere is the natural home for rational functions and Möbius transformations — a pole of a rational function, where |f(z)| → ∞, is simply the function's value at ∞ on the sphere, not a breakdown of the function. Möbius transformations become exactly the rotations of this sphere (up to the identification), which explains both why they form a group and why they are always bijections of the whole sphere to itself, never leaving a point undefined.

Formal Definition

Definition

Let the sphere S² = {(x,y,z) ∈ ℝ³ : x²+y²+z²=1} sit with its equator on the unit circle of the plane ℂ = {(x,y,0)}. Stereographic projection from the north pole N=(0,0,1) sends a point (x,y,z) ≠ N to the complex number where the line from N through (x,y,z) meets the plane z=0:

ζ=x+iy1z,(x,y,z)S2{N}\zeta = \frac{x+iy}{1-z}, \qquad (x,y,z) \in S^2 \setminus \{N\}
Stereographic projection (sphere to plane)
x=2Re(ζ)ζ2+1,y=2Im(ζ)ζ2+1,z=ζ21ζ2+1x = \frac{2\,\mathrm{Re}(\zeta)}{|\zeta|^2+1},\quad y = \frac{2\,\mathrm{Im}(\zeta)}{|\zeta|^2+1},\quad z = \frac{|\zeta|^2-1}{|\zeta|^2+1}
Inverse stereographic projection (plane to sphere)
N=(0,0,1)    ζ=,S=(0,0,1)    ζ=0N = (0,0,1) \;\longleftrightarrow\; \zeta = \infty, \qquad S = (0,0,-1) \;\longleftrightarrow\; \zeta = 0
North pole is the point at infinity; south pole is the origin
C^=C{}S2(as topological/Riemann surfaces)\hat{\mathbb{C}} = \mathbb{C} \cup \{\infty\} \cong S^2 \quad \text{(as topological/Riemann surfaces)}
The Riemann sphere

Worked Examples

  1. Compute |ζ|² for ζ=1+i.

    ζ2=12+12=2|\zeta|^2 = 1^2+1^2 = 2
  2. Apply the inverse stereographic projection formulas with |ζ|²=2, Re(ζ)=1, Im(ζ)=1.

    x=2(1)2+1=23,y=2(1)2+1=23,z=212+1=13x = \frac{2(1)}{2+1} = \frac{2}{3}, \quad y = \frac{2(1)}{2+1} = \frac{2}{3}, \quad z = \frac{2-1}{2+1} = \frac{1}{3}
  3. Verify the point lies on the unit sphere.

    (23)2+(23)2+(13)2=49+49+19=1\left(\tfrac{2}{3}\right)^2+\left(\tfrac{2}{3}\right)^2+\left(\tfrac{1}{3}\right)^2 = \tfrac{4}{9}+\tfrac{4}{9}+\tfrac{1}{9} = 1

Answer: ζ = 1+i corresponds to the point (2/3, 2/3, 1/3) on the Riemann sphere.

Practice Problems

Difficulty 5/10

Find the point on the Riemann sphere corresponding to ζ = 0 under inverse stereographic projection.

Difficulty 6/10

Find the point on the Riemann sphere corresponding to ζ = 1 (on the unit circle in the plane).

Difficulty 7/10

Explain why a rational function like f(z) = 1/z, which is undefined at z=0 as a map ℂ→ℂ, becomes a well-defined bijection of the Riemann sphere to itself.

Quiz

The Riemann sphere is formed by:
Under the standard stereographic projection convention used here, the north pole corresponds to:
On the Riemann sphere, a pole of a rational function (where the function's modulus blows up) is best understood as:

Summary

  • The Riemann sphere realizes ℂ ∪ {∞} as a compact surface via stereographic projection from the north pole of a sphere onto the plane.
  • Formulas: ζ = (x+iy)/(1-z) projects sphere to plane; the inverse uses |ζ|² to recover (x,y,z), e.g. ζ=1+i ↦ (2/3, 2/3, 1/3).
  • The south pole corresponds to ζ=0 and the north pole corresponds to ζ=∞; approaching the north pole from any direction sends |ζ| → ∞.
  • On the sphere, rational functions and Möbius transformations become genuine bijections of the whole surface, with poles simply mapping to the point ∞.

References