Stereographic Projection & The Riemann Sphere
Analytical Geometry Approach
This document contains detailed derivations using Analytical Coordinate Geometry (symmetric form of lines) rather than vector calculus. This method is often more intuitive for understanding the intersection of lines and spheres in 3D space.
The Core Concept: A line is drawn connecting the North Pole of a sphere to a point on the Complex Plane. The point where this line intersects the sphere is the “projection.”
Case 1: The Unit Sphere (Equatorial Plane)
Geometry Setup:
- Sphere: Radius \(R=1\), Center \((0,0,0)\). Equation: \(X^2 + Y^2 + Z^2 = 1\).
- Complex Plane: The plane \(Z=0\) (The equatorial plane).
- North Pole (N): The point \((0,0,1)\).
Summary of Formulas
1. Plane \(\to\) Sphere: $$ X = \frac{2x}{x^2 + y^2 + 1}, \quad Y = \frac{2y}{x^2 + y^2 + 1}, \quad Z = \frac{x^2 + y^2 – 1}{x^2 + y^2 + 1} $$ 2. Sphere \(\to\) Plane: $$ x = \frac{X}{1-Z}, \quad y = \frac{Y}{1-Z} $$We need the equation of a line passing through two points in 3D space:
- Point 1 (North Pole): \( (x_1, y_1, z_1) = (0, 0, 1) \)
- Point 2 (Complex Plane): \( (x_2, y_2, z_2) = (x, y, 0) \)
Using the symmetric form of a line equation \(\frac{X-x_1}{x_2-x_1} = \frac{Y-y_1}{y_2-y_1} = \frac{Z-z_1}{z_2-z_1}\):
$$ \frac{X – 0}{x – 0} = \frac{Y – 0}{y – 0} = \frac{Z – 1}{0 – 1} $$ $$ \frac{X}{x} = \frac{Y}{y} = \frac{Z – 1}{-1} = k \quad (\text{Let this constant be } k) $$From this ratio, we can express coordinates \(X, Y, Z\) in terms of \(k\):
$$ X = kx, \quad Y = ky, \quad Z = 1 – k $$ Step 2: Use the Sphere’s EquationThe point \((X, Y, Z)\) lies on the sphere \(X^2 + Y^2 + Z^2 = 1\). Substitute the expressions from Step 1:
$$ (kx)^2 + (ky)^2 + (1 – k)^2 = 1 $$ $$ k^2x^2 + k^2y^2 + (1 – 2k + k^2) = 1 $$Group the \(k^2\) terms:
$$ k^2(x^2 + y^2 + 1) – 2k + 1 = 1 $$ $$ k^2(x^2 + y^2 + 1) – 2k = 0 $$ Step 3: Solve for kFactor out \(k\):
$$ k [ k(x^2 + y^2 + 1) – 2 ] = 0 $$This gives two solutions. \(k=0\) corresponds to the North Pole (which we ignore). The other solution is:
$$ k(x^2 + y^2 + 1) = 2 \implies k = \frac{2}{x^2 + y^2 + 1} $$ Step 4: Determine Coordinates X, Y, ZSubstitute \(k\) back into the equations from Step 1:
For X: \( X = kx = \frac{2x}{x^2 + y^2 + 1} \)
For Y: \( Y = ky = \frac{2y}{x^2 + y^2 + 1} \)
For Z: $$ Z = 1 – k = 1 – \frac{2}{x^2 + y^2 + 1} $$ Take the LCM: $$ Z = \frac{(x^2 + y^2 + 1) – 2}{x^2 + y^2 + 1} = \frac{x^2 + y^2 – 1}{x^2 + y^2 + 1} $$
We form a line passing through the North Pole \((0,0,1)\) and the given point on the sphere \((X,Y,Z)\).
Using the symmetric form \(\frac{x’ – x_1}{x_2 – x_1} \dots\):
$$ \frac{x’ – 0}{X – 0} = \frac{y’ – 0}{Y – 0} = \frac{z’ – 1}{Z – 1} = \lambda $$Here, \(x’, y’, z’\) represent any variable point on this line.
Step 2: Intersect with the Complex PlaneWe want to find the specific point where this line hits the complex plane. The defining feature of the complex plane in this setup is \(z’ = 0\).
Using the Z-ratio from Step 1:
$$ \frac{0 – 1}{Z – 1} = \lambda $$ $$ \lambda = \frac{-1}{Z – 1} = \frac{1}{1 – Z} $$ Step 3: Solve for x and yNow use this value of \(\lambda\) to find the \(x’\) (which is our \(x\)) and \(y’\) (which is our \(y\)).
From the X-ratio: \(\frac{x}{X} = \lambda \implies x = \lambda X = \frac{X}{1 – Z}\)
From the Y-ratio: \(\frac{y}{Y} = \lambda \implies y = \lambda Y = \frac{Y}{1 – Z}\)
Case 2: The Tangent Sphere (Riemann Sphere)
Geometry Setup:
- Sphere: Radius \(R=1/2\), Center \((0,0,1/2)\).
Equation: \(X^2 + Y^2 + (Z – 1/2)^2 = 1/4\) which simplifies to \(X^2 + Y^2 + Z^2 = Z\). - Complex Plane: The plane \(Z=0\) (Tangent to the South Pole).
- North Pole (N): The point \((0,0,1)\).
Summary of Formulas
1. Plane \(\to\) Sphere: $$ X = \frac{x}{x^2 + y^2 + 1}, \quad Y = \frac{y}{x^2 + y^2 + 1}, \quad Z = \frac{x^2 + y^2}{x^2 + y^2 + 1} $$ 2. Sphere \(\to\) Plane: $$ x = \frac{X}{1-Z}, \quad y = \frac{Y}{1-Z} $$The geometry of the line is identical to Case 1. It passes through \(N(0,0,1)\) and \((x,y,0)\).
$$ \frac{X}{x} = \frac{Y}{y} = \frac{Z – 1}{-1} = k $$Expressing coordinates in terms of \(k\):
$$ X = kx, \quad Y = ky, \quad Z = 1 – k $$ Step 2: Use the Tangent Sphere’s EquationSubstitute these coordinates into the specific sphere equation \(X^2 + Y^2 + Z^2 = Z\):
$$ (kx)^2 + (ky)^2 + (1 – k)^2 = (1 – k) $$ $$ k^2(x^2 + y^2) + (1 – 2k + k^2) = 1 – k $$Subtract \(1\) from both sides and add \(k\) to both sides:
$$ k^2(x^2 + y^2) + k^2 – k = 0 $$ $$ k^2(x^2 + y^2 + 1) – k = 0 $$ Step 3: Solve for kFactor out \(k\): \( k [ k(x^2 + y^2 + 1) – 1 ] = 0 \).
Ignoring the North Pole (\(k=0\)), we solve for the intersection:
$$ k = \frac{1}{x^2 + y^2 + 1} $$ Step 4: Determine Coordinates X, Y, ZFor X: \( X = kx = \frac{x}{x^2 + y^2 + 1} \)
For Y: \( Y = ky = \frac{y}{x^2 + y^2 + 1} \)
For Z: $$ Z = 1 – k = 1 – \frac{1}{x^2 + y^2 + 1} $$ $$ Z = \frac{(x^2 + y^2 + 1) – 1}{x^2 + y^2 + 1} = \frac{x^2 + y^2}{x^2 + y^2 + 1} $$
Line through \(N(0,0,1)\) and sphere point \((X,Y,Z)\). Using symmetric form:
$$ \frac{x’ – 0}{X – 0} = \frac{y’ – 0}{Y – 0} = \frac{z’ – 1}{Z – 1} = \lambda $$ Step 2: Intersect with the Complex PlaneWe are looking for the point on the plane where \(z’ = 0\).
$$ \frac{0 – 1}{Z – 1} = \lambda \implies \lambda = \frac{1}{1 – Z} $$ Step 3: Solve for x and yThe algebra is identical to Case 1 because the similarity of the triangles (projection geometry) is the same, even though the sphere size changed.
From X-ratio: \( \frac{x}{X} = \lambda \implies x = \frac{X}{1-Z} \)
From Y-ratio: \( \frac{y}{Y} = \lambda \implies y = \frac{Y}{1-Z} \)
Practice Numericals
Problem 1: Find the coordinates \((X, Y, Z)\) on the unit sphere corresponding to the complex number \( z = 1 + i \).
Solution:
Given \(z = 1 + i\), we have \(x = 1\) and \(y = 1\).
First, calculate \(|z|^2 = x^2 + y^2 = 1^2 + 1^2 = 2\).
Using the Case 1 formulas:
$$ X = \frac{2x}{|z|^2 + 1} = \frac{2(1)}{2 + 1} = \frac{2}{3} $$ $$ Y = \frac{2y}{|z|^2 + 1} = \frac{2(1)}{2 + 1} = \frac{2}{3} $$ $$ Z = \frac{|z|^2 – 1}{|z|^2 + 1} = \frac{2 – 1}{2 + 1} = \frac{1}{3} $$Answer: The point is \( \left( \frac{2}{3}, \frac{2}{3}, \frac{1}{3} \right) \).
Check: \((2/3)^2 + (2/3)^2 + (1/3)^2 = 4/9 + 4/9 + 1/9 = 9/9 = 1\). (Correct)
Problem 2: A point on the unit sphere has coordinates \( P \left( \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) \). Find the corresponding complex number \( z \).
Solution:
We are given \(X = 1/3\), \(Y = 2/3\), \(Z = 2/3\).
Use the reverse mapping formula \( z = \frac{X + iY}{1 – Z} \):
$$ z = \frac{\frac{1}{3} + i\left(\frac{2}{3}\right)}{1 – \frac{2}{3}} $$Simplify the denominator: \( 1 – 2/3 = 1/3 \).
$$ z = \frac{\frac{1}{3}(1 + 2i)}{\frac{1}{3}} $$Cancel the \(1/3\) terms:
Answer: \( z = 1 + 2i \).
Problem 3: Map the purely imaginary number \( z = 2i \) onto the tangent sphere (Riemann Sphere).
Solution:
Given \(z = 0 + 2i\), we have \(x = 0\) and \(y = 2\).
Calculate \(|z|^2 = 0^2 + 2^2 = 4\).
Using the Case 2 formulas (note the absence of ‘2’ in the numerators):
$$ X = \frac{x}{|z|^2 + 1} = \frac{0}{4 + 1} = 0 $$ $$ Y = \frac{y}{|z|^2 + 1} = \frac{2}{4 + 1} = \frac{2}{5} $$ $$ Z = \frac{|z|^2}{|z|^2 + 1} = \frac{4}{4 + 1} = \frac{4}{5} $$Answer: The point is \( \left( 0, \frac{2}{5}, \frac{4}{5} \right) \).
Problem 4: Find the complex number corresponding to the point \( P \left( \frac{1}{3}, -\frac{1}{3}, \frac{2}{3} \right) \) on the tangent sphere.
Solution:
We are given \(X = 1/3\), \(Y = -1/3\), \(Z = 2/3\).
Use the reverse formula (which is the same for both cases):
$$ z = \frac{X + iY}{1 – Z} $$ $$ z = \frac{\frac{1}{3} – i\frac{1}{3}}{1 – \frac{2}{3}} $$Simplify denominator: \( 1 – 2/3 = 1/3 \).
$$ z = \frac{\frac{1}{3}(1 – i)}{\frac{1}{3}} $$Answer: \( z = 1 – i \).
Stereographic Projection & The Riemann Sphere
1. Formal Definitions & Framework
Based on the formal mathematical definitions, we define three distinct sets to establish the framework for the projection.
1.1 Definitions of the Sets
A sphere with radius \(1/2\) and center at \((0, 0, 1/2)\). $$ S_1 = \left\{ (X, Y, Z) \in \mathbb{R}^3 : X^2 + Y^2 + \left(Z – \frac{1}{2}\right)^2 = \frac{1}{4} \right\} $$
This set is the sphere \(S_1\) excluding the North Pole \(N(0, 0, 1)\). $$ S_2 = S_1 \setminus \{ (0, 0, 1) \} $$
This set represents the complex plane \(\mathbb{C}\) located at \(Z=0\). $$ S_3 = \mathbb{C} = \{ (x, y, 0) : x, y \in \mathbb{R} \} $$
1.2 The Mapping Principle
The Stereographic Projection defines the relationship between these sets:
- Forward Mapping: For every complex number \(z \in S_3\) (plane), there exists a unique point \(P \in S_2\) (sphere).
- Converse Mapping: Conversely, every point on the punctured sphere \(S_2\) corresponds to a unique point on the complex plane \(S_3\).
1.3 One-to-One Correspondence (Bijection)
There is a Bijective (one-to-one and onto) correspondence between \(S_2\) and \(S_3\).
Note on Infinity: While \(S_3\) (the plane) is infinite, the North Pole \((0,0,1)\) corresponds to the “Point at Infinity”. Thus, the full sphere \(S_1\) corresponds to the Extended Complex Plane: $$ S_1 \cong \mathbb{C} \cup \{\infty\} $$
Chordal Distance & The Metric Space
1. Definition
Chordal Distance \(\chi(z_1, z_2)\) is defined as the straight-line Euclidean distance between the projected points \(P\) and \(Q\) on the Riemann Sphere corresponding to complex numbers \(z_1\) and \(z_2\).
It represents the length of the “chord” cutting through the sphere’s interior connecting \(P\) and \(Q\).
If \(P = (X_1, Y_1, Z_1)\) and \(Q = (X_2, Y_2, Z_2)\) are points on the sphere: $$ \chi(z_1, z_2) = \sqrt{(X_1 – X_2)^2 + (Y_1 – Y_2)^2 + (Z_1 – Z_2)^2} $$
Key Properties [cite: 77, 78, 90]
1. [cite_start]Boundedness[cite: 77]:
For any two points \(z_1, z_2 \in \mathbb{C}^*\), the chordal distance satisfies:
$$ \chi(z_1, z_2) \leq \text{Diameter of Sphere} $$
Note: The document states \(\chi(z_1, z_2) \leq 1\). This is true for the specific sphere used in the text (Radius \(1/2\), Diameter \(1\)). If using the standard Unit Sphere (Radius \(1\)), the maximum distance would be \(2\).
2. [cite_start]Antipodal Points[cite: 78]:
If \(\chi(z_1, z_2) = \text{Diameter}\) (i.e., \(1\) for the tangent sphere), then the points are Antipodal Points. This means they are diametrically opposite on the sphere (e.g., \(0\) and \(\infty\)).
3. [cite_start]Metric Space[cite: 90]:
The chordal distance forms a valid Metric Space. The function \(d(z_1, z_2) = \chi(z_1, z_2)\) satisfies all standard metric properties:
- Non-negativity: \(d(x,y) \ge 0\)
- Identity: \(d(x,y)=0 \iff x=y\)
- Symmetry: \(d(x,y) = d(y,x)\)
- Triangle Inequality: \(d(x,z) \le d(x,y) + d(y,z)\)
Example Problem (GATE-21) [cite: 72, 73, 74, 75]
Question: Let \(z = 1+i\) be projected on the sphere \(x^2 + y^2 + (z – 1/2)^2 = 1/4\). What is the chordal distance \(\chi(1+i, \infty)\)?
Solution:
Step 1: Identify Coordinates of \(z = 1+i\) on the sphere.
Using the projection formulas for this sphere:
$$ X = \frac{x}{|z|^2+1}, \quad Y = \frac{y}{|z|^2+1}, \quad Z = \frac{|z|^2}{|z|^2+1} $$
Since \(z=1+i\), we have \(x=1, y=1\), and \(|z|^2 = 2\).
$$ X = \frac{1}{3}, \quad Y = \frac{1}{3}, \quad Z = \frac{2}{3} $$
[cite_start]So the point is \(P(\frac{1}{3}, \frac{1}{3}, \frac{2}{3})\)[cite: 74].
Step 2: Identify Coordinates of Infinity.
The North Pole corresponds to infinity: \(N(0, 0, 1)\).
Step 3: Calculate Euclidean Distance.
$$ \chi(z, \infty) = \sqrt{(X – 0)^2 + (Y – 0)^2 + (Z – 1)^2} $$
$$ \chi = \sqrt{\left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{2}{3} – 1\right)^2} $$
$$ \chi = \sqrt{\frac{1}{9} + \frac{1}{9} + \left(-\frac{1}{3}\right)^2} $$
$$ \chi = \sqrt{\frac{1}{9} + \frac{1}{9} + \frac{1}{9}} = \sqrt{\frac{3}{9}} = \frac{1}{\sqrt{3}} $$
Answer: \(\frac{1}{\sqrt{3}}\) [cite: 75]
Case 1: The Tangent Sphere (Radius \(R=1/2\))
This is the sphere used in the GATE problem above.
- Equation: \(X^2 + Y^2 + (Z – 1/2)^2 = 1/4\)
- Diameter: 1
- Infinity Point: \((0, 0, 1)\)
Case 2: The Unit Sphere (Radius \(R=1\))
This is the standard sphere centered at the origin.
- Equation: \(X^2 + Y^2 + Z^2 = 1\)
- Diameter: 2
- Infinity Point: \((0, 0, 1)\)
Comparison Summary
| Feature | Tangent Sphere (Your Doc) | Unit Sphere (Standard) |
|---|---|---|
| Radius | \(R = 1/2\) | \(R = 1\) |
| Max Distance (Diameter) | \(1\) | \(2\) |
| Formula for \(\chi(z, \infty)\) | \(\displaystyle \frac{1}{\sqrt{1+|z|^2}}\) | \(\displaystyle \frac{2}{\sqrt{1+|z|^2}}\) |
Symmetric (Inverse) Points
Based on Pages 9–12 of the provided document.
1. General Definition (Page 9)
The concept of “Symmetric Points” generalizes the idea of reflection across a mirror.
Definition: Let \(\gamma\) be a curve in the complex plane \(\mathbb{C}\). [cite_start]If \(\gamma\) is treated as a “mirror” and \(z’\) is the image of a point \(z\) in this mirror, then \(z\) and \(z’\) are called Symmetrical, Inverse, or Conjugate points with respect to \(\gamma\)[cite: 91, 92, 93].
Example (Real Axis): If the curve \(\gamma\) is the real axis (line \(y=0\)), then the symmetric point of \(z = x+iy\) is simply its complex conjugate \(\bar{z} = x-iy\)[cite: 94].
2. Symmetry with Respect to a Circle (Page 11)
This is the most critical definition in this section. [cite_start]Consider a circle in the complex plane defined by \( |z – a| = r \), where \(a\) is the center and \(r\) is the radius[cite: 114].
2.1 Geometric Definition
Two points \(P\) and \(Q\) (represented by \(z\) and \(z’\)) are symmetric with respect to the circle if:
-
[cite_start]
- Collinearity: Points \(P, Q\) and the Center \(C\) are collinear (lie on the same line)[cite: 125]. [cite_start]
- Direction: \(P\) and \(Q\) lie on the same side of the center \(C\)[cite: 125]. [cite_start]
- Distance Product: The product of their distances from the center equals the square of the radius[cite: 126].
2.2 Algebraic Formula
[cite_start]To find the symmetric point \(z’\) of \(z\) with respect to the circle \(|z – z_0| = r\)[cite: 117, 118]:
Alternatively written as: \( (z’ – z_0)(\bar{z} – \bar{z_0}) = r^2 \).
3. Key Properties (Pages 10-12)
- Self-Conjugate Points: If a point \(z\) lies on the curve \(\gamma\) itself, then its symmetric point is \(z\) itself. (i.e., The mirror surface reflects onto itself) [cite_start][cite: 144, 145]. [cite_start]
- Center & Infinity: The center of the circle is symmetric to the point at infinity (\(\infty\))[cite: 100].
- Symmetry w.r.t Lines: A line can be thought of as a circle with infinite radius. [cite_start]Symmetry w.r.t a line is standard perpendicular reflection[cite: 92].
4. Numerical Example (Page 12)
Problem:
[cite_start]Find the inverse (symmetric point) of \(z = i\) with respect to the circle \( |z – 1| = 1 \)[cite: 127, 131].
Solution:
1. Identify Circle Parameters:
Center \( z_0 = 1 \)
Radius \( r = 1 \)
2. Apply Formula:
$$ z’ = z_0 + \frac{r^2}{\bar{z} – \bar{z_0}} $$
Substitute \(z = i\) (so \(\bar{z} = -i\)) and \(z_0 = 1\) (so \(\bar{z_0} = 1\)):
$$ z’ = 1 + \frac{1^2}{-i – 1} $$
$$ z’ = 1 – \frac{1}{1 + i} $$
3. Simplify:
Multiply the fraction by conjugate \((1-i)/(1-i)\):
$$ z’ = 1 – \frac{1-i}{1^2 + 1^2} = 1 – \frac{1-i}{2} $$
$$ z’ = \frac{2 – (1 – i)}{2} = \frac{1 + i}{2} $$
Note: The document also mentions finding the inverse of \(z=1+i\). Since \(| (1+i) – 1 | = |i| = 1\), this point lies ON the circle. [cite_start]Therefore, its inverse is itself (\(1+i\))[cite: 140, 144].