The Cross Ratio 2

Ned, I would love to stand here and talk with you —but I’m not going to.” — Phil Connors, Groundhog Day.

The Stereographic Projection

Consider the unit sphere $S^2$ in $\mathbb{R}^3$ that is $S^2 = \{ (u, v, w) \in \mathbb{R}^3 | u^2 + v^2 + w^2 = 1 \}$. We identify the complex plane with the equitorial plane, that is, for coordinates (u, v, w) of $\mathbb{R}^3$, ℂ is the plane where w = 0. The North pole has coordinates N = (0, 0, 1). We will denote a complex number z by x + iy. Notice that the unit circle in ℂ coincides with the equator of $S^2$.

For any point $P \in S^2$, there is a unique line from N to P, which we denote by NP. This line intersects the complex plane in exactly one point $z \in \mathbb{C}$. The map from $SP: S^2 \setminus N \to \mathbb{C}$ which assigns to a point P = (u, v, w) the point $z \in \mathbb{C}$ given by the intersection NP ∩ $\mathbb{C}$ is called the stereographic projection where $SP(u, v, w) = \frac{u}{1-w} + i\frac{v}{1-w}$

If we identify, via stereographic projection, points in the complex plane with points in $S^2 \setminus N$ and further identify ∞ with the North POle N then we have a bijection between the extended complex plane $\mathbb{C}_{\infin}$ and the Riemann Sphere $S^2$. Under this identification $S^2$ is known as the Riemann sphere. Under this projection, circles on the sphere correspond exactly to generalized circles (circles or lines) in the plane.

Proposition. Let $z_1, z_2, z_3$, and $z_4$ be four distinct points in $\mathbb{C}_{\infin} = \mathbb{C} \cup \{ \infin \}$ (the Riemann sphere). Then, the cross ratio $(z_1, z_2, z_3, z_4)$ is a real number if and only if the four points $z_1, z_2, z_3$, and $z_4$ lie on a common circle or straight line (viewed as a “generalized circle”).

Möbius transformations preserve the class of “Generalized Circles” (which includes straight lines). If C is a generalized circle, then for any Möbius map T, the image $T(C)$ is also a generalized circle. We define the real line $\mathbb{R} \cup \{\infty\}$ as a “generalized circle” passing through $1, 0, \text{and } \infty$.

Proof

($\Rightarrow$): If $(z_1, z_2, z_3, z_4)$ is a real number, then $z_1, z_2, z_3, z_4$ lie on a circle.

$(z_1, z_2, z_3, z_4) = r \in \mathbb{R}$

Since Möbius transformations are 3‑point transitive on the Riemann sphere, there is a unique Möbius map $S: \mathbb{C} \cup \{ \infin \} \to \mathbb{C} \cup \{ \infin \}$ that maps $z_2 \to 1$, $z_3 \to 0$, $z_4 \to \infty$. This transformation can be conveniently described using the cross-ratio, S(z) = $(z, z_2, z_3, z_4)$.

By definition, $r = S(z_1)$. So S maps $z_1$ to the real line. Therefore, S maps all four points $\{z_1, z_2, z_3, z_4\}$ onto the real extended line $\mathbb{R} \cup \{ \infin \}$ (a “generalized circle”).

Consider the inverse map $S^{-1}$. Since S is a Möbius transformation, $S^{-1}$ is also a Möbius transformation.

A fundamental property of Möbius maps is that they send (generalized) circles to (generalized) circles. Therefore, the inverse image $S^{-1}(\mathbb{R} \cup \{ \infin \})$ is also a generalized circle in the z-plane. This inverse image contains the four original points $z_1, z_2, z_3, z_4$ (they are all pre-images of real numbers). Hence, they must all lie on some circle or line $\Gamma$.

($\Leftarrow$): Suppose $z_1, z_2, z_3, z_4$ lie on a common circle (or line) $\Gamma$, then their cross ratio is real.

Again choose a Möbius map S sending $z_2 \to 1$, $z_3 \to 0$, $z_4 \to \infty$. By its definition, S sends $\Gamma$ (a generalized circle through $z_2, z_3, z_4$) to the real line (three distinct points of $\Gamma$ map to $1, 0, \infin$, which lie on $\mathbb{R} \cup \{ \infin \}$). Since Möbius transformations send circles/lines to circles/lines, the whole circle $\Gamma$ must map to the real line $\mathbb{R}_{\infty}$ (because a generalized circle is uniquely determined by 3 points).

Because $z_1$ is also on $\Gamma$, its image $S(z_1)$ lies on the real line. But by contruction $S(z_1) = (z_1, z_2, z_3, z_4)$. Therefore, the cross ratio $(z_1, z_2, z_3, z_4)$ is a real number.

Proposition. The pre-image of the real line under a Möbius map is also a circle (or line).

Proof.

We want to find the shape of all points w such that $T(w)$ is real, $\{ w \in \mathbb{C}: T(w) \in \mathbb{R} \}$. Let $T(w) = \frac{aw+b}{cw+d}, a, b, c, d \in \mathbb{C}, ad-bc \ne 0$ and show it is always either a straight line or a circle (a generalized circle).

Recall: A number $\zeta$ is real if and only if $\zeta = \bar{\zeta}$ (it equals its conjugate).

T(w) is real exactly when it equals its complex conjugate: $T(w) = \overline{T(w)}$

$\frac{aw+b}{cw+d} = \overline{\left( \frac{aw+b}{cw+d} \right)} = \frac{\bar{a}\bar{w}+\bar{b}}{\bar{c}\bar{w}+\bar{d}}$

Multiply both sides to clear the fractions: $(aw+b)(\bar{c}\bar{w}+\bar{d}) = (\bar{a}\bar{w}+\bar{b})(cw+d)$

Expand the terms: $a\bar{c}w\bar{w} + a\bar{d}w + b\bar{c}\bar{w} + b\bar{d} = \bar{a}cw\bar{w} + \bar{a}dw + \bar{b}c\bar{w} + \bar{b}d$

Move everything to the left side and group like terms: $\underbrace{(a\bar{c} - \bar{a}c)}_{\text{Coeff of } |w|^2} |w|^2 + \underbrace{(a\bar{d} - \bar{b}c)}_{\text{Coeff of } w} w + \underbrace{(b\bar{c} - \bar{a}d)}_{\text{Coeff of } \bar{w}} \bar{w} + \underbrace{(b\bar{d} - \bar{b}d)}_{\text{Constant}} = 0$

Recall that $w\bar{w} = |w|^2$.

Let’s name these coefficients to make it readable: $A = a\bar{c} - \bar{a}c, B = a\bar{d} - \bar{b}c, C = b\bar{d} - \bar{b}d$ and notice that A and C are pure imaginary numbers or zero and the coefficient of $\bar{w}$ is equal to $-\bar{B}$. Thereofre, the equation can be written or expressed as: $A|w|^2 + Bw - \bar{B}\bar{w} + C = 0$

For any complex number Z, $Z - \bar{Z} = 2i\text{Im}(Z)$.

There are two cases:

Case 1: The Linear Case (A = 0). If A = 0, then the $|w|^2$ term vanishes. We are left with: $Bw - \bar{B}\bar{w} + C = 0$.

Write w = x + iy and $B=\beta_1+i\beta_2, C=\gamma_1+i\gamma_2$ (real/imag parts). Expanding all terms shows that the equation reduces to a linear equation in (x,y): $\alpha x + \beta y + \delta =0,$ hence it is a straight line (geometrically this is the case where the preimage of the real axis is a line).

Case 2: The Circular Case ($A \neq 0$)

If $A \neq 0$, we have a quadratic in x, y, which represents a circle. To see the radius and center clearly (transform it into the standard centered-circle form), we must “complete the square.”

Divide the whole equation by A: $|w|^2 + \frac{B}{A}w - \frac{\bar{B}}{A}\bar{w} + \frac{C}{A} = 0$

We want to force this into the form $|w - center|^2 - radius^2 = 0$, taking into account that $(w - \text{center})(\bar{w} - \overline{\text{center}}) = |w|^2 - w\overline{\text{center}} - \bar{w}\text{center} + |\text{center}|^2$. Matching the terms, let the center be $-w_0 = -\frac{\bar{B}}{A} \implies w_0 = \frac{\bar{B}}{A}$. (Note: Since A is pure imaginary, $\bar{A} = -A$).

So we add $|\frac{B}{A}|^2$ to both sides to complete the square: $\left| w - \frac{\bar{B}}{A} \right|^2 = \left| \frac{B}{A} \right|^2 - \frac{C}{A}$

We need to simplify the Right Hand Side (RHS) to prove it is a positive real number (a valid radius), $\text{RHS} = \frac{|B|^2}{A\bar{A}} - \frac{C}{A}$

Since $A$ is pure imaginary, $A\bar{A} = |A|^2$ and $\bar{A} = -A$, hence $\text{RHS} = \frac{|B|^2}{A\bar{A}} - \frac{C(\bar{A})}{A\bar{A}} = \frac{|B|^2}{A\bar{A}} - \frac{C(\bar{A})}{A\bar{A}} = \frac{|B|^2 -C\bar{A}}{|A|^2} = \frac{|B|^2 + AC}{|A|^2}$

Let’s look at the numerator $N = |B|^2 + AC$. Substitute the definitions of A, B, C:

$$ \begin{aligned} N = |a\bar{d} - \bar{b}c|^2 + (a\bar{c} - \bar{a}c)(b\bar{d} - \bar{b}d) &=(a\bar{d} - \bar{b}c)(\bar{a}d -b\bar{c}) + (a\bar{c} - \bar{a}c)(b\bar{d} - \bar{b}d) \\[2pt] &=a\bar{d}\bar{a}d -a\bar{d}b\bar{c} -\bar{b}c\bar{a}d +\bar{b}cb\bar{c} + a\bar{c}b\bar{d} -a\bar{c}\bar{b}d-\bar{a}cb\bar{d}+\bar{a}c\bar{b}d\\[2pt] &=|a|^2|d|^2 -ab\bar{d}\bar{c}-\bar{a}\bar{b}cd + |b|^2|c|^2 + ab\bar{c}\bar{d} -a\bar{b}\bar{c}d-\bar{a}bc\bar{d}+\bar{a}\bar{b}cd\\[2pt] &=|a|^2|d|^2+ |b|^2|c|^2-a\bar{b}\bar{c}d-\bar{a}bc\bar{d}\\[2pt] &=(ad -bc)(\bar{a}\bar{d}-\bar{b}\bar{c})\\[2pt] &=|ad - bc|^2 \end{aligned} $$

So the radius equation becomes: $\left| w - \text{center} \right|^2 = \frac{|ad - bc|^2}{|a\bar{c} - \bar{a}c|^2}$ Since the RHS is a positive real number (absolute value squared), this defines a valid Circle.

The set of points $w$ that map to the real line is always a Circle (Case 2) or a Line (Case 1). The pre-image is always a generalized circle.

Proposition. The following two statements are equivalent:

1. Möbius transformations send generalized circles to generalized circles.
2. For every Möbius transformation T, the pre-image of the real line $T^{-1}(\mathbb{R}_{\infin})$ is a generalized circle.

Proof.

⇒ If Möbius maps send generalized circles to generalized circles, then the pre-image of the real line under a Möbius map is a generalized circle.

Let T be any Möbius transformation. The inverse of a Möbius transformation, $T^{-1}$, is also a Möbius transformation. Let $\mathbb{R}_{\infty}$ be the Real Line (which is a generalized circle). Since $T^{-1}$ is a Möbius transformation, it must map the generalized circle $\mathbb{R}_{\infty}$ to another generalized circle (either a circle or a line). So, saying “Möbius maps preserve generalized circles” automatically implies “the pre-image of the real line under any Möbius transformation is a generalized circle.”

⇐ Assume that for every Möbius transformation M, the pre-image $M^{-1}(\mathbb{R}_{\infin})$ is a generalized circle. We show that every Möbius transformation sends generalized circles to generalized circles.

Step 1. We want to prove that if we assume “For any Möbius map M, the pre-image of the real line is a generalized circle,” then it must be true that “Any Möbius map T sends the real line to a generalized circle.”

Let $T$ be any Möbius transformation. Let $M = T^{-1}$. Since $T$ is a Möbius transformation (of the form $\frac{az+b}{cz+d}$), its inverse $M$ is also a Möbius transformation (of the form $\frac{dw-b}{-cw+a}$).

As we have already stated that we know that “the pre-image of the real line under a Möbius map is a circle,” M is a Möbius map, then $\text{Shape } \Gamma = M^{-1}(\mathbb{R}_{\infty})$ is a generalized circle. Since $M = T^{-1}$, then $M^{-1} = (T^{-1})^{-1} = T$. Hence, $\Gamma = T(\mathbb{R}_{\infty})$ is a generalized circle.

Step 2. Images of arbitrary generalized circles

Let $C$ be any circle or line we want to map. There exists a simple Möbius map $R$ that moves the Real Line onto $C$. So $C = R(\mathbb{R}_{\infty})$.

We want to know the shape of $T(C)$. Substitute C and we get: $T(C) = T( R(\mathbb{R}_{\infty}) ) = (T \circ R) (\mathbb{R}_{\infty})$.

Since $T$ and $R$ are Möbius maps, their combination $(T \circ R)$ is again a Möbius map, and we know from Step 1 $(T \circ R)$ sends the real line to a generalized circle. Therefore, T(C) is a generalized circle.

Proposition. Let $C$ be any circle or line we want to map. There exists a simple Möbius map $R$ that moves the Real Line onto $C$.

Proof.

Recall the Fundamental Theorem in Möbius geometry, there is a unique Möbius transformation that maps any three distinct points $(z_1, z_2, z_3)$ to any other three distinct points $(w_1, w_2, w_3)$.

Pick three points on the Real Line (typically, we use the three more famous points on the Real Line: 0, 1, and $\infin$) and on the circle C (let’s call them $z_1, z_2, z_3$; Note: If C is a straight line, one of these points might be $\infty$).

According to the Fundamental Theorem, there exists a Möbius transformation R such that: $R(0) = z_1, R(1) = z_2$, and $R(\infty) = z_3$

We know R maps three specific points of the Real Line to three specific points of $C$. But does it map the rest of the line to the rest of the circle?

Yes, it does because (1) Möbius maps send generalized circles to generalized circles; (2) The Real Line is a generalized circle determined by the three points $\{0, 1, \infty\}$; (3) The image under R must be a generalized circle determined by the three image points $\{z_1, z_2, z_3\}$; (4) Since there is only one generalized circle passing through any three distinct points, the image $R(\mathbb{R} \cup \{\infty\})$ must be exactly $C$.

• Find a Möbius map $R$ that moves the Real Line onto the Unit Circle, $|z|=1$.

Pick three points on the Real Line (typically, we use the three more famous points on the Real Line: 0, 1, and $\infin$) and on the unit circle, $z_1 = -1, z_2 = -i$, and $z_3 = 1$.

We want $R(0)=-1, R(1)=-i, R(\infty)=1$. A Möbius transformation has the form: $R(z)=\frac{az+b}{cz+d}$​, where $ad−bc\ne 0$. The cross-ratio is invariant under Möbius transformations, $(R(z), R(z_2​), R(z_3​), R(z_4​) ) =(z, z_2​, z_3​, z_4​).$

For points $z_1, z_2, z_3, z_4$, the cross-ratio is $(z_1, z_2, z_3, z_4) = \frac{(z_1-z_3)(z_2-z_4)}{(z_1-z_4)(z_2-z_3)}$. In our case, the cross-ratio condition is: $(R(z), -1, -i, 1) = (z, 0, 1, \infin)$

Compute the right-hand side. For $(z, 0, 1, \infin)$, since $\infin$ is involves, the cross-ratio simplifies to: $(z, 0, 1, \infin) = \frac{z_1-z_3}{z_2-z_3} = \frac{z - 1}{0-1} = 1-z$

Compute the left-hand side. $(R(z), -1, -i, 1) = \frac{(R(z) + i)(-1 - 1)}{(R(z) - 1)(-1 + i)} = \frac{-2(R(z) + i)}{(R(z) - 1)(-1 + i)} = \frac{-2(R(z) + i)}{(i - 1)(R(z) - 1)}$

Set the two cross-ratios equal:

$$ \begin{aligned} \frac{-2(R(z) + i)}{(i - 1)(R(z) - 1)} = 1 - z &\implies[\text{Multiply both sides}] \\[2pt] &-2(R + i) = (1 - z)(i - 1)(R - 1)\\[2pt] &[(1−z)(i−1)(R−1) = [(i−1)−(i−1)z](R−1) = (i−1)R−(i−1)−(i−1)zR+(i−1)z] \\[2pt] &\implies −2R−2i = R[(i−1)−(i−1)z]+[(i−1)z−(i−1)] \\[2pt] &\implies R[-2-(i-1)(1-z)] = (i-1)(z-1)+2i \\[2pt] &\implies R[−2−(i−1)+(i−1)z] = (i−1)z−(i−1)+2i \\[2pt] &\implies R((i−1)z−(1+i)) = (i−1)z+(1+i) \\[2pt] &\implies R = \frac{(i−1)z+(1+i)}{(i−1)z-(1+i)​} \\[2pt] &\implies R = \frac{z+\frac{1+i}{i-1}}{z-\frac{1+i}{i-1}​} = \frac{z-i}{z+i} \end{aligned} $$

and this is the Cayley transform, the canonical map from $\mathbb{R}_{\infin}$ to the unit circle.

Note: $\frac{1+i}{i-1}=\frac{(1+i)(-i-1)}{(i-1)(-i-1)}=\frac{-2i}{2} = -i$