Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the universe trying to build bigger and better idiots. So far, the universe is winning, Rick Cook

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The Cross Ratio

Imagine we want to build a Möbius transformation $T(z)$ that behaves like a specific sorting machine. We want to feed it three specific points ($z_2, z_3, z_4$) and get out three standard results:

Input $z_2 \to$ Output 1.
Input $z_3 \to$ Output 0. An easy way to get zero is to put $(z - z_3)$ in the numerator.
Input $z_4 \to$ Output $\infty$. An easy way to get infinity is to divide by zero, so we put $(z - z_4)$ in the denominator.

Current Draft: $T(z) \approx \frac{z - z_3}{z - z_4}$. How do we handle the “1” (normalization)? Let’s plug $z_2$ into our current draft and see what happens: $\frac{z_2 - z_3}{z_2 - z_4}$. This isn’t our desired result 1. To fix this, we need to multiply our function by the reciprocal (inverse) of that error, $\frac{z_2-z_4}{z_2-z_3}$

Let $z_2, z_3, z_4 \in \mathbb{C}_{\infin}$ be distinct points. Define $T(z) := \bigr(\frac{z-z_3}{z-z_4} \bigl)\bigr(\frac{z_2-z_4}{z_2-z_3} \bigl)$ where $T(z_2)=1, T(z_3) = 0, T(z_4) = \infin$.

Handling Infinity (The “Cancel Out” Rule)

If $z_2 = \infty$. We look at the second part (the correction factor): $\frac{z_2 - z_4}{z_2 - z_3}$. As $z_2$ gets huge, $z_4$ and $z_3$ become irrelevant. The fraction becomes $\frac{\infty}{\infty} \to 1$. The term $z_2$ essentially “cancels out” or disappears from the formula. So, we simply remove the terms with $z_2, T(z) = \frac{z - z_3}{z - z_4}$.
If $z_3 = \infty$. We look at the terms containing $z_3, \frac{z - z_3}{z_2 - z_3}$. As $z_3 \to \infty$, this looks like $\frac{-z_3}{-z_3} = 1$. So, we remove the terms with $z_3, T(z) = \frac{z_2 - z_4}{z - z_4}$
If $z_4 = \infty$. We look at the terms containing $z_4, \frac{z_2 - z_4}{z - z_4}$. As $z_4 \to \infty$, this looks like $\frac{-z_4}{-z_4} = 1$. So, we remove the terms with $z_4, T(z) = \frac{z - z_3}{z_2 - z_3}$.
The Rule of Thumb: If a term $z_i$ goes to infinity, any fraction containing $z_i$ approaches 1 (or -1) and essentially cancels out or disappears from the formula.

In any of the above case, $T(z_2) = 1, T(z_3) = 0, T(z_4) = \infin$ and T is the unique Möbius transformation with this property.

The Cross Ratio

Definition. Given four distinct points $z,z_2,z_3,z_4$ in the extended complex plane $\mathbb{C}_{\infin}=\mathbb{C} \cup \{ \infty \}$, the cross ratio is defined as the value T(z), where T is the unique transformation that maps or takes $z_2$ to 1, $z_3$ to zero, and $z_4$ to infinity, $(z,z_2,z_3,z_4) = \bigr(\frac{z-z_3}{z-z_4} \bigl)\bigr(\frac{z_2-z_4}{z_2-z_3} \bigl).$

It basically asks: If I set up a coordinate system where $z_2, z_3, z_4$ are my standard markers or reference frame (1, 0, infinity), where does z land? In other words, where does the point z land in this normalized coordinate system where $z_3$ is the origin, $z_2$ is the unit marker, and $z_4$ is the point at infinity.

Examples

$(z_2, z_2, z_3, z_4) = 1$. If I plug $z_2$ into the machine that maps $z_2 \to 1$, what do I get? Obviously 1.
$(z, 1, 0, \infin) = z, \forall z \in \mathbb{C}_{\infin}$. If my reference points are already at the standard positions ($z_2=1, z_3=0, z_4=\infty$), then the transformation is just the identity function $T(z) = z$. So the output is just z.

Proposition. Möbius transformations preserve the ‘geometry’ of points. If $z_2, z_3, z_4 \in \mathbb{C}_{\infin}$ are distinct and T is any Möbius transformation, then the cross ratio $(z, z_2, z_3, z_4) = (T_z, T_{z_2}, T_{z_3}, T_{z_4}) =[\text{Notation}](T(z), T(z_2), T(z_3), T(z_4))$

If you take four points and calculate their cross ratio, then run them all through a Möbius transformation (twist, slide, zoom, or invert them) and calculate the cross ratio of the new points, you get the exact same number.

Proof

Let $S$ be the Möbius transformation (a specific “machine” or black box) that maps or takes $z_2$ to 1, $z_3$ to zero, and $z_4$ to infinity, the cross ratio of the originals $(z, z_2, z_3, z_4)$ is defined by the unique map S and $\text{LHS} = S_z = S(z) = (z,z_2,z_3,z_4)$.

To find the cross ratio of the transformed points $(T(z), T(z_2), T(z_3), T(z_4))$, we need to find a new Möbius transformation or machine (let’s call it $M$) that standardizes these specific points. M must take $T(z_2) \to 1, T(z_3) \to 0$, and $T(z_4) \to \infty$. If we find such a map M, then by definition: $\text{RHS} = M(T(z))$

Consider the composite function: $M = S \circ T^{-1}$, where both $T^{-1}$ and composition are Möbius transformations. We feed it the transformed points $T(z_2), T(z_3), T(z_4)$ and see what comes out.

$M(T(z_2)) = S(T^{-1}(T(z_2)))$ =[The $T^{-1}$ and T cancel out:] $S(z_2) = 1.$
$M(T(z_3)) = S(T^{-1}( T(z_3) ) ) = S(z_3) = 0$
$M(T(z_4)) = S( T^{-1}( T(z_4) ) ) = S(z_4) = \infty$

Because Möbius transformations are unique, if we find any Möbius transformation that sends those three points to $1, 0, \infty$, it must be the one that defines the cross ratio, $\text{RHS} = M(T(z))$. Substitute our definition of $M, \text{RHS} = S( T^{-1}( T(z) ) ) = S(z)$

Since $\text{LHS} = S(z)$ and $\text{RHS} = S(z)$, they are equal. $(z, z_2, z_3, z_4) = (T_z, T_{z_2}, T_{z_3}, T_{z_4}) =[\text{Notation}](T(z), T(z_2), T(z_3), T(z_4))$

3-Point Theorem for Möbius transformations. If $z_2, z_3, z_4$ are distinct points in $\mathbb{C}_{\infin} = \mathbb{C} \cup \{ \infin \}$ and $w_2, w_3, w_4$ are also distinct, then there is one and only one Möbius transformation such that $S_{z_2}=w_2, S_{z_3}=w_3, S_{z_4}=w_4$.

It tells us exactly how much freedom we have: Three points determine everything.

Proof.

We have two sets of three distinct points: $z_2, z_3, z_4$ (start set) and $w_2, w_3, w_4$ (target set). We want to find a single Möbius transformation S that connects them perfectly: $z_2 \to w_2, z_3 \to w_3, z_4 \to w_4$.

Let T(z) be the unique Möbius transformation that maps the starting points to the standard configuration: $T(z) = (z, z_2, z_3, z_4), T(z_2) = 1, T(z_3) = 0, T(z_4) = \infty$ and M(w) be the unique Möbius transformation that maps the target points to the standard configuration: $M(w) = (w, w_2, w_3, w_4), M(w_2) = 1, M(w_3) = 0, M(w_4) = \infty$

Let’s build the bridge (S) because we want to go from z to w: (i) We use T to get from z to standard, (ii) We use the inverse $M^{-1}$ to get from standard back to w, $S(z) = (M^{-1} \circ T)(z)$

Does it work?

$S(z_2) = M^{-1}( T(z_2) ) = M^{-1}(1) =[M(w_2) = 1] w_2$.
$S(z_3) = M^{-1}( T(z_3) ) = M^{-1}(0) [M(w_3) = 0] w_3$.
$S(z_4) = M^{-1}( T(z_4) ) = M^{-1}(\infty) [M(w_4) = \infin] w_4$

So, the transformation $S = M^{-1} \circ T$ definitely exists. Now we must prove that S is the only map that does this.

Assume there is another cap (a “fake copycat”) R that also does the job, $R(z_2) = w_2, R(z_3) = w_3, R(z_4) = w_4$. Consider the composition: $L = R^{-1} \circ S$ and check what it does to our starting points ($z_2, z_3, z_4$):

$(R^{-1} \circ S)(z_2) = R^{-1}(S(z_2)) = R^{-1}(w_2) =[R(z_2) = w_2] z_2$ (it fixes $z_2$).
$(R^{-1} \circ S)(z_3) = R^{-1}(S(z_3)) = R^{-1}(w_3) =[R(z_3) = w_3] z_3$ (it fixes $z_3$).
$(R^{-1} \circ S)(z_4) = R^{-1}(S(z_4)) = R^{-1}(w_4) =[R(z_4) = w_4] z_4$ (it fixes $z_4$).

Now we could apply the Fixed Point Theorem The only Möbius transformation with 3 or more fixed points is the Identity map. We have a Möbius transformation L that has three fixed points ($z_2, z_3, z_4$). Therefore, L = Identity, $R^{-1} \circ S = Id \implies R \circ (R^{-1} \circ S) = R \circ \text{Id} \implies R = S$. Any “other” map R turns out to be exactly $S$. Thus, S is unique.