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

The cross ratio is one of the most important quantities in Möbius geometry. It is the basic invariant of Möbius transformations: if you apply any Möbius transformation to four points, their cross ratio does not change.

It is the right tool for answering the question: If I choose three points as my reference frame, where does a fourth point land in that normalized coordinate system? That is exactly what the cross ratio measures.

Imagine we want to build a specific Möbius transformation $T(z)$ that behaves like a sorting machine. We want to feed it three specific points ($z_2, z_3, z_4$) and map them to the “standard” basis of the complex plane: $1, 0,$ and $\infty$.

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: $T(z) \approx \frac{z - z_3}{z - z_4}$.

This successfully maps $z_3 \to 0$ and $z_4 \to \infty$. But what about the final point, $z_2$? We want it to map to $1$ (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 must scale our function by multiplying it by the reciprocal (inverse) of that exact error: $\frac{z_2 - z_4}{z_2 - z_3}$.

Final Machine: Let $z_2, z_3, z_4 \in \mathbb{C}_{\infin}$ be distinct points. Define $T(z) = \left( \frac{z - z_3}{z - z_4} \right) \left( \frac{z_2 - z_4}{z_2 - z_3} \right) $.

By construction, this unique Möbius transformation satisfies exactly $T(z_2) = 1$, $T(z_3) = 0$, and $T(z_4) = \infty$.

The Formal Definition

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 $(z, z_2, z_3, z_4)$ is defined as the image of $z$ under the unique Möbius transformation that maps $z_2 \to 1$, $z_3 \to 0$, and $z_4 \to \infty$.

Algebraically, it is written as $(z,z_2,z_3,z_4) = \bigr(\frac{z-z_3}{z-z_4} \bigl)\bigr(\frac{z_2-z_4}{z_2-z_3} \bigl).$

It basically asks: If I set up a or redefine my universe (coordinate system) so that $z_2, z_3, z_4$ are become my standard markers or reference frame (1, 0, infinity), where does z land or what are the new coordinates of the point z? 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. It measures the relative position of four points, stripping away the absolute coordinate system.

Handling Infinity (The “Cancel Out” Rule)

Because we operate in the extended complex plane $\mathbb{C}_\infty$, any of our given points might be $\infty$. How does the formula adapt? We use limits.

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.

Trivial Examples

$(z_2, z_2, z_3, z_4) = 1$. If I plug $z_2$ (the “1-marker”) into the machine that maps $z_2 \to 1$, what do I get? Obviously 1.
$(z, 1, 0, \infin) = \frac{z - z_3}{z_2 - z_3}|_{z_2=1, z_3=0, z_4=\infty} = \frac{z-0}{1-0} = z, \forall z \in \mathbb{C}_{\infin}$. 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.

Invariance Under Möbius Transformations

One of the most beautiful properties of the cross ratio is that it is a projective invariant.

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 of the original points equals the cross ratio of the transformed points: $(z, z_2, z_3, z_4) = (T_z, T_{z_2}, T_{z_3}, T_{z_4}) =[\text{Notation}](T(z), T(z_2), T(z_3), T(z_4))$

If you take four points and calculate their cross ratio, then run them all through a complxe Möbius transformation (twisting, sliding, zooming, or inverting the plane) and calculate the cross ratio of the new landing spots, you will get the exact same complex number.

Proof

Let $S$ be the Möbius transformation (a specific “machine” or black box) that maps the original point to the standard basis, $z_2$ 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 inverse $T^{-1}$ 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 a Möbius transformation is uniquely determined by three points, $M(w)$ must be the exact function that defines the cross ratio for the transformed points, $\text{RHS} = M(T(z))$. 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. $(z, z_2, z_3, z_4) = (T_z, T_{z_2}, T_{z_3}, T_{z_4}) =[\text{Notation}](T(z), T(z_2), T(z_3), T(z_4))$

The 3-Point Theorem

The invariance of the cross ratio provides us with the ultimate tool for constructing Möbius transformations.

The 3-Point Theorem (Fundamental Theorem of Möbius Geometry). Given any three distinct starting points $(z_2, z_3, z_4)$ and any three distinct target points $(w_2, w_3, w_4)$ in $\mathbb{C}_\infty$, there exists exactly one unique Möbius transformation $S(z)$ that maps $z_2 \to w_2$, $z_3 \to w_3$, and $z_4 \to w_4$.

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.

The Ultimate Calculation Formula

The proof of existence gives us the most practical formula in all of complex geometry. Because the transformation connecting the points is $w = S(z) = M^{-1}(T(z))$, we can apply $M$ to both sides to get $M(w) = T(z)$.

Substituting the definitions of $M$ and $T$ gives us the Cross Ratio Equation: $\frac{w - w_3}{w - w_4} \cdot \frac{w_2 - w_4}{w_2 - w_3} = \frac{z - z_3}{z - z_4} \cdot \frac{z_2 - z_4}{z_2 - z_3}$.

To find the specific Möbius transformation mapping any three points to any three points, you simply equate their cross ratios and algebraically solve for $w$!

Practical Example

Find the Möbius transformation that maps $z: (0, 1, \infty)$ to $w: (-1, -i, 1)$ (This is a classic mapping from the upper half-plane to the unit disk).

We set up the cross ratio equation: $(w, -1, -i, 1) = (z, 0, 1, \infty)$.

Left Side ($w$): $$ \frac{w - (-i)}{w - 1} \cdot \frac{-1 - 1}{-1 - (-i)} = \frac{w+i}{w-1} \cdot \frac{-2}{-1+i} $$
Right Side ($z$): Because $z_4 = \infty$, the rule of thumb says the terms containing it cancel out. $$ \frac{z - z_3}{z_2 - z_3}|_{z_2=1, z_3=0, z_4=\infty} = \frac{z - 0}{1 - 0} = z $$
Equate and Solve: $$ \frac{w+i}{w-1} \cdot \frac{-2}{-1+i} = z \implies w = \frac{z-i}{z+i} $$ From here, you simply use basic algebra to isolate $w$ in terms of $z$, yielding the final transformation function $w = T(z)$.

Notice that a Möbius transformation $T(z)=\frac{az+b}{cz+d},\quad ad-bc\neq 0,$ such that $T(0)=-1,\quad T(1)=-i,\quad T(\infty )=1.$

From these:

From T(0) = -1: $\frac{b}{d}=-1\; \Rightarrow \; b=-d$.

From $T(\infty ) = 1: \frac{a}{c}=1\; \Rightarrow \; a=c.$
From T(1)=-i: $\frac{a+b}{c+d}=-i.$
Substitute a = c, b = -d: $\frac{c-d}{c+d}=-i$. We can scale all coefficients, so set c = 1. Then: $\frac{1-d}{1+d}=-i$. Solve for d: 1 - d = -i(1 + d)= -i-id
1 - d +i +id = 0
(1 + i) + d(-1+i) = 0
d(-1+i) = -(1 +i )
$d = \frac{-(1+i)}{-1+i}=\frac{1+i}{1-i}.$
Multiply numerator and denominator by 1+i: $d=\frac{(1+i)^2}{(1-i)(1+i)}=\frac{1+2i+i^2}{1-i^2}=\frac{1+2i-1}{2}=i.$
So: $c=1,\quad d=i,\quad b=-i,\quad a=1$ .Therefore, $\boxed{T(z)=\frac{z-i}{z+i}}$.