# Cross-ratio relations

The way I define the cross-ratio in projectve geometry:

Let $P_0,P_1,P_2,P_3$ being four points on a projective line G, such that $P_0,P_1,P_2$ are pairwise distinct. Let $\pi:\mathbb KP^1\rightarrow G$ the unique projective map $\pi([1:0])=P_0,\pi([0:1])=P_1, \pi([1:1])=P_2, \pi([x_0:x_1])=P_3$

The cross-ratio $CR(P_0,P_1,P_2,P_3):=\frac{x_1}{x_0}$

Question: Is there a fast way showing the relations $CR(P_1,P_0,P_2,P_3)=\frac{1}{CR(P_0,P_1,P_2,P_3)}=CR(P_0,P_1,P_3,P_2)$

and $CR(P_2,P_1,P_0,P_3)=1-CR(P_0,P_1,P_2,P_3)=CR(P_0,P_3,P_2,P_1)$

-

# Different but equivalent definition

I'd rather use this more general definition:

$$CR(P_0,P_1;P_2,P_3) := \frac{[P_0,P_2][P_1,P_3]}{[P_0,P_3][P_1,P_2]}$$

The brackets on the right hand side denote $2\times2$ determinatnts. This is equivalent to your definition: you can easily show that if you take $\pi$ to be the identity transformation, then the computed result will be the same.

$$CR(P_0,P_1;P_2,P_3) = \frac {\begin{vmatrix}1&1\\0&1\end{vmatrix}\cdot \begin{vmatrix}0&x_0\\1&x_1\end{vmatrix}} {\begin{vmatrix}1&x_0\\0&x_1\end{vmatrix}\cdot \begin{vmatrix}0&1\\1&1\end{vmatrix}} = \frac{x_0}{x_1}$$

Furthermore, the above definition is invarinat under a projective transformation, as the determinant of its matrix will cancel.

$$CR(\pi P_0,\pi P_1;\pi P_2,\pi P_3) = \frac{[\pi P_0,\pi P_2][\pi P_1,\pi P_3]}{[\pi P_0,\pi P_3][\pi P_1,\pi P_2]} = \frac{\det(\pi)^2[P_0,P_2][P_1,P_3]}{\det(\pi)^2[P_0,P_3][P_1,P_2]}$$

So now I hope you are convinced that the two definitions are equivalent.

# Effect of permutations

Using the above definition, you can simply plug in the special coordinates of your four points, but in an arbitrary order, and compute the resulting value.

For example, the case of swapping $P_0$ and $P_2$:

$$(P_2,P_1;P_0,P_3) = \frac {\begin{vmatrix}1&1\\1&0\end{vmatrix}\cdot \begin{vmatrix}0&x_0\\1&x_1\end{vmatrix}} {\begin{vmatrix}1&x_0\\1&x_1\end{vmatrix}\cdot \begin{vmatrix}0&1\\1&0\end{vmatrix}} = \frac{x_0}{x_0-x_1} \neq 1-\frac{x_0}{x_1} = 1-(P_0,P_1;P_2,P_3)$$

So one of the equations you wanted to show cannot be shown, since it is wrong.

# All permutations

Here is a complete list of all permutations of your four points, and the cross ratios resulting from each of them.

\begin{align*} (P_0,P_1;P_2,P_3)=(P_1,P_0;P_3,P_2)=(P_3,P_2;P_1,P_0)=(P_2,P_3;P_0,P_1) &=\lambda\\ (P_0,P_1;P_3,P_2)=(P_1,P_0;P_2,P_3)=(P_2,P_3;P_1,P_0)=(P_3,P_2;P_0,P_1) &=\tfrac1\lambda\\ (P_0,P_2;P_1,P_3)=(P_2,P_0;P_3,P_1)=(P_3,P_1;P_2,P_0)=(P_1,P_3;P_0,P_2) &=1-\lambda\\ (P_0,P_2;P_3,P_1)=(P_2,P_0;P_1,P_3)=(P_1,P_3;P_2,P_0)=(P_3,P_1;P_0,P_2) &=\tfrac{1}{1-\lambda}\\ (P_0,P_3;P_1,P_2)=(P_3,P_0;P_2,P_1)=(P_2,P_1;P_3,P_0)=(P_1,P_2;P_0,P_3) &=1-\tfrac1\lambda\\ (P_0,P_3;P_2,P_1)=(P_3,P_0;P_1,P_2)=(P_1,P_2;P_3,P_0)=(P_2,P_1;P_0,P_3) &=\tfrac{\lambda}{\lambda-1} \end{align*}

-