For any four distinct points $P_{1},P_{2},P_{3},P_{4}\in P^{1}(\mathbb{R})$ we have $(P_{1}P_{3};P_{2}P_{4})=1-(P_{1}P_{2};P_{3}P_{4})$.

Question

Show that the cross-ratio has the following property: for any four distinct points $P_{1},P_{2},P_{3},P_{4}\in P^{1}(\mathbb{R})$ we have $(P_{1}P_{3};P_{2}P_{4})=1-(P_{1}P_{2};P_{3}P_{4})$.

What is a good way to prove this problem? Thanks a lot.

Answer 1

If you have established that cross ratios are invariant under projective transformations, you can choose three of the points without loss of generality. E.g. make three of them $0,1,\infty$. Then you'll only have to deal with a single real parameter. If you haven't established this invariance yet, you may start by doing so.

Another attempt would be writing variables as coordinates of homogeneous coordinate vectors. You can simplify the situation by assuming that the second coordinate of each homogeneous coordinate vector is equal to $1$. That choice would again be without loss of generality if you know about the invariance under projective transformations, but if not, you may have to handle that case separately.

A third attempt would be addressing this at the level of $2\times 2$ determinants, which I'll write as $[\cdot,\cdot]$. One usually deines the cross ratio in terms of such determinants, plugging in homogeneous coordinate vectors. So you have

$$(P_1,P_3;P_2,P_4)+(P_1,P_2;P_3,P_4)= \frac{[P_1,P_2][P_3,P_4]}{[P_1,P_4][P_3,P_2]}+ \frac{[P_1,P_3][P_2,P_4]}{[P_1,P_4][P_2,P_3]}= \frac{[P_1,P_2][P_3,P_4]-[P_1,P_3][P_2,P_4]}{[P_1,P_4][P_3,P_2]}= -\frac{[P_1,P_4][P_2,P_3]}{[P_1,P_4][P_3,P_2]}= 1$$

The penultimate transformation step there makes use of the Grassmann-Plücker relation

$$[P_1,P_2][P_3,P_4] - [P_1,P_3][P_2,P_4] + [P_1,P_4][P_2,P_3] = 0$$

which you might have to establish first. Suitable ways to show this are similar to the ones described above: choose points in suitable ways without loss of generality in order to reduce the number of variables in a symbolic computation.