0

I am using the formula for the cross ratio found here:

http://en.wikipedia.org/wiki/Cross-ratio#Definition

Let us focus our attention on the cross ratio of points which are in an arithmetic sequence. For example the real-line points $(z_1,z_2,z_3,z_4)=(0,1,2,3)$ have cross ratio equal to $4/3$.

Now set $z_1=0$, $z_2=1$, $z_3=10$. If I fix the cross ratio of $(z_1,z_2,z_3,z_4)$ to be $4/3$, I would expect $z_4$ to be some large number, since the cross ratio is preserved by projective transformations. However, when I calculate $z_4$, I get $z_4=-5$. That means the order of the points is now $\color{red}{z_4},z_1,z_2,z_3$.

How can this be?

wircho
  • 1,431

1 Answers1

1

Remember that a projective line contains a point "at infinity" which is approached by points moving toward either end of the ordinary, non-projective line. What happened in your example is that, as $z_3$ increased from the original $2$ to the final $10$, the corresponding $z_4$ moved from the original $3$ to larger and larger values until it reached the point at infinity, after which it kept moving in the same direction from the other end of the ordinary line, starting at very negative values and gradually increasing until it reached $-5$.

Andreas Blass
  • 71,833