Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the cross-ratio of four lines $L_1,L_2,L_3,L_4$ if they are parallel? What is the cross-ratio if $L_4$ is the line at infinity?

share|cite|improve this question
I know what the cross ratio of points. What is a cross ratio of lines? – William Jun 27 '12 at 16:05

The fact of being parallel is an affine notion. If you projectivize your affine plane, then this notion is indeed restated as intersecting at infinity. However the cross-ratio is a purely projective notion, it doesn't matter if your projective plane comes from (is the projectivization of) an affine plane or not. Therefore the cross-ratio of four lines crossing at infinity has nothing special compared to the cross-ratio of four lines crossing somewhere else.

share|cite|improve this answer

You obtain the cross ratio of four lines $\ell_i$ going through the same point $M$ by intersecting these lines with any line $g$ not going through $M$. In this way you obtain $4$ points $A_i$ on $g$, and the cross-ratio of these four points is the cross-ratio of the $\ell_i$.

Your four lines go through the same point $M$ on the line at infinity, so their cross-ratio makes sense. When they are given, e.g., in the form $$\ell_i:\quad y=mx + a_i\qquad(1\leq i\leq4)\ ,$$ you can take the $y$-axis as your $g$. In this way the numbers $a_i$ can be viewed as admissible (i.e., projective) coordinates of the $4$ points $A_i$. Therefore the cross-ratio $C$ of the $\ell_i$ is the cross-ratio $C$ of the $a_i$.

In order to obtain the correct value when $\ell_4$ is the line at infinity, let $a_4\to\infty$ in the expression $C$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.