1
$\begingroup$

I am currently trying to understand the cross ratio in projective geometry more. I wondered about the following and appreciate any answers:


Assume four lines $l_1, l_2, l_3, l_4 \in \mathbb{RP}^3$. Also assume two skew lines $s, t$ intersect all four lines such that the cross ratio of the points of intersection is the same, namely $$(l_1 \cap s, l_2 \cap s, l_3 \cap s, l_4 \cap s) = (l_1 \cap t, l_2 \cap t, l_3 \cap t, l_4 \cap t)$$

Is that possible? (I think it is not) Why is it not possible? Does follow that the four lines are concurrent?

$\endgroup$
  • $\begingroup$ Yes, this is true, but I can prove it only in real projective plane $\endgroup$ – Aqua Jun 10 '18 at 15:27
  • $\begingroup$ @ChristianF So if the cross ratios are equal it follows that $s,t$ are coplanar? Why? $\endgroup$ – user526159 Jun 10 '18 at 15:37
  • $\begingroup$ I didn't say that. But on the other hand I don't know how to define cross ratio if all those lines are not in the same plane. $\endgroup$ – Aqua Jun 10 '18 at 15:38
  • $\begingroup$ @ChristianF Ahh my mistake. I thougt for the cross ratio (of points) it is only necessary that those four points lie on the same line, which they clearly do ($s, t$ respectively). Now it is generally possible that four lines are intersected by two skew lines (just choose four different points on each one of two skew lines and take the lines through four pairs of these points). So basically i wonder if these points could be chosen in such a way that the cross ratio of the points on one of the skew lines equals the one of the other points on the other skew line. $\endgroup$ – user526159 Jun 10 '18 at 15:50
1
$\begingroup$

Think about it the other way round. Start with two skew lines. Pick four points on one, and three on the other. Then there exists a unique point on the second such that the cross ratios are the same. Now you can connect corresponding points on both skew lines to get your lines $l_i$. So it certainly appears possible. And since the lines don't have a plane in comon, there is no place whee the lines could pass through a single point, so no, I don't think they need to be concurrent.

$\endgroup$
  • $\begingroup$ ok thank you, makes sense! Now a lot of new questions came up - if i may. So by the eight points on two lines respectively i have already defined a projective transformation? More general does the equalty of cross ratios of two times four points (lines, planes,...) define such a transformation already? If i take another line that intersects $l_1, l_2, l_3$ - does it intersect $l_4$ aswell? Does it do so because the four lines define a projective transformation (unsure if this is how one would formulate it)? $\endgroup$ – user526159 Jun 10 '18 at 17:11
  • $\begingroup$ @user526159: Knowing 3 points and their images defines a projective transformation of a line. So you have a transformation between a basis on one line and one on the other line, and the fourth point is transformed consistent with that transformation. But that's not saying anything about the 3-space containing the lines. A genera 3d projective transformation is uniquely defined by 5 points and their images, and those 5 points have to be in general position, i.e. no 4 coplanar. You don't have enough info for uniqueness there. Existence should be possible, though. $\endgroup$ – MvG Jun 10 '18 at 22:03
  • $\begingroup$ @user526159: For your second question about whether a line intersecting $l_1,l_2,l_3$ will always intersect $l_4$ I have no good argument in mind either way just now, and too little time to really think about it. Feel free to post as a separate question, mentioning your question here as motivation. Not sure whether the answer will even depend on the lines preserving the cross ratio. $\endgroup$ – MvG Jun 10 '18 at 22:08
  • $\begingroup$ thank you for your time and effort! posted another question. And will get back to studying projectivities more tomorrow. $\endgroup$ – user526159 Jun 10 '18 at 22:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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