# Can skew lines preserve cross ratio?

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?

• Yes, this is true, but I can prove it only in real projective plane Jun 10, 2018 at 15:27
• @ChristianF So if the cross ratios are equal it follows that $s,t$ are coplanar? Why? Jun 10, 2018 at 15:37
• 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. Jun 10, 2018 at 15:38
• @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. Jun 10, 2018 at 15:50

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.
• 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)? Jun 10, 2018 at 17:11
• @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.