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?
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.