**Problem.** How to partition R^3 into pairwise non-parallel lines?


A possible solution is to stack infinitely many ``concentric'' hyperboloids, by increasing radius and decreasing slope. And don't forget the line on the $z$ axis at the center. The prototype hyperboloid [looks like this](http://www-lm.ma.tum.de/archiv/sos04/la2lb04/Regel_Hyperboloid.gif).


I heard a talk to which I didn't understand a lot ; a solution was given using Hopf fibration. I'm not familiar to these notions, and at the end it went like ``Tadaa! And here is our partition!''. The speaker could not describe what the partition looks like.


I would be very glad to: (1) understand the math he did (article, book?), (2) see what his solution looks like, and (3) know what kind of solutions exist.


Thanks in advance!



How to partition R^3 into pairwise non-parallel lines?
Take the complex lines in ℂ2, and intersect with a copy of ℝ3 not containing the origin. This gives a foliation of ℝ3 by lines, which is the projection (from the origin) of the Hopf fibration of the unit sphere in ℂ2 (which is the foliation of S3 by intersections with complex lines). 


One may easily write this down in coordinates, thinking of ℝ3 =ℝ x ℂ =(1+ iℝ) x ℂ ⊂ ℂ2. 
Then for a fixed z ∈ ℂ, the line is given by (t, (1+it) z), t ∈ ℝ. When z=0, you get a vertical axis. For |z|=r, you get a hyperboloid which is obtained by rotating the line through (0,r) about the axis. I think this is probably the foliation by lines described in the talk you attended. 


Another remark is that since you're interested in partitions rather than foliations, on each hyperboloid there is two foliations by lines (which are mirror images). So you can "flip" the foliation on each hyperboloid independently to obtain uncountably (actually 2|ℝ|) many partitions of ℝ3 into non-parallel lines. 


