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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.

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!

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That link really helps clarify. – Anton Geraschenko Oct 19 '09 at 15:58
up vote 12 down vote accepted

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.

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The central projection from the origin should not be called "stereographic": this term is usually used for central projection from a point of the sphere. A genuine stereographic projection would map all but one tori supporting the circles to tori, not to hyperboloids. – Benoît Kloeckner Jul 7 '10 at 7:51
You're right, Benoit, I changed it to just projection. This projection is really the Hopf fibration mapped to projective space by the double cover, and then restricted to $R^3$. So each line in the foliation is $1/2$ of a circle of the Hopf fibration, since $S^3\to RP^3$ is a double cover, as can also be seen from the fact that the projection is from the origin, so it only captures points in a hemisphere. – Ian Agol Jul 7 '10 at 16:26
The stereographic projections are quite interesting, see and the "external link" at the end with animated graphics. – Will Jagy Jul 7 '10 at 19:14

The video series "Dimensions"...

[Available for free download, viewing on-line, or purchase on DVD.]

Episodes 7 and 8 on fibrations contain computer graphics intended to help in visualization of such things.

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First off, I'm not sure the Hopf-fibration-based solution, whatever it is, is necessarily different from the concentric hyperboloid ones you describe. The Hopf fibration contains hyperboloids galore, when looked at in various ways, although of course I don't know if this is really relevant since I'm not sure what the specific construction is that the speaker used to build a partition from the fibration.

The Hopf fibration itself is an amazing map from S^3 to S^2 (the three dimensional and two dimensional spheres, respectively). The inverse image of each point in S^2 is a circle. Therefore, if you think of S^3 as R^3 with an added point using the standard stereographic projection, the fibers (=inverse images of points) are all circles except for one circle, the one passing through the "north pole" of the projection, which becomes a straight line.

It could be the case that by varying the north pole, those straight lines form a partition of the kind you describe (you'll need to avoid double-counting lines coming from antipodic points on S^3, but otherwise those lines are distinct). This is just a wild guess really.

[Note: this is not a complete answer to the question, but it's really hard to pack a few paragraphs with links like this into a comment, so I tend to prefer the "Answer" format - if mo-etiquette dictates otherwise, just let me know!]

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There is a solution using the axiom of choice, similar to this. Enumerate the points of space as $\{p_\alpha:\alpha<\phi\}$ where $\phi$ is the least ordinal of cardinality continuum. This means that we have a well ordering of $\mathbb{R}^3$ such that each element is preceded by less than continuum many elements. We are going to choose, by transfinite recursion on $\alpha$, a line $L_\alpha$ throu $p_\alpha$ such that these lines are not parallel and cover the space. Actually, for some $\alpha$ we do not choose $L_\alpha$.

Assume that we have made the choices for all $\beta<\alpha$ and we have to treat $p_\alpha$. We do nothing, if some earlier $L_\beta$ covers $p_\alpha$. Otherwise, draw a sphere $S$ around $p_\alpha$. We have to select a line $L_\alpha$ throu $p_\alpha$ and this is the same as to choose a point of $S$. Those points are disqualified which give rise to lines parallel to some earlier lines, this gives a set of less than continuum points on $S$. Further, to any given line $L_\beta$ ($\beta<\alpha$) we cannot choose any line that intersects $L_\beta$. The directions corresponding to these bad choices form a great circle on $S$. We have, therefore, a collection of less than continuum many points and great circles on $S$. An easy argument shows that they cannot cover $S$ [choose a great circle $C$ different from them, then the pints/great circles can cover at most two points of $C$ each], so we can make the choice of the diection of $L_\alpha$.

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If I understand correctly, your proof also works to cover by disjoint, non parallel lines the set $\mathbb{R}^3 \setminus \{0\}$, or more generally, to cover the complement of any set $F$ of less than continuum points of $\mathbb{R}^3$. We just have to include among the disqualified points in $S$ those that would connect $p_\alpha$ to a forbidden point. Am I correct? – Pietro Majer Sep 28 '11 at 13:44

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