7
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ That link really helps clarify. $\endgroup$ – Anton Geraschenko Oct 19 '09 at 15:58
  • 1
    $\begingroup$ I think that Gluck and Warner, Great circle fibrations of the 3-sphere, worked out the complete description of all foliations of 3-dimensional Eucldean space by pairwise nonparallel lines; but it might not be quite the same problem. $\endgroup$ – Ben McKay Aug 9 at 17:41
18
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Benoît Kloeckner Jul 7 '10 at 7:51
  • $\begingroup$ 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. $\endgroup$ – Ian Agol Jul 7 '10 at 16:26
  • $\begingroup$ The stereographic projections are quite interesting, see en.wikipedia.org/wiki/Hopf_fibration#Geometry_and_applications and the "external link" at the end with animated graphics. $\endgroup$ – Will Jagy Jul 7 '10 at 19:14
9
$\begingroup$

The video series "Dimensions"... http://www.dimensions-math.org/

[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.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

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!]

| cite | improve this answer | |
$\endgroup$
6
$\begingroup$

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 $L_\alpha$ through $p_\alpha$ 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 direction of $L_\alpha$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 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? $\endgroup$ – Pietro Majer Sep 28 '11 at 13:44
  • $\begingroup$ See also the answer to mathoverflow.net/questions/92919/… $\endgroup$ – Dale Aug 9 at 14:43
2
$\begingroup$

Each foliation of 3-dimensional Euclidean space by pairwise nonparallel lines extends to a foliation of real projective 3-space by nonintersecting projective lines. Hence lifts to a foliation of the 3-sphere by nonintersecting great circles. So the complete classification of all foliations of 3-dimensional Euclidean space by pairwise nonparallel lines is the classification of great circle fibrations of the 3-sphere, which is worked out very clearly in Gluck, Warner, Great circle fibrations of the 3-sphere, Duke Math. J., Volume 50, Number 1 (1983), 107-132.

To give some idea of the method: each great circle spans an unique oriented 2-plane through the origin of $\mathbb{R}^4$, which is spanned by an oriented orthonormal basis $u,v$. The 2-vector $\zeta=u\wedge v$ is uniquely determined. It splits into a self-dual and an anti-self-dual part, say $\zeta_+,\zeta_-$, each of the same length, which we rescale to be unit length.

For a great circle fibration, every choice of $\zeta_-$ occurs uniquely, i.e. for a unique circle fiber, so $\zeta_+$ can be written as a function of $\zeta_-$. This functions turns out to be smooth, a smooth map from the unit sphere in the anti-self-dual 2-vectors to the unit sphere in the self-dual. This map is strictly contracting in the usual metric on those spheres. Moreover, any strictly contracting map arises in this way. All of this is proven in great detail in the paper.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The Hopf fibration example explained above is just one in an infinite dimension modulo space. $\endgroup$ – Ben McKay Aug 9 at 17:47
  • 1
    $\begingroup$ The Hopf fibration corresponds to a constant map $\zeta_+=$constant. The choice of constant is a choice of self-dual symplectic form turning $\mathbb{R}^4$ into a complex vector space. $\endgroup$ – Ben McKay Aug 9 at 18:00
  • $\begingroup$ Forgive the silly question—what is the duality with respect to which $\zeta_+$ and $\zeta_-$ are (anti-)self-dual? $\endgroup$ – LSpice Aug 9 at 18:14
  • $\begingroup$ @LSpice: In a 4-dimensional real vector space with orientation and positive definite inner product, the Hodge star operator (of the Euclidean metric) acting on translation invariant 2-forms has a 3-dimensional (-1)-eigenspace, and a 3-dimensional (+1)-eigenspace. These are the anti-self-dual and self-dual translation invariant 2-forms. $\endgroup$ – Ben McKay Aug 9 at 20:56
1
$\begingroup$

Treating the problem more geometrically: There is a well-known partition of $\mathbb{R}^3$ into nested one-sheet hyperboloids of revolution, along with their common axis of symmetry; https://en.wikipedia.org/wiki/Skew_lines has an image. Using the $z$ axis as the axis of symmetry, the equation of a hyperboloid of revolution is $\frac{x^2+y^2}{a^2} - \frac{z^2}{c^2} = 1$ with $a \neq 0, c \neq 0$.

This is a family with two parameters and almost all pairs of members intersect. To obtain a family that partitions $\mathbb{R}^3$, fix a relationship between $a$ and $c$ so that every choice of $x^2+y^2 \neq 0$ and $z^2$ admits exactly one choice of $a$ and $c$, thus reducing the family to one parameter.

Each hyperboloid can be ruled in two ways. All of the lines of a ruling have the same slope relative to the $z$ axis, which is $\pm a/c$, with $+$ for one ruling and $-$ for the other. Clearly, all of the lines in a ruling are mutually skew, and to ensure the lines in different hyperboloids are mutually skew, the chosen relationship between $a$ and $c$ must not admit two solutions with the same $a/c$. A trivial choice is to fix $c=1$.

With the relationship between $a$ and $c$ fixed, there is an additional (uncountable) choice of which ruling of each hyperboloid.

This family of solutions to the problem can be extended by noting that all of the required properties are preserved under affine transformations of $\mathbb{R}^3$, as well as preserving the component hyperboloids as hyperboloids (but not necessarily as hyperboloids of revolution).

Projective geometry allows further extension: Skewness of lines is preserved by projective transformations. (In projective geometry, parallel lines become lines whose intersection is a point at infinity, so the union of parallel lines and intersecting lines is projectively invariant, despite that neither set alone is.) The extension of the above construction fills all of projective 3-space except the "equatorial" line which is the intersection of the $z=0$ plane of symmetry with the plane at infinity, because none of the lines are parallel to the $x=y=0$ plane. So the equatorial line must be added to the partition, providing a dual of some sort to the $z$ axis.

Any projective transformation can be applied to the extended construction, followed by restriction to $\mathbb{R}^3$, leaving a partition into skew lines (and possibly one transformed line is contained in the plane at infinity). In particular a projective transformation can be applied which maps the equatorial line into a line in $\mathbb{R}^3$ while mapping the $z$ axis into a line still in $\mathbb{R}^3$. (My memory is that given two pairs of skew lines, there is a projective transformation that maps the first pair into the second pair.)

This constructs a solution that is difficult to visualize. It has two skew "axis" lines, one mapped from the $z$ axis and one mapped from the equatorial line (the line at infinity with $z=0$). The hyperboloids are mapped into quadratic surfaces (because projective transformations preserver quadratics). Each axis is surrounded by nested one-sheet hyperboloids. I suspect that both families, going outward from their axis, expand more quickly on the side away from the other axis, and the families almost partition $\mathbb{R}^3$, with the limiting surface between them being a hyperbolic paraboloid.

I'm sure this was worked out in the 19th century, and I have a vague memory I've seen an image of this structure. Does anyone know a reference?

I suspect that looking at projective 3-space using homogeneous coordinates (which is more or less $\mathbb{R}^4$ modulo scalar multiplication, intersected with a 3-plane that does not contain the origin) is isomorphic to some of the above constructions using $\mathbb{C}^2$ intersected with $\mathbb{R}^3$.

| cite | improve this answer | |
$\endgroup$

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.