How to partition R^3 into pairwise non-parallel lines? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T01:05:48Z http://mathoverflow.net/feeds/question/1194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines How to partition R^3 into pairwise non-parallel lines? subshift 2009-10-19T10:18:26Z 2010-07-07T16:22:04Z <p><strong>Problem.</strong> How to partition R^3 into pairwise non-parallel lines?</p> <p>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 <a href="http://www-lm.ma.tum.de/archiv/sos04/la2lb04/Regel%5FHyperboloid.gif" rel="nofollow">looks like this</a>.</p> <p>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.</p> <p>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.</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines/1214#1214 Answer by Gerald Edgar for How to partition R^3 into pairwise non-parallel lines? Gerald Edgar 2009-10-19T13:25:52Z 2009-10-19T13:25:52Z <p>The video series "Dimensions"... <a href="http://www.dimensions-math.org/" rel="nofollow">http://www.dimensions-math.org/</a> </p> <p>[Available for free download, viewing on-line, or purchase on DVD.]</p> <p>Episodes 7 and 8 on fibrations contain computer graphics intended to help in visualization of such things.</p> http://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines/1285#1285 Answer by Alon Amit for How to partition R^3 into pairwise non-parallel lines? Alon Amit 2009-10-19T20:58:12Z 2009-10-19T20:58:12Z <p>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.</p> <p>The <a href="http://en.wikipedia.org/wiki/Hopf%5Ffibration" rel="nofollow">Hopf fibration</a> 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. </p> <p>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.</p> <p><em>[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!]</em></p> http://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines/3714#3714 Answer by Agol for How to partition R^3 into pairwise non-parallel lines? Agol 2009-11-01T22:56:55Z 2010-07-07T16:22:04Z <p>Take the complex lines in &#8450;<sup>2</sup>, and intersect with a copy of &#8477;<sup>3</sup> not containing the origin. This gives a foliation of &#8477;<sup>3</sup> by lines, which is the projection (from the origin) of the Hopf fibration of the unit sphere in &#8450;<sup>2</sup> (which is the foliation of S<sup>3</sup> by intersections with complex lines). </p> <p>One may easily write this down in coordinates, thinking of &#8477;<sup>3</sup> =&#8477; x &#8450; =(1+ i&#8477;) x &#8450; &sub; &#8450;<sup>2</sup>. Then for a fixed z &isin; &#8450;, the line is given by (t, (1+it) z), t &isin; &#8477;. 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. </p> <p>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<sup>|&#8477;|</sup>) many partitions of &#8477;<sup>3</sup> into non-parallel lines. </p> http://mathoverflow.net/questions/1194/how-to-partition-r3-into-pairwise-non-parallel-lines/30861#30861 Answer by Péter Komjáth for How to partition R^3 into pairwise non-parallel lines? Péter Komjáth 2010-07-07T06:21:41Z 2010-07-07T06:21:41Z <p>There is a solution using the axiom of choice, similar to <a href="http://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles/28650#28650" rel="nofollow">this</a>. Enumerate the points of space as <code>$\{p_\alpha:\alpha&lt;\phi\}$</code> 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 $L_\alpha$. </p> <p>Assume that we have made the choices for all $\beta&lt;\alpha$ 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 $L_\beta$ ($\beta&lt;\alpha$) 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 $L_\alpha$.</p>