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Let $\operatorname{frac}(x) = x - \lfloor x \rfloor$ be the fractional part of $x$. Then, for $\alpha$ irrational, $\operatorname{frac}(n \alpha)$, $n=1,2,\ldots$, distributes randomly in $[0,1)$, which can be viewed as the circle $\mathbb{S}^1$. Weyl's Equidistribution Theorem establishes the uniformity of the distribution.
           WeylCircle
           $\operatorname{frac}(n \pi)$ for $n=1,2,\ldots,100$.

Is there an analog to $\operatorname{frac}(n \alpha)$ for $\mathbb{S}^2$? Is there a function $f(n,x)$ with the behavior that, for some point $x \in \mathbb{S}^2$, $f(n,x)$ for $n=1,2,\ldots$ fills $\mathbb{S}^2$ randomly and uniformly?

A pointer would suffice if this is well known. Thanks!

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  • $\begingroup$ One natural analog is the answer to this previous question mathoverflow.net/questions/63574/… $\endgroup$ May 14, 2012 at 13:24
  • $\begingroup$ @Gjergji: Yes, that's what got me thinking in this direction. But I thought there might be a more direct analog. Perhaps a vain hope... $\endgroup$ May 14, 2012 at 13:34
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    $\begingroup$ What would happen if one took a pair of irrationals, $(\alpha_1 ,\alpha_2)$ acted on a torus, and then took an orbit. Then pick a two meridians of the torus, collapse them to a point, then took the image of an orbit in the resulting 2-sphere? $\endgroup$ May 15, 2012 at 3:35

2 Answers 2

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If you take two random elements in $SO(3),$ (this is sort of like irrationals in $SO(2)$) the group they generate is both free and equidistributed in $SO(3),$ so the orbit of the north pole (or any other point you favor) will be equidistributed in $\mathbb{S}^2.$ You may complain that a free group will need two indices, but since free groups are orderable, you can make the two indices into one.

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  • $\begingroup$ Nice, Igor---Thanks! [And I would never so complain. :-)] $\endgroup$ May 14, 2012 at 15:15
  • $\begingroup$ This may reveal my ignorance, but is not the composition of two rotations itself a rotation, so that there will be nonuniform clustering near the axis of that rotation? $\endgroup$ May 14, 2012 at 17:31
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    $\begingroup$ The composition of two rotations is a rotation, but different words in the group will have different axes... $\endgroup$
    – Igor Rivin
    May 14, 2012 at 17:34
  • $\begingroup$ Equivalently you can take $a\mathbb{Z}+b\mathbb{Z}$ dense in the torus $\mathbb{C}/\mathbb{Z}[i]$ and let $\mathbb{S}^2 = \mathbb{C}/\mathbb{Z}[i]/(z = -z)$. $\endgroup$
    – reuns
    Sep 9, 2017 at 19:33
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Kuipers and Niederreiter, Uniform Distribution of Sequences, give these references for questions of uniform distribution of sequences on a sphere:

V I Arnol'd, A L Krylov, Uniform distribution of points on a sphere and some ergodic properties of solutions of linear ordinary differential equations in a complex region, Dokl Akad Nauk SSSR 148 (1963) 9-12; English translation, Soviet Math Dokl 4 (1963) 1-5.

P Gerl, Gleichverteilung auf der Kugel, Arch der Math 24 (1973) 203-207.

Those of us whose familiarity with kugel extends only to the side dish commonly served on the Shabbes and other holidays (http://en.wikipedia.org/wiki/Kugel) may find some amusement in the title of Gerl's paper.

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