$\mathbb{S}^2$ equivalent to frac$(n \alpha)$ equidistribution on $\mathbb{S}^1$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T16:00:41Z http://mathoverflow.net/feeds/question/96890 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96890/mathbbs2-equivalent-to-fracn-alpha-equidistribution-on-mathbbs1 $\mathbb{S}^2$ equivalent to frac$(n \alpha)$ equidistribution on $\mathbb{S}^1$ Joseph O'Rourke 2012-05-14T12:48:01Z 2012-05-14T23:14:58Z <p>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 $\mathbb{S}^1$. <a href="http://en.wikipedia.org/wiki/Equidistribution_theorem" rel="nofollow">Weyl's Equidistribution Theorem</a> establishes the uniformity of the distribution. <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://cs.smith.edu/~orourke/MathOverflow/WeylCircle.jpg" alt="WeylCircle" /> <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sub>$\operatorname{frac}(n \pi)$ for $n=1,2,\ldots,100$.</sub> <br /></p> <blockquote> <p>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?</p> </blockquote> <p>A pointer would suffice if this is well known. Thanks!</p> http://mathoverflow.net/questions/96890/mathbbs2-equivalent-to-fracn-alpha-equidistribution-on-mathbbs1/96900#96900 Answer by Igor Rivin for $\mathbb{S}^2$ equivalent to frac$(n \alpha)$ equidistribution on $\mathbb{S}^1$ Igor Rivin 2012-05-14T14:08:01Z 2012-05-14T14:15:21Z <p>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.</p> http://mathoverflow.net/questions/96890/mathbbs2-equivalent-to-fracn-alpha-equidistribution-on-mathbbs1/96953#96953 Answer by Gerry Myerson for $\mathbb{S}^2$ equivalent to frac$(n \alpha)$ equidistribution on $\mathbb{S}^1$ Gerry Myerson 2012-05-14T23:14:58Z 2012-05-14T23:14:58Z <p>Kuipers and Niederreiter, Uniform Distribution of Sequences, give these references for questions of uniform distribution of sequences on a sphere: </p> <p>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> <p>P Gerl, Gleichverteilung auf der Kugel, Arch der Math 24 (1973) 203-207. </p> <p>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. </p>