MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A ball rolls down an inclined plane, encountering horizontal obstacles, at which it rolls left/right with equal probability. There are regularly spaced staggered gaps that let the ball roll down to the next, lower obstacle. The pattern resembles a binary tree:
Rolling Obstacles
Suppose the vertical and horizontal rolls have equal length $\delta$. Tracing out the roll contact point on the ball surface we see a random walk, with each step a geodesic arc of length $\delta$, and $90^\circ$ turns. I expected that for rational (multiples of $\pi$) $\delta$, the trace would not fill the surface, but the experiment below for $\delta=\pi/16$ (for 10, $10^2$, $10^3$, $10^4$ downhill steps) indicates otherwise.

For which $\delta$ will this trace fill the sphere surface?

          Sphere Paths
Thanks for any insights!

Answer: The surface will be filled for every $\delta$ except $\pi/2$ and $\pi$. See Scott Carnahan's answer below, and Dylan Thurston's simplification. I find this answer remarkable!

share|cite|improve this question
There was a related problem about the density of a group generated by two spherical rotations, both of angular measure pi/8. I thought the group was finite, but turned out not to be. I suspect Bill Thurston's answer there may be relevant here. If I recall the link I will post it. Gerhard "Too Many Comments To Remember" Paseman, 2011.04.30 – Gerhard Paseman May 1 '11 at 2:35
Here it is: "Maneuvering with limited moves on $S^2$, . You are right, Gerhard, it does seem related... – Joseph O'Rourke May 1 '11 at 3:09
Given that the finite subgroups of $SO(3)$ follow the ADE classification, it shouldn't be too surprising that you are restricted to a very small class of admissible rotations. – S. Carnahan May 1 '11 at 5:15
I would just like to take this opportunity to say I absolutely love your questions. – B. Bischof May 2 '11 at 19:31
@BB: I am truly honored! :-) – Joseph O'Rourke May 2 '11 at 21:23
up vote 44 down vote accepted

Let $A = \begin{pmatrix} \cos \delta & -\sin \delta & 0 \\ \sin \delta & \cos \delta & 0 \\ 0 & 0 & 1 \end{pmatrix}$, and let $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \delta & -\sin \delta \\ 0 & \sin \delta & \cos \delta \end{pmatrix}$ be rotation by $\delta$ along the $z$ and $x$ axes, respectively. In suitable coordinates, a progression down one step in the tree is either $AB$ or $AB^{-1}$.

The trace will fill (a.s.) a dense subset of the surface if and only if the closure of the group generated by $AB$ and $AB^{-1}$ is not a subgroup of $SO(3)$ of dimension zero or one.

The dimension zero closed subgroups of $SO(3)$ are either cyclic, dihedral, or symmetries of Platonic solids, and the dimension one closed subgroups are conjugates of $SO(2)$ and $O(2)$. Therefore, it suffices to determine which values of $\delta$ yield a pair of elements in either a conjugate of $O(2)$, or a conjugate of one of the three Platonic groups (isomorphic to $A_4$, $S_4$, and $A_5$).

In order for $AB$ and $AB^{-1}$ to both lie in a conjugate of $O(2)$ it is necessary and sufficient that they have a common eigenvector with eigenvalue $\pm 1$ - this eigenvector is the axis of rotation. Writing this requirement explicitly yields a polynomial identity in $\sin \delta$ and $\cos \delta$ (whose solutions I haven't enumerated yet). Edit: Some straightforward case elimination with the $z$ coordinate of a common eigenvector shows that $\delta$ must be an integer multiple of $\pi/2$.

For the Platonic solutions, we can narrow down the solution set using the criterion that the rotation $(AB^{-1})^{-1}(AB) = B^2$ lies in the group, and Platonic solids have rotational symmetries of order at most 5. This means $\delta$ is a multiple of $\pi/3$, $\pi/4$ or $\pi/5$.

Since the traces of $AB$ and $AB^{-1}$ are both $(\cos \delta)(2 + \cos \delta)$, we can compare with character table entries to see if that number is the trace of an element in a Platonic group. It was pretty easy to eliminate candidates by eyeball in SAGE.

Conclusion: The only values of $\delta$ where the image is not dense are $0$, $\pm \pi/2$, and $\pi$.

share|cite|improve this answer
Beautifully clear reasoning, and a remarkable conclusion! – Joseph O'Rourke May 1 '11 at 11:08
You can make the calculations slightly easier by cutting the step size in two. Let A be the matrix you wrote down, and let C by rotation by π/2 around the y axis. Then you can alternately take the steps to be AC and AC−1. The trace of AC is then just $cos\delta$. If we have a Platonic solid group, this must be 1+2cosα for α∈{±2π/3,±π/2,±2π/5,±4pi/5}. (Every element in a Platonic group is a rotation by one of these angles.) 2δ is also in this set, and it's easy to find the solutions by hand, using $\cos\delta=(\cos\alpha-1)/2$. – Dylan Thurston May 1 '11 at 11:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.