Billiards in polyhedra: a method
to convert 2-dimensional uniformity
to 3-dimensional uniformity

For rational polygons where every angle is a rational multiple of $\pi$, we have the following fundamental result of Kerckhoff, Masur and Smillie [4] in 1986.

Given any initial direction, a point $\mathbf{p}_{0}\in P$ is called a pathological starting point for this direction if the half-infinite billiard orbit starting from $\mathbf{p}_{0}$ and with this direction hits a singularity of $P$. Otherwise the point $\mathbf{p}_{0}$ is called a non-pathological starting point for this direction. It is easy to see that for any given direction, almost every point in $P$ is a non-pathological starting point.

The proof of Theorem A consists of essentially three steps. The first step is to establish the ergodicity of the corresponding interval exchange transformation. The second step is to use the well known Birkhoff ergodic theorem. The final step is to extend ergodicity to unique ergodicity.

Our aim is to convert Theorem A to a result concerning equidistribution of $3$-dimensional billiard in some polyhedra. However, we need to restrict our discussion to rational polygonal right prisms. A rational polygonal right prism is a region in $3$-dimensional cartesian space of the form

where $P$ is a rational polygon and $I=[0,z_{0}]$ is an interval.

As the rational polygonal right prism $M=P\times I$ is integrable in the direction of the interval $I$, our extension is somewhat limited.

For illustration, we consider a special case where $P$ is a right triangle. It is well known that the right triangle billiard with angle $\pi/4$ and the right triangle billiard with angle $\pi/6$ are the only right triangle billiards that are integrable, exhibiting stable and predictable behaviour. Perhaps the simplest non-integrable billiard is the right triangle billiard with angle $\pi/8$. It is also well known that unfolding in the spirit of König and Szücs [5] leads to a $16$-fold covering of the triangle and shows that this billiard is equivalent to geodesic flow on the regular octagon surface $\mathcal{P}$ where parallel edges are identified in pairs. On the other hand, unfolding also leads to a $2$-fold covering of the interval $I=[0,z_{0}]$. Thus billiard in the rational polygonal right prism $M=P\times I$, where $P$ is the right triangle with angle $\pi/8$ and $I=[0,z_{0}]$, is equivalent to geodesic flow in the translation $3$-manifold $\mathcal{M}=\mathcal{P}\times\mathcal{I}$, where $\mathcal{P}$ is the regular octagon translation surface and $\mathcal{I}=[0,2z_{0}]$, treated as a torus. Figure 1 illustrates that $\mathcal{M}=\mathcal{P}\times\mathcal{I}$ gives a $32$-fold covering of the rational polygonal right prism $M=P\times I$. It has $2$ octagonal faces which are identified with each other, and $8$ rectangular faces, with pairs of parallel ones identified with each other, analogous to the edge identification of the regular octagon translation surface $\mathcal{P}$.

Suppose that a half-infinite geodesic

in $\mathcal{M}$ has a non-pathological starting point $\mathcal{L}_{S_{0},\mathbf{v}}(0)=S_{0}=(s_{1},s_{2},s_{3})$, and direction given by the unit vector

with arc-length parametrization. The coordinates $(s_{1}+v_{1}t,s_{2}+v_{2}t)$ are modulo $\mathcal{P}$ and the coordinate $s_{3}+v_{3}t$ is modulo $\mathcal{I}$.

We may assume without loss of generality that $v_{3}>0$. Then the geodesic $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, hits the octagon face of $\mathcal{M}$ for the very first time at time $t=t_{0}$, where $s_{3}+v_{3}t_{0}=2z_{0}$, so that $t_{0}=(2z_{0}-s_{3})/v_{3}$. Indeed, the geodesic hits the octagon face of $\mathcal{M}$ for the $(k+1)$-th time at time $t=t_{k}$, where

with parameters xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

This gives rise to an arithmetic progression

with common gap $\theta$ between consecutive terms.

We need the following result on arithmetic progressions; see [1, Theorem 2.2.2].

We have following immediate consequence.

Meanwhile, every point in $\mathcal{M}$ is of the form $(x,y,z)$, where $(x,y)\in\mathcal{P}$ and $z\in\mathcal{I}$. We consider the projection

Then the image of the geodesic $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, under this projection is a half-infinite geodesic xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

on the regular octagon translation surface $\mathcal{P}$. Clearly the key parameters $s_{3}$ and $v_{3}$, particularly concerning the hitting times given in (2.1), are lost under this projection (2.3). However, we know the arithmetic progression (2.1) of the time instances when the geodesic $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, hits the octagon face of $\mathcal{M}$. This gives rise to an infinite sequence of points $\mathcal{H}_{S_{0},\mathbf{v}}(t_{k})$, $k=0,1,2,3,\ldots,$ on $\mathcal{P}$. For any $S_{0}$ and $\mathbf{v}$, if we can show that this sequence of points is uniformly distributed on $\mathcal{P}$, then the half-infinite geodesic $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, is uniformly distributed in $\mathcal{M}$.

Consider a typical half-infinite geodesic $\mathcal{H}_{\mathbf{w}}(\tau)$, $\tau\geqslant 0$, on $\mathcal{P}$, with direction given by the unit vector $\mathbf{w}=(w_{1},w_{2})\in\mathbb{R}^{2}$. Suppose that this geodesic is the image on $\mathcal{P}$ of $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, under the projection (2.3). Then

In view of the different parametrizations of (2.4) and (2.5), we have

Corresponding to the arithmetic progression $t_{k}$, $k=0,1,2,3,\ldots,$ of hitting times given by (2.1) is the arithmetic progression

with parameters xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

By the geodesic analogue of Theorem A, for almost every direction $\mathbf{w}$ and for every non-pathological starting point for this direction, the geodesic $\mathcal{H}_{\mathbf{w}}(\tau)$, $\tau\geqslant 0$, is uniformly distributed on $\mathcal{P}$. Let $\operatorname{RP}$ denote an arbitrary polygon on $\mathcal{P}$ where all the vertices have rational coordinates, and let the measurable set

denote the set of time instances when this geodesic visits $\operatorname{RP}$. The uniformity of the geodesic then implies that $\mathcal{S}$ has asymptotic density

The uniformly distributed geodesic $\mathcal{H}_{\mathbf{w}}(\tau)$, $\tau\geqslant 0$, is clearly the image on $\mathcal{P}$ under the projection (2.3) of infinitely many different geodesics $\mathcal{L}_{S_{0},\mathbf{v}}(t)$, $t\geqslant 0$, in the $3$-manifold $\mathcal{M}$, as there are only two requirements, namely $\mathcal{H}_{\mathbf{w}}(0)=(s_{1},s_{2})$ concerning the starting point, where $S_{0}=(s_{1},s_{2},s_{3})$, and $w_{1}v_{2}=w_{2}v_{1}$ concerning equality of the relevant directions. Applying Lemma 2.2, we see that for every $\ell\in\mathbb{Z}$ and for almost every pair $\theta,\lambda$ of the form (2.6) satisfying $2^{\ell}\leqslant\theta<2^{\ell+1}$ and $0\leqslant\lambda<\theta$, the inequality (2.2) with $\mathcal{S}$ given by (2.7) holds for every sufficiently large positive integer $n$.

This means that for almost every pair of starting point $S_{0}$ and unit direction vector $\mathbf{v}$, the infinite sequence

of points, where the sequence $t_{k}$, $k=0,1,2,3,\ldots,$ of time instances is given by (2.1), is uniformly distributed on $\mathcal{P}$ relative to the single test set $\operatorname{RP}$.

The set of all polygons $\operatorname{RP}$ on $\mathcal{P}$ where all the vertices have rational coordinates is countable. On the other hand, a countable union of sets of measure zero has measure zero. It follows that for almost every pair of starting point $S_{0}$ and unit direction vector $\mathbf{v}$, the infinite sequence (2.8) of points is uniformly distributed on $\mathcal{P}$ relative to every polygon $\operatorname{RP}$ on $\mathcal{P}$ where all the vertices have rational coordinates. This guarantees uniformity in general, in the classical Weyl sense, and completes the proof of Theorem 2.

Throughout the proof, the set $\mathcal{S}\subset\mathbb{R}$ is measurable.

Let a non-negative integer $\ell$ be chosen and fixed.

Consider an infinite sequence $N_{1},N_{2},N_{3},\ldots$ of positive integers satisfying

and another infinite sequence $M_{1},M_{2},M_{3},\ldots$ of positive integers satisfying

both to be specified later in terms of the parameter $h$ and the chosen integer $\ell$. For every positive integer $h$, let $S(h)\subset[0,1]$ denote the contraction of $\mathcal{S}\cap[0,N_{h}]$ to the unit interval, so that

Since the characteristic function

defined over $[0,1]$ and extended periodically over the whole real line with period $1$, is measurable, we can consider its Fourier series

with Fourier coefficients $a_{j}$, $j\in\mathbb{Z}$. In particular,

The Parseval formula gives

so that

Lemma 2.1 for the chosen value $\ell$ concerns the arithmetic progression $k\theta+\eta$, $k\geqslant 0$, where $2^{\ell}\leqslant\theta<2^{\ell+1}$ and $0\leqslant\eta<\theta$. The contraction (3.1) leads to a new arithmetic progression $k\alpha+\beta$, $k\geqslant 0$, where $2^{\ell}/N_{h}\leqslant\alpha<2^{\ell+1}/N_{h}$ and $0\leqslant\beta<\alpha$.

For any $\alpha\in[2^{\ell}/N_{h},2^{\ell+1}/N_{h})$, let $K(\alpha)$ be the unique integer satisfying

Using the Fourier series (3.2), we have

for every $\alpha$ and $\beta$ satisfying $2^{\ell}/N_{h}\leqslant\alpha<2^{\ell+1}/N_{h}$ and $0\leqslant\beta<\alpha$.

To study (3.5), we consider the integral

To obtain a bound on this integral, we observe that for $\gamma\in[-2^{\ell-1}/M_{h},2^{\ell-1}/M_{h}]$, the inequality $2(1-|\gamma|M_{h}/2^{\ell})\geqslant 1$ holds. It then follows that

where for integers $N$ and $M$ satisfying $1<M<N$,

At the end of this section, we establish the following bound for this integral.

Combining (3.6), (3.7) and (3.9), we deduce that

The inequality (3.10) is a quadratic average result, from which we can derive information concerning the majority of pairs

Let xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

denote the collection of pairs $(\alpha,\gamma)$ satisfying (3.11) such that

Then it follows from (3.10) that

where $\lambda_{2}$ denotes $2$-dimensional Lebesgue measure.

Next, note that as we move from (3.5) to (3.6), we replace the parameter $\beta$ over a short interval $[0,\alpha)$ by a parameter $\gamma\in[-2^{\ell-1}/M_{h},2^{\ell-1}/M_{h}]$ over a longer interval. For any pair xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

there are at least $N_{h}/2M_{h}$ values of $\gamma\in[-2^{\ell-1}/M_{h},2^{\ell-1}/M_{h}]$ where $\{\gamma/\alpha\}\alpha=\beta$. For each of these values of $\gamma$, consider the two arithmetic progressions

and xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Since xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

the arithmetic progression (3.15) is obtained by simply advancing the arithmetic progression (3.16) by $[\gamma/\alpha]$ terms. More precisely, the arithmetic progression (3.15) is given by

Let xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Then the above argument leads to the inequality

Combining (3.13) and (3.20), we obtain the upper bound

Combining this with (3.5), it is not difficult to see that apart from a set of measure $\lambda_{2}(A(N_{h};M_{h}))$, every pair $(\alpha,\beta)$ such that (3.14) holds satisfies the inequality

Next, note that the two expressions

differ by at most $2$, due to the possibility that $(K(\alpha)-1)\alpha+\beta>1$ and the difference $|K(\alpha)-1/\alpha|<1$, in view of (3.4). It follows that on reversing the contraction, we see that apart from a set of measure at most

every pair $(\theta,\eta)$ such that $2^{\ell}\leqslant\theta\leqslant 2^{\ell+1}$ and $\eta\in[0,\theta)$ satisfies the inequality

Since xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

we conclude that for almost every pair $(\theta,\eta)$ such that $2^{\ell}\leqslant\theta\leqslant 2^{\ell+1}$ and $\eta\in[0,\theta)$, the inequality (3.21) holds for all sufficiently large positive integers $h$.

Finally, we specify the integers $N_{h}$ and $M_{h}$ in terms of the parameter $h\geqslant 1$ and the chosen integer $\ell$. Choosing them to satisfy

ensures that the two dominant terms on the right hand side of (3.21) have the same order of magnitude in terms of $h$. For an arbitrary sufficiently integer $n$, we choose $h$ to satisfy $N_{h}\leqslant n<N_{h+1}$. Then it follows from (3.21) and (3.22) that

provided that $n$, and hence also $h$, is sufficiently large.

This completes the proof of Lemma 2.1.