Proof that Polygonal Billiards Have Zero Entropy

Question

I'm reading this paper "Billiards In Polygons" by Boldrighini et al. They say that polygonal billiards have zero measure-theoretic entropy, because a given element of the configuration space is almost surely determined by its forward hit-sequence(the sequence of sides it collides with). Why does being so determined imply zero entropy?

Answer 1

First, the billiard flow is a special flow over the shift $\sigma$ determined by the hit sides, hence by Abramov the entropy of the flow is proportional to that of the shift. Next, this shift has zero entropy because the fact that you state is equivalent to the shift having a one-sided generator: if $\alpha$ is the partition determined by the cylinder sets at the zeroth coordinate, then the ambient sigma-algebra $\mathcal{B}$ is up to a measure zero set equal to the sub-sigma-algebra corresponding to the "past partition"

$$\alpha_1^\infty=\prod_{k\geq1}\sigma^{-k}(\alpha)$$

(the product signifies the coarsest common refiner, often denoted by $\bigvee$). Thus $\alpha$ is a (two-sided) generator too, whence the entropy of $\sigma$ is the entropy of $\alpha$ conditioned on $\alpha_1^\infty=_{æ}\mathcal{B}$, which is zero. See e.g. Petersen's Ergodic Theory, pp.244-245 for more details.