When T is an irrational rotation, and are Bernoulli: explicit isomorphisms

Abstract

Let $\theta $ be an irrational real number. The map $T_\theta : y \mapsto (y+\theta ) \,\mod \!\!\: 1$ from the unit interval $\mathbf {I} = [0,1[$ (endowed with the Lebesgue measure) to itself is ergodic. In 2002, Rudolph and Hoffman showed in [Uniform endomorphisms which are isomorphic to a Bernoulli shift. Ann. of Math. (2) 156(1) (2002), 79–101] that the measure-preserving map $[T_\theta ,\mathrm {Id}]$ is isomorphic to a one-sided dyadic Bernoulli shift. Their proof is not constructive. A few years before, Parry [Automorphisms of the Bernoulli endomorphism and a class of skew-products. Ergod. Th. & Dynam. Sys.16 (1996), 519–529] had provided an explicit isomorphism under the assumption that $\theta $ is extremely well approached by the rational numbers, namely, $$ \begin{align*}\inf_{q \ge 1} q^44^{q^2}~\mathrm{dist}(\theta,q^{-1}\mathbb{Z}) = 0.\end{align*} $$ Whether the explicit map considered by Parry is an isomorphism or not in the general case was still an open question. In Leuridan [Bernoulliness of $[T,\mathrm {Id}]$ when T is an irrational rotation: towards an explicit isomorphism. Ergod. Th. & Dynam. Sys.41(7) (2021), 2110–2135] we relaxed Parry’s condition into $$ \begin{align*}\inf_{q \ge 1} q^4~\mathrm{dist}(\theta,q^{-1}\mathbb{Z}) = 0.\end{align*} $$ In the present paper, we remove the condition by showing that the explicit map considered by Parry is always an isomorphism. With a few adaptations, the same method works with $[T,T^{-1}]$ .