An endomorphism whose square is Bernoulli

Abstract

One of the corollaries of Ornstein's isomorphism theorem is that if $(Y,S,\nu)$ is an invertible measure-preserving transformation and $(Y,S^2,\nu)$ is isomorphic to a Bernoulli shift, then $(Y,S,\nu)$ is isomorphic to a Bernoulli shift. In this paper we show that non-invertible transformations do not share this property. We do this by exhibiting a uniformly two-to-one endomorphism $(X,\sigma,\mu)$ which is not isomorphic to the one-sided Bernoulli two shift. However, $(X,T^2,\mu)$ is isomorphic to the one-sided Bernoulli four shift.