Let $T$ be an arbitrary Lebesgue measure-preserving automorphism of the unit interval $I$. Let $R_{\alpha}$ denote rotation by $\alpha$, i.e. $R_{\alpha}(x)=x+\alpha \pmod{1}$ for $x \in I$ and $\alpha \in \mathbb{R}$. Is it true that the composition $R_{\alpha} \circ T$ is ergodic for (Lebesgue) almost every $\alpha \in \mathbb{R}$?

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    $\begingroup$ Quite a nice question - can you say something about the motivation? $\endgroup$ Jan 27, 2014 at 2:15

1 Answer 1


No, for trivial reasons. Take as your $T$ the map $x\mapsto-x\mod 1$. Than the map $x\mapsto\alpha-x\mod 1$ is a reflection around $\alpha/2$, and is never ergodic.

  • $\begingroup$ Thanks! Good point! In arxiv.org/abs/1210.5013, the authors prove that the answer to my question is "yes" in the case that $T$ is an interval exchange map. I wonder what it is about interval exchange maps that makes it true.... $\endgroup$
    – klindsey
    Jan 27, 2014 at 4:12
  • $\begingroup$ @Klindsey: it's true that "most" IETs are uniquely ergodic. If you compose an IET with a rotation, of course you get another IET, so it's not completely unexpected that you should almost always get an ergodic transformation. $\endgroup$ Jan 27, 2014 at 5:57

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