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Let $T$ be an invertible ergodic transformation on a Lebesgue space $X$ and $O$ be the dyadic odometer on $(0,1)$. Is it true that $T\times O$ is ergodic if and only if $T^{2^n}$ is ergodic for every $n \geq 0$ ?

In fact, I am pretty sure this is true, because, unless I'm wrong, I have just realized that I have shown this result during my own research some time ago, but with an usual point of view (too long to explain here), and I would like to be sure I don't misinterpret.

I see how to prove: $$ T\times O \text{ ergodic} \quad \implies \quad T^{2^n} \text{ ergodic}. $$ For $n=0$, just say that the first factor of an ergodic product is ergodic. For $n=1$ it suffices to look at the transformation induced by $T \times O$ on $X \times [0, \frac12[$. It is clearly isomorphic to $T^2 \times O$. Then this transformation is ergodic when $T \times O$ is ergodic, and then the first factor $T^2$ is ergodic as well. For $n=2$, it suffices to look at the transformation induced by $T \times O$ on $X \times [0, \frac14[$, and so on.

But how to prove the converse ?

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    $\begingroup$ Same proof as I showed you the other day. $O$ has eigenvalues $e^{2\pi im/2^n}$ for each $m$ and $n$. If $T$ has any of these eigenvalues then you have non-trivial invariant functions. If not, then $T\times O$ is ergodic for the same reason, by taking the same decomposition of $L^2(X)$ into continuous and discrete spectrum. $\endgroup$ – Anthony Quas Jan 29 '16 at 1:38
  • $\begingroup$ Ok, thank you @AnthonyQuas. I see: saying that $e^{2\pi im/2^n}$ is an eigenvalue is the same as saying that $T^{2^n}$ is not ergodic. I still have to study your answer to the other post. I link it here by the way: mathoverflow.net/questions/229435/… $\endgroup$ – Stéphane Laurent Jan 29 '16 at 2:17
  • $\begingroup$ So the general statement is that if $S$ and $T$ are ergodic, then $S\times T$ is ergodic unless $S$ and $T$ have a common eigenvalue (other than 1). $\endgroup$ – Anthony Quas Jan 29 '16 at 4:38

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