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I am interested in the ergodic (invertible) transformations $T$ such that $T\times R_\theta$ is ergodic where $R_\theta$ is the rotation on $S^1$ with a given irrational angle $\theta$ (not all $R_\theta$, only this one).

This includes all weakly mixing transformations, because of the two following results which can be found in Rudolph's book:

If $T$ is ergodic and has no factor isomorphic to an isometry on a compact space, then $T \times S$ is ergodic for every ergodic transformation $S$.

and

If $T$ is weakly mixing then it has no factor isomorphic to an isometry on a compact space.

There are other suitable $T$: every rotation $R_\alpha$ fulfills the desideratum when $\alpha$ and $\theta$ are rationally independent.

Can we expect a characterization of the suitable $T$ ? Or say something more ? Rotations $R_\alpha$ are not suitable when $\alpha$ and $\theta$ are not rationally independent. Are there other transformations which are not suitable ?

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$T\times R_\theta$ is ergodic if and only if $T$ is ergodic and $e^{2\pi im\theta}$ is not an eigenvalue of $T$ for any $m\in \mathbb Z\setminus\{0\}$.

For an idea of the proof, let $T\colon X\to X$. Then $L^2(X)$ can be decomposed as an orthogonal direct sum $V_c\oplus V_d$, the functions with continuous spectrum and the functions with discrete spectrum. $V_d$ is spanned by the eigenfunctions of $T$. Similarly $L^2(S^1)$ has the standard Fourier basis, with $e_k(x)=e^{2\pi ikx}$ being an eigenfunction of $R_\theta$ with eigenvalue $e^{2\pi ik\theta}$.

Now suppose there's an invariant function $F(x,t)$ for the product transformation. It can be decomposed as $\sum_k (g_k+h_k)\otimes e_k$, where $g_k\in V_c$ and $h_k\in V_d$. Since $F$ is invariant, we have $\frac 1N\sum_{i=0}^{N-1}F\circ(T\times R)^i=F$.

On the other hand, $\frac 1N\sum_{i=0}^{N-1}(g_k\otimes e_k)\circ (T\times R)^i\to 0$ in $L^2$ by some spectral theory. Also $h_k\otimes e_k$ is a sum of eigenfunctions of $T\times R$. Any eigenfunction with eigenvalue other than 1 disappears under averaging.

Hence if $F$ is invariant, it must be of the form $\sum_k h_k\otimes e_k$ where $h_k\otimes e_k$ is an invariant function of $T\times R$, which implies $h_k$ must satisfy $h_k(Tx)=e^{-2\pi ik\theta}h_k(x)$. Since by assumption $e^{2\pi i k\theta}$ is not an eigenvalue of $T$ for any $k$, all of the $h_k$ must be 0 other than the constant term. So we deduce any invariant function for the product map is constant.

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  • $\begingroup$ That looks nice, thank you. But could you provide an idea of the proof ? $\endgroup$ – Stéphane Laurent Jan 27 '16 at 19:58
  • $\begingroup$ yes. i was trying to edit it to do this. i will try and write something down later. $\endgroup$ – Anthony Quas Jan 27 '16 at 21:29
  • $\begingroup$ Thank you. I definitely need to learn more about spectral theory. $\endgroup$ – Stéphane Laurent Jan 27 '16 at 21:54
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I wanted to comment on Anthony's remark (but couldn't because of reputation issues) by saying that Keane's ergodic multiplier theorem (See for example Aaronson's book page 81) gives an if and only if condition as to when $T\times S$ is ergodic for $S$ is an invertible probability preserving transformation (like an irrational rotation) and $T$ is an invertible ergodic and conservative (satisfies Poincare reccurrence theorem) quasi invariant (preserves the $\sigma$ ideal of zero measure sets) transformation.
If $T,S$ are probability measure preserving transformation then the condition from Keane's multiplier theorem (which was proven earlier by Furstenberg in his multiple recurrence paper) says that $S\times T$ is ergodic if and only if $S$ and $T$ do not share a common eigenvalue.

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  • $\begingroup$ Isn't it exactly @AnthonyQuas's proof ? He treats the case when the eigenvalues are the ones of the rotation ($e^{2\pi im\theta}$), but I thought this proof works in the general situation (but I have not really read it yet). $\endgroup$ – Stéphane Laurent Jan 29 '16 at 16:54
  • $\begingroup$ In the quasi invariant (similarly to the case of transformations preserving an infinite measure) case there could be more complicated eigenvalue groups. For example eigenvalue groups of positive Hausdorff dimension (see the book of Nadkarni on spectral analysis or the papers of Aaronson-Nadkarni or Host-Mela-Parreau). For this reason it involves an extra step and a slightly different use of the spectral theorem for unitary operators. $\endgroup$ – user78465 Jan 30 '16 at 19:29
  • $\begingroup$ Ah ok, this is something more general. Thank you for this complement. $\endgroup$ – Stéphane Laurent Feb 13 '16 at 13:02

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