Ergodicity of irrational rotation of product

ergodic-theorymeasure-theoryskew-mapping

I need some help with the following.

Suppose $(X,\mathcal{B},\mu,R)$ is an ergodic measure preserving dynamical system. Consider the torus $Y=\mathbb{R}/\mathbb{Z}\times\mathbb{R}/\mathbb{Z}$ and the map $S:Y\to Y$ given by $Sy=y+c$ mod $1$ where $c\in\mathbb{R}\setminus \mathbb{Q}$.

This is known to be ergodic wrt Lebesgue measure. Let $T:X\times Y\to X\times Y$ be defined by $T(x,y)=(Rx,Sy)$. Is $T$ ergodic w.r.t the product measure $\mu\times Leb$?

Naturally, I would expect so. However there are doubts. I found the following related question (which takes $X$ to be the torus with another rotation) (Ergodic Rotation of the Torus) which gives necessary and sufficient conditions of ergodicity, which leads me to believe that ergodicity may fail. But I can not come with anything concrete.

Remark: In the case $c$ is rational, ergodicity of $T$ does not hold because $S$ is not ergodic.

Edit: From (Product of ergodic transformations), ergodicity fails when $X=Y$ is also the irrational rotation by $c$. So I guess the answer to my original question is 'not necessarily'. Can this be generalised? i.e., will ergodicity always fail when $R$ is ergodic and
$S$ is an irrational rotation?

Thanks in advance!

Best Answer

Fix irrational $\alpha, \beta \in (0,1)$. Let $R$ be the rotation by $\alpha$ and $S$ the rotation by $\beta$. $R$ and $S$ are ergodic. The obtained $T$ is ergodic iff $\alpha$ and $\beta$ are linearly independent over $\mathbb{Q}$. To see why, take a $T$-invariant $f \in L^2(\mathbb{R}/\mathbb{Q})$. We may write $$f(x_1,x_2) = \sum_{k_1,k_2 \in \mathbb{Z}} c_{k_1,k_2}e^{2\pi i [k_1x_1+k_2x_2]}.$$ $T$-invariance implies that $$f = \sum_{k_1,k_2} c_{k_1,k_2} e^{2\pi i [k_1(x_1+\alpha)+k_2(x_2+\beta)]} = \sum_{k_1,k_2} e^{2\pi i (k_1\alpha+k_2\beta)}c_{k_1,k_2}e^{2\pi i [k_1x_1+k_2x_2]}.$$ By uniqueness of fourier coefficients, we must have that, for each $(k_1,k_2) \in \mathbb{Z}^2$, either $c_{k_1,k_2} = 0$ or $e^{2\pi i (k_1\alpha+k_2\beta)} = 1$. If $\alpha,\beta$ are independent over $\mathbb{Q}$, or equivalently over $\mathbb{Z}$, then this means that all $c_{k_1,k_2}$ are $0$ except for $c_{0,0}$. So $f$ is constant in this case, and therefore $T$ is ergodic.

The converse is similarly easy to prove. Indeed, you can show that if $\alpha k_1 + \beta k_2 = 0$ for $(k_1,k_2) \not = (0,0)$, then $e^{2\pi i [k_1x_1+k_2 x_2]}$ is a non-constant $T$-invariant function.

Related Question