2

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!

Mr Martingale
  • 952

1 Answers1

3

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.

mathworker21
  • 35,247
  • 1
  • 34
  • 88
  • This is not quite what I had asked - I only want to suppose $R$ is ergodic - it is not necessarily an irrational rotation! – Mr Martingale Jul 16 '18 at 11:32
  • @MrMartingale Yes it is. The only remaining question is the one you concluded with. "Will ergodicity always fail when $R$ is ergodic and $S$ is an irrational rotation?" I answered this. – mathworker21 Jul 16 '18 at 11:43
  • I probably did not phrase this great. What I mean is fix $R$ ergodic (so this may or may not be an irrational rotation.) If it is, then you argument deals with this. If it is not an irrational rotation, can we deduce ergodicity still fails? – Mr Martingale Jul 16 '18 at 15:19
  • @MrMartingale I also showed for some irrational rotations it is ergodic. To expect that it is never ergodic if R is not an irrational rotation would be absurd. And what would you like for an answer then? I would have to check every single $R$ and show whether the resulting $T$ is ergodic or not? – mathworker21 Jul 16 '18 at 20:52
  • Is it a typo that you take $\mathbb{R}/\mathbb{Q}$ instead of $\mathbb{R}/\mathbb{Z}$ inside $L^2$? – テレビ スクリーン Dec 22 '22 at 09:16
  • 1
    @テレビスクリーン haha, yep. sorry about that – mathworker21 Dec 22 '22 at 14:25