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Consider the measure preserving dynamical system $(\mathbb{R}^2 / \mathbb{Z}^2, \mathcal{B} \otimes \mathcal{B}, \lambda \otimes \lambda, R_{(\alpha, \beta)})$. This is the torus with the borel $\sigma$-algebra, the Lebesgue measure, and the rotation defines as

$$ R_{(\alpha, \beta)}(x,y) = (x + \alpha \ (mod1), y + \beta \ (mod1)) $$

I'm searching for a necessary and suficient condition for this system to be ergodic.

I have come by using Fourier Analysis to a sufficient condition for the system to be ergodic:

$$ \forall (k_1, k_2) \in \mathbb{Z}^2 \backslash \{(0,0)\}, \ k_1 \alpha + k_2 \beta \notin \mathbb{Z} $$

And by a more geometric analysis to a necessary condition for the system to be ergodic:

$$ \forall \lambda > 0, \ \lambda \cdot (\alpha, \beta) \neq 0 $$

But I haven't been able to unify them into one.

Any help or good references on this topic?

Thanks in advance.

Jarana
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1 Answers1

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Note that the system is ergodic if and only if the orbit of each point is equidistributed in $\mathbb{R}^2 / \mathbb{Z}^2$.

The first condition $$\forall (k_1, k_2) \in \mathbb{Z}^2 \backslash \{(0,0)\}, \ k_1 \alpha + k_2 \beta \notin \mathbb{Z}$$ is a necessary and sufficient condition. You can use Weyl criterion for equidistribution to show this. The Weyl criterion says that a sequence $x:\mathbb{N}\to(\mathbb{R}/\mathbb{Z})^d$ is equidistributed in $(\mathbb{R}/\mathbb{Z})^d$ if and only if $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}e(k\cdot x(n))=0 \text{ for all } k\in\mathbb{Z}^d\backslash\{0\},$$ where $e(y):=e^{2\pi iy}$ and $(k_1,\ldots,k_d)\cdot(x_1,\ldots,x_d)=k_1x_1+\ldots+k_dx_d$. It is not hard to see that the first condition is equivalent to the above formula.

For a reference, you can find the proof of Weyl criterion in Terence Tao's book "Higher order Fourier analysis".

Siming Tu
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  • Thank you for your help. Still I have one small doubt. Why would the first condition be equivalent to the limit of the sum being 0? – Jarana Mar 30 '14 at 14:23
  • @Jarana: In your example, you can directly calculate the sum in the limit, since it is a sum of geometric progression. – Siming Tu Mar 30 '14 at 14:43