# When are circle maps ergodic?

It is well known that the map $$M: S^1 \rightarrow S^1$$ given by $$x \mapsto x + \alpha$$ (mod 1) is ergodic whenever $$\alpha$$ is irrational. What about maps $$M_2 : S^1 \times S^1 \rightarrow S^1 \times S^1$$ given by $$(x_1,x_2) \mapsto (x_1 + \alpha_1, x_2 + \alpha_2)$$? Or more generally, the maps $$M^n : S^1 \times \dots \times S^1 \rightarrow S^1 \times \dots \times S^1$$ given by $$(x_1, \dots, x_n) \mapsto (x_1 + \alpha_1, \dots, x_n + \alpha_n)$$?

I have managed to notice that for $$M_2$$, if we take $$\alpha_1 = \alpha_2$$, then the map fails to be ergodic since $$e^{2 \pi i (x_1 - x_2)}$$ is invariant but non-constant. So I conjecture that the maps $$M^n$$ are ergodic when $$\alpha_1,\dots,\alpha_n$$ are distinct irrational numbers.

• Commented Nov 3, 2023 at 18:47

First of all, I assume that when you say ergodic you mean ergodic with respect to the Haar/Lebesgue measure. As mentioned, the translation on the torus is ergodic if and only if the list of real numbers $$(\alpha_1, \ldots, \alpha_n)$$ is rationally independent. This a particular case for something more general, see [Theorem 6.3.8, Foundations of Ergodic Theory, M. Viana and K. Oliveira]:

Theorem. Let $$G$$ be a compact metrizable topological group, $$\mu_G$$ its Haar measure, $$g \in G$$ and $$L_g$$ the left translation. The following are equivalent:

(i) $$L_g$$ is uniquely ergodic.

(ii) $$L_g$$ is ergodic with respect to $$\mu_G$$.

(iii) The subgroup generated by $$g$$ is dense in $$G$$.

In the case $$G = \mathbb{T}^n$$, it is a good exercise to show that if $$g = (\alpha_1, \ldots, \alpha_n)$$ is rationally independent, then the set $$\langle g \rangle = \{(k \alpha_1, \ldots, k \alpha_n) \mod \mathbb{Z}^n\}_{k \in \mathbb{Z}}, \quad \text{is dense on}\; \mathbb{T}^n.$$

You would need them to be linearly independent as well. Otherwise, assuming $$\sum_{j=1}^n c_j \alpha_j=0$$, you would get that

$$f(x_1,...,x_n)= e^{2\pi \sum_{j=1}^n c_j x_j}$$

is also an invariant but non constant function. You can see that the orbit would then be dense by the simultaneous version of the Dirichlet approximation theorem. Then, you can approximate the indicator of any measurable set, like in this thread.