Proof when the circle map is ergodic

Question

Let $E=[0,1)$ with Lebesgue measure. For $a \in E$ consider the mapping

$\theta_a:E \rightarrow E, \ \ \theta_a(x) = (x+a) \mod \ 1$.

a) Show that $\theta_a$ is not ergodic when $a$ is rational.

b) Show that $\theta_a$ is ergodic when a is irrational.

I think that $\theta_a$ is isomorphic to the map $\theta_a=e^{2\pi i(x+a)}$ but not sure how this would help solve the problem. For part b) the hint, consider $a_n=\int_E f(x)e^{2 \pi i nx}dx$ to show that every invariant function is constant, is given.

Strong hints would be greatly appreciated.

Answer 1

Hint For part (a): We know that $\theta_a$ is ergodic if and only if the validity of the condition $m(\theta_a(X)\Delta X)=0$ implies that $X=[0,1[$ or $X=\emptyset$, where $m$ denotes a linear Lebesgue measure on $[0,1[.$ If $a$ is rational and admits a representation $a=l+k/n$ with $l \in Z, k,n \in N, 0<k<n$, then for a set $X_0=[0/n,0/n+1/(nk)[\cup [1/n,1/n+1/(nk)[\cup \cdots [(n-1)/n+1/(nk)[$ we have $m(\theta_a(X_0)\Delta X_0)=0$ because $\theta_a(X_0)=X_0$ but $X_0\neq [0,1[$ and $X_0 \neq \emptyset$. Hence $\Theta_a$ is not ergodic. If $a$ is integer then the validity of the condition (a) is obvious because for each subset $X \subset [0,1[$ we have $\theta_a(X)=X$. Hence setting $X_1=[0,1/2[$ we get $m(\theta_a(X_1)\Delta X_1)=0$ but $X_1\neq [0,1[$ and $X_1 \neq \emptyset$.

Answer 2

For part (b) Take the Fourier transform of the indicator function, $f$, of the fixed set of $\theta_a$.

Then the map is represented on the unit circle by multiplication by $e^{2\pi ia}$

Since $L^2(S^1)$ is an inner product space, you get that, almost everywhere:

$$f(x)=\sum_{n=-\infty}^\infty a_ne^{nx}=\sum_n a_ne^{n(x+a)}$$

Since Fourier series are unique and $a$ is irrational, the only possibility is that $a_n=0$ for $n\ne 0$, hence $f$ is constant almost everywhere, i.e. $\theta_a$ is ergodic by definition.