Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a book, Ergodic problems of classical mechanics by Arnold/Avez, and in it they prove that rotation $Tx = x+a \pmod 1$ of the circle $M=\{x \pmod 1\}$ is Ergodic if and only if a is irrational. In it they use an earlier corollary that a system is ergodic if and only if any invariant measurable absolutely integrable function is constant a.e. So the start proof goes like this:

Suppose $a$ is rational, then write $a=p/q$, $p,q$ coprime. Since $e^{2\pi qx}$ is nonconstant and measurable, $T$ is not ergodic.

This is fine, but then how come I can't make a similar argument for irrational $a$? As in: Suppose $a$ is irrational, then $e^{2\pi x/a}$ is nonconstant, and it seems to be measurable and absolutely integrable.

What did I do wrong?

share|improve this question
1  
$e^{2 \pi i a x}$ doesn't give a well-defined function on the line mod $1$ if $a$ isn't an integer. –  Qiaochu Yuan May 9 '12 at 4:38

1 Answer 1

up vote 4 down vote accepted

You mean $e^{2 \pi i q x}$. The point is that this is a nonconstant measurable function on the circle that is invariant under $T$. But if $a \in (0,1)$ is irrational, $e^{2 \pi i x/a}$ is not invariant under $T$: if $Tx = x + a - 1$, as it is when $x + a > 1$, then $$e^{2 \pi i T(x)/a} = e^{2 \pi i (x+a-1)/a} = e^{2 \pi i x/a} e^{-2 \pi i/a} \ne e^{2 \pi i x/a}$$

share|improve this answer
    
Oh, thank you! I knew the function was suppose to be invariant, I just forgot to include it. I just mistakenly thought the function with the irrational a was invariant. –  JLA May 9 '12 at 5:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.