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Here is my reasoning. I use multiplicative notation.

Let $f \in L^2$ then it has a Fourier series $$f(x,y)=\sum a_{n,m}x^ny^m$$ because $x^ny^m$ are characters and so they form an orthonormal basis. Now $$f(T(x,y))=\sum a_{n,m}x^{n+m}y^m.$$ If $f$ is $T$ invariant then it must be that $a_{n,m}=a_{n+m,m}=a_{n+2m,m}=...$ Thus unless $a_{n,m}$ are zero this the coefficents are not square summable. The series $$\sum a_{n,m}x^{n+m}y^m$$ is a Fourier series as $x^{n+m}y^m$ are disinct charcters for all $n,m$.

Where is my mistake? My books says this map is not ergodic.

Edit: Is $m=0$ perhaps the problem here? since $a_{n,0}$ might still be non-zero. If so, I demand a refund of my time i spent staring at this

Sorfosh
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1 Answers1

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Assuming you mean with respect to Lebesgue measure: $f(x,y) = x$ is a non-constant $T$-invariant function.

In the Fourier series for this $f$ you have all $a_{n,m}=0$ except for $a_{1,0}=1$. So you were correct when you guessed that the coefficients with m=0 create the problem with your original proof

Adam
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  • I know I am wrong. I just do not see where I am wrong – Sorfosh Feb 25 '21 at 20:26
  • In the Fourier series for this $f$ you have all $a_{n,m} = 0$ except for $a_{1,0} = 1$. So you were correct when you guessed that the coefficients with $m = 0$ create the problem with your original proof – Adam Feb 25 '21 at 20:34
  • Follow up question, is the map $T(x,y)=(xe^{2\pi i \alpha},xy)$ ergodic for $\alpha$ non integer? Using this argument + the change in the first coordinate allows us to conclude $a_{n,0}=0$ for all $n>0$ – Sorfosh Feb 25 '21 at 20:38
  • Yes it is, in fact it is uniquely ergodic with Lebesgue measure as the unique invariant measure. This is a result originally due to Furstenberg and you can find a nice proof in (for example) Einsiedler and Ward's book – Adam Feb 25 '21 at 20:40
  • Thank you very much Adam, that was very helpful. Could you put the comment in the answer? So that it is complete. – Sorfosh Feb 25 '21 at 20:44