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I am having trouble proving that if $\alpha_1,\alpha_2$ are irrational, then the sequence $(\alpha_1n,\alpha_2n^2)$ is equidistributed in $\mathbb{T}^2$. It is straightforward to use Furstenberg's skrew product when $\alpha_1=\alpha_2$ but I can't see a way of doing the general case. Any hints?

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  • $\begingroup$ Did you try to find a map of the $2$-torus that is related? $\endgroup$
    – John B
    Feb 27, 2016 at 10:26
  • $\begingroup$ Yep, the map that sends $(x_1,x_2)$ to $(x_1+\alpha_1,x_2+2\frac{\alpha_2}{\alpha_1}x_1-\alpha_2)$.I can finish the question if I can prove that this is ergodic. $\endgroup$
    – Mathmo
    Feb 27, 2016 at 10:29
  • $\begingroup$ However, a Fourier analysis approach doesn't seem to lead anywhere. In fact I'm not even sure that $x_1\rightarrow{\frac{\alpha_2}{\alpha_1}}x_1$ is continuous. $\endgroup$
    – Mathmo
    Feb 27, 2016 at 10:30
  • $\begingroup$ Well, that's what I had in mind. :) Sometimes the computations are complicated but eventually they lead somewhere. $\endgroup$
    – John B
    Feb 27, 2016 at 11:39
  • $\begingroup$ I am slightly confused by cty. I am not sure the multiplication by $\frac{\alpha_1}{\alpha_2}$ is continuous - it seems to get messed up near 0 (or 1). $\endgroup$
    – Mathmo
    Feb 27, 2016 at 12:08

1 Answer 1

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Hints:

The usual approach is to first find a measurable map of the $2$-torus such that under iteration you get precisely $(\alpha_1 n,\alpha_2 n^2)\bmod1$.

Then you should show that Lebesgue measure is ergodic, most probably by showing that all but one Fourier coefficient vanish.

After that you should apply Birkhoff's ergodic theorem to an arbitrary continuous function on the $2$-torus. For example, by considering a continuous function depending only on the second component you are actually looking at the uniform distribution of $\alpha_2 n^2\bmod1$.

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