# Using Furstenberg's skew-product for $\alpha n^2$ equidistributed.

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?

• Did you try to find a map of the $2$-torus that is related? Feb 27, 2016 at 10:26
• 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. Feb 27, 2016 at 10:29
• 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. Feb 27, 2016 at 10:30
• Well, that's what I had in mind. :) Sometimes the computations are complicated but eventually they lead somewhere. Feb 27, 2016 at 11:39
• 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). Feb 27, 2016 at 12:08

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