# Equidistribution theorem in two dimension

If $$\alpha\in \mathbb R$$ is an irrational, then $$\{ \langle k\alpha \rangle \}_{k \ge 1}$$ is dense in $$[0,1]$$, where $$\langle x \rangle$$ denote the fractional part of $$x$$. Moreover, $$\{ \langle k\alpha \rangle \}_{k \ge 1}$$ is equidistributed in $$[0,1]$$, which means that for $$0 \le a < b \le 1$$, $$\lim_{n \to \infty} \frac{\# \{ 1\le k \le n: a \le \langle k \alpha \rangle \le b \} }{n} = b-a.$$

Does the result holds for higher dimension?

Let $$\alpha,\beta\in \mathbb R$$ are two irrationals such that $$1,\alpha,\beta$$ are linearly independent over $$\mathbb{Q}$$. How to prove that $$\{(\langle k\alpha \rangle, \langle k\beta \rangle)\}_{k \ge 1}$$ is dense in $$[0,1]^2$$ and is equidistributed in $$[0,1]^2$$?

As a generalization of Weyl's criterion, $$\{u_k\}_{k\geqslant 1}$$ is equidistributed in $$[0,1]^2$$ if and only if $$\forall \ell\in\mathbb{Z}^2\setminus\{0,0\},\lim\limits_{n\rightarrow +\infty}\frac{1}{n}\sum_{k=0}^{n-1}e^{2i\pi\ell\cdot u_k}=0$$ But for all $$\ell\in\mathbb{Z}^2\setminus\{0,0\}$$, we have $$\sum_{k=0}^{n-1}e^{2i\pi\ell\cdot(\langle k\alpha\rangle,\langle k\beta\rangle)}=\sum_{k=0}^{n-1}e^{2i\pi\ell_1\langle k\alpha\rangle}e^{2i\pi\ell_2\langle k\beta\rangle}=\sum_{k=0}^{n-1}\left(e^{2i\pi\ell_1\alpha+2i\pi\ell_2\beta}\right)^k$$ Thus $$\left|\frac{1}{n}\sum_{k=0}^{n-1}e^{2i\pi\ell\cdot(\langle k\alpha\rangle,\langle k\beta\rangle)}\right|\leqslant\frac{2}{n|1-e^{2i\pi\ell_1\alpha+2i\pi\ell_2\beta}|}\underset{n\rightarrow +\infty}{\longrightarrow}0$$ because $$\ell_1\alpha+\ell_2\beta\notin\mathbb{Z}$$ since $$1$$, $$\alpha$$ and $$\beta$$ are linearly independent. As for the density, it follows from the fact that $$\alpha\mathbb{Z}+\mathbb{Z}$$ and $$\beta\mathbb{Z}+\mathbb{Z}$$ are dense in $$\mathbb{R}$$ since $$\alpha,\beta\notin\mathbb{Q}$$

• Is there any other condition for $1,\alpha,\beta$ to be $\mathbb Q$-linearly independent? Commented Sep 21, 2020 at 12:16
• I don't follow the argument about the density. The way you state it, it would also supposedly be true for $\alpha=\beta$, no? Or does it somehow rely on the previous proof of equidistribution? Commented Sep 17, 2023 at 7:36