Question: Does there exist a $d$-tuple $\alpha = (\alpha_1,\dots,\alpha_d) \in \mathbb{R}^d$ (with $1,\alpha_1, \dots,\alpha_d$ linearly independent over $\mathbb{Q}$) and an algebraic variety $V \subsetneq \mathbb{R}^d$ such that $$(\{n \alpha_1\},\{n \alpha_2\},\dots,\{n \alpha_d\}) \in V$$ for infinitely many integers $n$? [As usual, $\{x\} \in [0,1)$ denotes the fractional part of $x$.]

Rationale: The linear independence condition ensures (via Kronecker's theorem) that the fractional parts $\{n\alpha\} = (\{n \alpha_1\},\{n \alpha_2\},\dots,\{n \alpha_d\})$ are equidistributed in $[0,1)^d$. Since $V$ has measure $0$, the Banach density of the set of $n$ such that $\{n\alpha\} \in V$ is certainly $0$, and one could reasonably have the intuition that such a set has to be finite - unless there is some algebraic coincidence explaining why it should be infinite. It may also be worth pointing out that if $d = 1$ then $V$ is finite, so $\{n \alpha\} \in V$ at most finitely often (assuming that $\alpha$ is irrational).

On the other hand, it is easy to construct a generalised polynomial $g(n)$ (a polynomial-like expression involving also the fractional parts) such that $g(n)$ vanishes on an infinite set with Banach density $0$. One such example is: $ g(n) = \lfloor n \{n \varphi \} \rfloor$, where $\varphi$ is the golden ratio and $\lfloor x \rfloor = x - \{x\}$ (amusingly, $g(n) = 0$ for every second Fibonacci number, and not much besides). Using the work of Bergelson and Leibman on bounded generalised polynomials, one can now constuct a nilpotent Lie group $G$ together with a discrete, cocompact subgroup $\Gamma < G$, group element $a \in G$ and a set $S \subset G/\Gamma$ which is given by polynomial equations in the natural coordinates of $G/\Gamma$ such that $a^n \Gamma \in S$ for an infinite set of $n$'s with Banach density $0$. When $G = \mathbb{R}^d$ and $\Gamma = \mathbb{Z}^d$ this is exactly the situation from the question (with $V = S$ and $\alpha = a$). Hence, algebraic coincidendes of the type explained above can happen, and the question is whether they can occurr in the simplest possible situation.

  • $\begingroup$ $V$ is closed in the Euclidean topology, and as you observe the set of fractional values is dense. It follows $V$ contains $[0,1]^d$. $\endgroup$
    – Wojowu
    Jan 12, 2019 at 10:44
  • $\begingroup$ @Wojowu, I'm not sure what you mean. The two statements that you make are of course correct, but I don't see how it follows that V has to contain the whole unit cube. Note that {n\alpha} are dense when n runs over all integers, but I'm only asking for {\alpha} to be in V for infinitely many n (the set of such n's will necessarily be rather sparse). $\endgroup$ Jan 12, 2019 at 10:49
  • $\begingroup$ Ah, I have misread the question. Never mind me, then... $\endgroup$
    – Wojowu
    Jan 12, 2019 at 10:49
  • $\begingroup$ No worries, thanks anyways :) $\endgroup$ Jan 12, 2019 at 10:50


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