# Can the orbit with respect to a rotation on the torus hit an algebraic variety infinitely often?

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.

• $V$ is closed in the Euclidean topology, and as you observe the set of fractional values is dense. It follows $V$ contains $[0,1]^d$. Jan 12, 2019 at 10:44
• @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). Jan 12, 2019 at 10:49
• Ah, I have misread the question. Never mind me, then... Jan 12, 2019 at 10:49
• No worries, thanks anyways :) Jan 12, 2019 at 10:50