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Let $\mathbb{T}^3$ be the three-dimensional flat torus. I wanna think of it as being $\mathbb{R}^3$ with each coordinate taken modulo $1$. Let $(x, y, z)$ be such coordinates, and let $\partial_x, \partial_y, \partial_z$ be the corresponding vector fields.

For any choice of constants $\alpha_1,\alpha_2,\alpha_3 \in \mathbb{R} $ we can look at the vector field

$$V_{\alpha} = \alpha_1\partial_x + \alpha_2\partial_y + \alpha_3\partial_z.$$

My question is: when is the flow of $V_{\alpha}$ periodic? I wanna say it's when all $\alpha_i$ are rational, but I can't prove it and don't have a reference. Either a proof or reference would be appreciated.

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    Maybe true under milder assumption that the values $\alpha_j$ are rationally independent. – 2017-01-23

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The integral curves are the images of the lines $$ x(t) = (x_0, y_0, z_0) + (\alpha_1 t, \alpha_2 t, \alpha_3 t). $$ An integral curve is periodic if and only if there exists an integer vector $(n_1, n_2, n_3)$ and a $t_0 \neq 0$ such that $x(t_0) = x(0) + (n_1, n_2, n_3)$, i.e., $$ (\alpha_1 t_0, \alpha_2 t_0, \alpha_3 t_0) = (n_1, n_2, n_3). $$ Expressing $t_0$ in terms of the $\alpha_i$ and $n_i$ establishes your conjecture.

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    In light of your answer, I think that my conjecture is wrong. For instance, if $\alpha_1 = \alpha_2 = 0$, then the flow is periodic no matter what. What you seem to be proving is that the "angles" $\alpha_i / \alpha_j$, whenever $\alpha_j \neq 0$, all have to be rational. Do you agree? – 2017-01-23
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    As you say, you want the vector $(\alpha_1/\alpha_i,\alpha_2/\alpha_i,\alpha_3/\alpha_i) \in \mathbb{Q}^3$ where $\alpha_i$ is non-zero. This is equivalent to, but perhaps easier to check than, Andrew's answer. This is also sometimes called 'rational dependence' if you think about the set of vectors $\{\alpha_1,\alpha_2,\alpha_3\}$ in the vector space $\mathbb{R}$ over the field $\mathbb{Q}$. – 2017-01-23
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    @L..: As Daniel notes, you're right; I'd implicitly read in "up to scaling", i.e., some scalar multiple of the generating vector has rational components, which is equivalent to the condition each of you mentions. – 2017-01-23