The Kronecker-Weyl theorem asserts the following: fix real numbers $\theta_1,\dots,\theta_d$, and consider the infinite ray $t(\theta_1,\dots,\theta_d)$ $(t\in\Bbb R)$ inside the $d$-dimensional torus $(\Bbb R/\Bbb Z)^d$. Then there exists a subtorus $A$ such that the limiting distribution of $t(\theta_1,\dots,\theta_d)$ is uniform on $A$. (In fact, $A$ is the torus defined by any $\Bbb Q$-linear relations among the $\theta_j$. In other words, the ray is "obviously" confined to this subtorus, and the theorem says that there aren't any other restrictions on where the ray goes.)
I am looking for a reference for this theorem and its proof. It seems to be one of these theorems that is often stated, and called "classical" and "well-known", but almost always without citation. I have been unable to find the theorem proved in this full generality; a proof under the assumption that the $\theta_j$ are linearly independent over the rational numbers will not suffice for me.
I am most interested in a source from the research literature or from a research monograph. But I would also find useful a fully written proof from someone's course notes or the like. Even a published source where this general version was carefully stated (preferably with a definition of "uniformly distributed") might be of some slight use, even if they don't include a proof.
Best Answer
Why not just prove the result from scratch? It's only a couple of pages and involves only basic Fourier analysis for locally compact abelian groups.
Let $$\mathbb{T}^n = \left\{(z_1,\ldots,z_n) \in \mathbb{C}^n : |z_l| = 1 \text{ for all $1 \leq l \leq n$}\right\}$$ be the $n$-torus. Let $t_1, \ldots, t_n$ be arbitrary real numbers, and let $H$ be the topological closure in $\mathbb{T}^n$ of the subgroup $$\widetilde{H} = \left\{\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \in \mathbb{T}^n : y \in \mathbb{R}\right\}.$$ The Kronecker--Weyl theorem states that
(1) $H$ is a closed connected subgroup of $\mathbb{T}^n$, namely an $r$-dimensional subtorus of $\mathbb{T}^n$, where $0 \leq r \leq n$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and that
(2) for any continuous function $h : \mathbb{T}^n \to \mathbb{C}$, we have that $$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} = \int_{H}{h(z) \: d\mu_H(z)},$$ where $\mu_H$ is the normalised Haar measure on $H$.
To prove this, we begin by observing that $H$ is a closed connected subgroup of $\mathbb{T}^n$ as it is the topological closure of $\widetilde{H}$, which is a subgroup of the compact abelian group $\mathbb{T}^n$, being the image of the continuous group homomorphism $\phi : \mathbb{R} \to \mathbb{T}^n$ given by $\phi(y) = \left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right)$.
Next we recall that a character $\chi : \mathbb{T}^n \to \mathbb{T}$ is of the form $$\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$$ for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$. Conversely, for any $(k_1,\ldots,k_n) \in \mathbb{Z}^n$, the function $\chi : \mathbb{T}^n \to \mathbb{T}$ defines a character of $\mathbb{T}^n$. In particular, the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence every character $\chi : \mathbb{Z}^n \to \mathbb{T}$ is of the form $$\chi(k_1,\ldots,k_n) = z_1^{k_1} \cdots z_n^{k_n}$$ for some $(z_1,\ldots,z_n) \in \mathbb{T}^n$.
We claim that the annihilator $H^{\perp}$ of $H$ (namely the set of characters of $\mathbb{T}^n$ that are trivial on $H$) is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_1 k_n = 0\right\}$, and consequently $H$ is isomorphic to a torus $\mathbb{T}^r$ for some $0 \leq r \leq n$. Indeed, each character $\chi \in H^{\perp}$ is of the form $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ for some $(k_1,\ldots,k_n) \in \mathbb{Z}^n$ with the property that for all $y \in \mathbb{R}$, $$1 = \chi\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) = e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y},$$ and hence $t_1 k_1 + \cdots + t_n k_n = 0$. Conversely, if $t_1 k_1 + \cdots + t_n k_n = 0$, then the homomorphism $\chi(z_1,\ldots,z_n) = z_1^{k_1} \cdots z_n^{k_n}$ satisfies $\chi \vert_{H} = 1$. Now by construction, $H^{\perp}$ is isomorphic to $V \cap \mathbb{Z}^n$ for some vector subspace $V$ of $\mathbb{Q}^n$ of dimension $n - r$, where $r$ is the dimension over $\mathbb{Q}$ of the span of $t_1, \ldots, t_n$, and so $H^{\perp} \cong \mathbb{Z}^{n-r}$. Consequently, $\widehat{H} \cong \mathbb{Z}^n / H^{\perp} \cong \mathbb{Z}^r$, as the dual group of $\mathbb{T}^n$ is isomorphic to $\mathbb{Z}^n$, and hence $H \cong \mathbb{T}^r$, thereby proving (1).
For (2), we require the Fourier transform $\widehat{h} : \mathbb{Z}^n \to \mathbb{C}$ of a continuous function $h : \mathbb{T}^n \to \mathbb{C}$, defined by $$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \: dz}.$$ The Poisson summation formula for a closed subgroup of $\mathbb{T}^n$ states that $$\int_{H}{h(z) \: d\mu_H(z)} = \int_{H^{\perp}}{\widehat{h}(\chi) \: d\mu_{H^{\perp}}(\chi)},$$ where $\mu_H$ is the normalised Haar measure on $H$ and $\mu_{H^{\perp}}$ is the induced Haar measure on $H^{\perp}$, which is the counting measure as $H^{\perp}$ is discrete. Now if $h : \mathbb{T}^n \to \mathbb{C}$ is a trigonometric polynomial, which is to say a function of the form $$h(z) = \sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n}$$ for $z = (z_1, \ldots, z_n) \in \mathbb{T}^n$, where all but finitely many of the coefficients $c_k \in \mathbb{C}$ are zero, we claim that $$\lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \, dy} = \int_{H}{h(z) \: d\mu_H(z)},$$ where $\mu_H$ is the normalised Haar measure on $H$. From this, we may easily obtain the result in the general case where $h$ is merely a continuous function by the density of the trigonometric polynomials in the space of continuous complex-valued functions on $\mathbb{T}^n$ with regards to the supremum norm. This yields (2).
To prove the claim, we let $\chi : \mathbb{T}^n \to \mathbb{T}$ be a character corresponding to $\widetilde{k} \in \mathbb{Z}^n$. Then $$\widehat{h}(\chi) = \int_{\mathbb{T}^n}{h(z) \overline{\chi(z)} \, dz} = \int_{\mathbb{T}} \hspace{-0.1cm} \cdots \hspace{-0.1cm} \int_{\mathbb{T}}{\sum_{k \in \mathbb{Z}^n} c_k z_1^{k_1} \cdots z_n^{k_n} \overline{z_1^{\widetilde{k_1}} \cdots z_n^{\widetilde{k_n}}} \, dz_1 \cdots dz_n}.$$ We may interchange the order of summation and integration, as there are only finitely many nonzero members in this sum, and evaluate this integral in order to find that $\widehat{h}(\chi) = c_{\widetilde{k}}$. Recalling that $H^{\perp}$ is isomorphic to $\left\{k \in \mathbb{Z}^n : t_1 k_1 + \cdots + t_n k_n \in \mathbb{Z}\right\}$, so that the Haar measure $\mu_{H^{\perp}}$ on $H^{\perp}$ is simply the counting measure, we therefore obtain by the Poisson summation formula that $$\int_{H}{h(z) \: d\mu_H(z)} = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k.$$ On the other hand, \begin{align*} \lim_{Y \to \infty} \frac{1}{Y} \int^{Y}_{0}{h\left(e^{2\pi i t_1 y}, \ldots, e^{2\pi i t_n y}\right) \: dy} & = \lim_{Y \to \infty} \frac{1}{Y} \sum_{k \in \mathbb{Z}^n} c_k \int^{Y}_{0}{e^{2\pi i (t_1 k_1 + \cdots + t_n k_n) y} \: dy} \\ & = \sum_{\substack{k \in \mathbb{Z}^n \\ t_1 k_1 + \cdots + t_n k_n = 0}} c_k \end{align*} as required, where we justify the interchanging of order of summation and integration by noting that there being only finitely many nonzero members in this sum.