Reference for Kronecker-Weyl theorem in full generality

Question

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.

Answer 1

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.

Answer 2

Let me try to reinstate honor to the solution that proposed the basis change, by reducing the general case to the independent ("generic") case via a basis change, as opposed to proving it from scratch. This time I am treating the continuous version. (as in Peter Humphries' solution)

Let $\vec\theta=(\theta_1,\dots,\theta_k)$. Let $S$ be the subspace of vectors $\vec x\in\mathbb R^k$ such that $$\langle \vec a,\vec x\rangle=0$$ for all rational vectors $\vec a\in\mathbb Q^k$ for which $$\langle \vec a,\vec\theta\rangle=0$$ The integer points in $S$ form a lattice $L$ (a discrete subgroup of $\mathbb R^n$), which can be written as $L=\{\,g_1\vec b_1+\dots+g_r\vec b_r\mid g_i\in\mathbb Z\,\}$ for some generating basis vectors $\vec b_1,\dots,\vec b_r\in\mathbb{Z}^k$. Since $S$ is defined by rational equations, this basis spans $S$.
Let $(\theta'_1,\dots,\theta_r')$ be the coordinates of the point $\vec\theta\in S$ with respect to this basis: $$\vec\theta=\theta_1'\vec b_1+\dots+\theta_r'\vec b_r$$ Then $(\theta'_1,\dots,\theta_r')$ is independent over the rationals (see below (*) for a proof). Thus, we can apply the generic continuous Kronecker-Weyl Theorem, and $(t(\theta_1',\dots,\theta_r'))_{t\in \mathbb{R}}$ is uniformly distributed modulo 1 in the $r$-torus $[0,1)^r$. Transforming back to the original coordinates, this means that $(t\vec \theta)_{t\in \mathbb{R}}$ is uniformly distributed modulo $L$ in the fundamental region $$ F = \{\,\lambda_1\vec b_1+\dots+\lambda_r\vec b_r \mid 0\le\lambda_i<1\,\}$$ of the lattice.
Now we map $F$ back into the standard torus $[0,1)^k$ by taking all coordinates modulo 1. No two points of $F$ are mapped to the same point (otherwise we would have a nonzero integer point inside $F$), but "opposite" boundary points are mapped to the same point because they differ by a basis vector $b_i\in\mathbb{Z}^k$. So $F$ forms a nice $r$-dimensional subtorus of $[0,1)^k$.

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(*) Here is the proof that $(\theta'_1,\dots,\theta_r')$ is independent over the rationals. It is not so obvious as I thought. Consider a rational relation $c_1\theta_1'+\dots+c_r\theta'_r=0.$ We can choose inside $S$ a rational vector $\vec a$ such that $\langle \vec a,\vec b_i\rangle = c_i$ for $i=1,\dots,r$. (The vector $\vec a$ is uniquely determined by these equations.) Then $$\langle \vec a, \vec\theta\rangle= \langle \vec a, (\theta_1'\vec b_1+\dots+\theta_r'\vec b_r)\rangle =c_1\theta_1'+\dots+c_r\theta'_r=0 $$ Thus, by the definition of $S$, $\vec a$ should be orthogonal to $S$. Therefore, $\vec a=\vec 0$, and all $c_i$ are $0$.

Answer 3

Maybe my answer is beside the point, because the question speaks about the "limit distribution of $t(\theta_1,\dots,\theta_d)$" (for real $t$?). Shouldn't it be $n(\theta_1,\dots,\theta_d)$ for $n=1,2,\ldots$? See https://mathworld.wolfram.com/Kronecker-WeylTheorem.html. The long answer "from scratch" by Peter Humphries proves a different theorem that seems to be more in line with real parameters $t$. Also, the sequence $n(\theta_1,\dots,\theta_d)$ can fill a disconnected manifold, which wouldn't properly be called a "subtorus".

If the question is indeed about the sequence $n(\theta_1,\dots,\theta_d)$, the book by Kuipers and Niederreiter (Uniform Distribution of Sequences, 1974), contains the generic version of what seems to be the Kronecker-Weyl Theorem as Example 6.1 on p. 48. The proper condition is the real numbers $1,\theta_1,\dots,\theta_d$ are linearly independent over the rationals.

The notes on p.51 mention that

a discussion of the exceptional case in this example was also carried out by Weyl,

referring to his classical paper: Hermann Weyl. Über die Gleichverteilung von Zahlen mod. Eins. Mathematische Annalen, 77:313–352, 1916. DOI:10.1007/BF01475864

Indeed, §5 of that paper, "Die Ausnahmefälle" (the exceptional cases), contains a Theorem 18 (pp. 340-341). It deals with the more general case where each coordinate is not just a linear function of $n$ but an arbitrary polynomial. The conclusion is that the points cover a finite number of affine-linear $r$-dimensional manifolds, (possibly with different integer multiplicities), all these manifolds are parallel, and each of them is filled with uniform density. The theorem specifies how to determine $r$ and the multiplicities.

Here is a statement of this theorem (with different notation), specialized and reformulated for the case of an arithmetic progression $n(\theta_1,\dots,\theta_d)$ as opposed to arbitrary polynomials. The multiplicities are then not necessary.

Let $\vec\theta=(\theta_1,\dots,\theta_d)$. Let $C$ be the set of vectors $\vec x\in\mathbb R^d$ such that $$\langle \vec a,\vec x\rangle\equiv b \pmod 1$$ for all integer vectors $\vec a\in\mathbb Z^d$ and rational numbers $b$ for which the equation $$\langle \vec a,\vec\theta\rangle= b$$ holds. Then all numbers $b$ appearing in these equations have a least common denominator $g$. The sequence $n\vec\theta$ is uniformly distributed modulo 1 in the disjoint union of parallel subspaces $C\cup 2C\cup \dots \cup gC$.

The denominator $g$ is the smallest number $g\ge1$ for which $gC$ contains an integer point (or equivalently, for which $gC$ modulo 1 contains the origin and is therefore equivalent to its corresponding linear subspace $C-C$). If rational dependencies exist only among the numbers $\theta_1,\dots,\theta_d$ and not with the number 1, then the right-hand side $b$ is always $0$, and we set $g=1$.

It is clear that $n\vec\theta$ cycles through the $g$ sets $C,2C,3C,...$. Thus it suffices to look at the generating vector $g\vec\theta$ and prove that its multiples are uniformly distributed modulo one in $gC=C-C$. It is an exercise to reduce this case to the independent case, along the lines of the reduction that I gave for the continuous version. (In fact many of the arguments in that proof appear in Weyl's proof already.)

Answer 4

Maybe I am missing something, but as far as I understand it, you can always assume the $\theta_j$ to be linearly independent over $\mathbb{Q}$, this is just a base change. A precise formulation of the theorem (under this assumption) is e.g. Theorem 444 in Hardy-Wright, "An Introduction to the Theory of Numbers". They give, I think, 3 different proofs.