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.

  • $\begingroup$ First place I'd look would be the Kuipers and Niederreiter book, though I don't know whether it's in there in the form you want. $\endgroup$ – Gerry Myerson Apr 9 '14 at 12:57
  • $\begingroup$ If you just wish for a complete proof, I can send you a copy of my honours thesis, where I prove the result from scratch. But I doubt that'd suffice for a reference should you wish for a citation. That being said, the reason that I wrote out the proof in full is that at the time (2010) I couldn't find a reference where the result is proved for the case where the $\theta_j$ may not be linearly independent. $\endgroup$ – Peter Humphries Apr 9 '14 at 14:50
  • $\begingroup$ One way to get that would be to use Ratner's theorem (the topological one), which is stated as you would like, but that seems like an overkill to me, as you just dealing with a nilflow. If I recall correctly, such a statement as you wish (and even in a quantitative manner) appears in one of the Green-Tao papers (and therefore, the purely qualitative theorem is probably appearing in one of Berglson's papers). $\endgroup$ – Asaf Apr 9 '14 at 15:34
  • $\begingroup$ Another place to get your theorem is through Furstenberg's proof of the density/equidistribution theorem (depends on your favourite category of dynamics), as that the statement for extensions of minimal systems (topological dynamics) would conclude that the resulting extension is semi-simple (that is, union of tori). $\endgroup$ – Asaf Apr 9 '14 at 15:35
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    $\begingroup$ Asaf, did you check my reference? Witte Morris' book discusses the Kronecker-Weyl theorem directly in the first three sections of the first chapter (as a motivating case of Ratner's theorems), which is what Greg Martin was asking about. I didn't say that Witte Morris' book actually gave the complete proof of Ratner's theorems. In any case, the proof of the Kronecker-Weyl theorem is really a theorem of abelian Fourier analysis, so even the nilpotent case of Ratner's theorems is overkill. $\endgroup$ – Peter Humphries Apr 9 '14 at 18:18

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.

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  • $\begingroup$ I don't think at this point that there is a suitable reference in the literature, so I'm accepting this answer in recognition of all the trouble Peter went to. $\endgroup$ – Greg Martin Apr 15 '14 at 19:01

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.

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    $\begingroup$ This is incorrect; the theorem depends on the linear relations between the $\theta_j$, because this defines the subtorus! $\endgroup$ – Peter Humphries Apr 9 '14 at 14:50
  • $\begingroup$ For example, if $d=2$ and $\theta_2=2\theta_1$, no change of variables is going to make the ray fill up the entire two-dimensional torus. $\endgroup$ – Greg Martin Apr 9 '14 at 19:44
  • $\begingroup$ Sounds right to me. I don't understand how Peter and Greg think their comments are responsive. $\endgroup$ – Ben Wieland Apr 9 '14 at 20:03
  • $\begingroup$ Oh, I think I get it; you mean a base change such that $\theta_1,\ldots,\theta_r$ are linearly independent over $\mathbb{Q}$ and $\theta_{r+1},\ldots,\theta_n = 0$. But this base change is essentially equivalent to part (1) of the theorem in my answer, and then at the end for part (2) you're going to need to apply the inverse of this base change to get the result. $\endgroup$ – Peter Humphries Apr 9 '14 at 20:34
  • $\begingroup$ I checked Hardy-Wright. In addition to only giving the proof in the special case where the $\theta_j$ are linearly independent, which I specifically said does not suffice for my purposes, they only prove density, not uniform distribution. $\endgroup$ – Greg Martin Apr 10 '14 at 23:14

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