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. $\endgroup$ Apr 9, 2014 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. $\endgroup$ Apr 9, 2014 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, 2014 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, 2014 at 15:35
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 9, 2014 at 18:18

4 Answers 4


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.

  • $\begingroup$ 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. $\endgroup$ Apr 15, 2014 at 19:01
  • $\begingroup$ @Peter Humphries Thank you for the nicely explained answer. Could you kindly provide a reference for the Poisson summation formula for closed subgroups of $\mathbb T^n$ that you are using here? I am not very comfortable with the theory of topological groups and have been unable to find a reference. Thank you for your time. $\endgroup$
    – asrxiiviii
    May 30 at 10:46
  • 1
    $\begingroup$ @asrxiiviii I wrote this proof up in Appendix A of my masters thesis, which is available on my website. From that, I see that the reference that I used for the Poisson summation formula was Theorem 3.6.3 of "Principles of Harmonic Analysis" by Deitmer and Echterhoff. $\endgroup$ May 30 at 13:57

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 $[0,1)^k$.

(*) 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 $0$.
  • $\begingroup$ This isn't very clear; can you write it out in (a lot) more detail? Already the beginning of the second paragraph is quite ambiguous without more precise notation. $\endgroup$ Oct 22, 2020 at 2:15
  • $\begingroup$ I have expanded, for example by adding the definition of a lattice. Unfortunately, the English wikipedia treats only lattices of full rank $r$, The German wikipedia <de.wikipedia.org/wiki/Gitter_(Mathematik)> is more accurate. I looks like some motives are shared between your proof and mine, but we are using different languages. What I call a linear equation (with integer coefficients) seems to correspond to what you call a character. $\endgroup$ Oct 22, 2020 at 8:06
  • $\begingroup$ Thanks, this is much more readable (to me, at least). $\endgroup$ Oct 22, 2020 at 8:08
  • $\begingroup$ I found that the continuous version is already in Weyl's 1916 paper (Theorem 5, p.319-320) without detailed proof, for the rationally independent case, and also the monographs contain separate chapters about it. Weyl writes (after the related Theorem 6, on p.321): "Es würde keine Schwierigkeiten machen, die möglichen Ausnahmefälle, [...], vollständig durchzudiskutieren." (It would not pose any difficulties to discuss the potential exceptional cases completely.) $\endgroup$ Oct 29, 2020 at 5:36

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.)

  • 1
    $\begingroup$ There are indeed two versions of the Kronecker-Weyl theorem: the discrete version, for $n \in \mathbb{N}$, and the continuous version, for $t \in \mathbb{R}$. My answer above is for the continuous version; the proof is reproduced from Section 4.3 of my honours thesis drive.google.com/open?id=1YoQpDCO4wvyD9EFxfExEVNmtJkjDxBdv. A proof for the discrete version is given in Appendix A of my masters thesis drive.google.com/open?id=1KW0wdc4Ydh_pupHp9sM3zXPBBnG5gJPp. $\endgroup$ Oct 21, 2020 at 22:26
  • $\begingroup$ @A.Bailleul has also recently given written up nice proofs of both versions: arxiv.org/abs/2007.05763 $\endgroup$ Oct 21, 2020 at 22:28
  • 1
    $\begingroup$ strange that @A.Bailleul writes "It seems a reference to a proof of the discrete version of the Kronecker-Weyl theorem is hard to find in a published form". totally unaware of the classical Kuipers-Niederreiter book, or the more recent monograph by Drmota and Tichy from 1997, or by Hlawka 1984. (Or why not the proof by Weyl himself?) $\endgroup$ Oct 21, 2020 at 23:11
  • $\begingroup$ Bailleul is referring to the discrete version of the Kronecker-Weyl theorem on $\mathbb{T}^n$ without the assumption of linear independence. Where does this appear in these references? $\endgroup$ Oct 22, 2020 at 2:10
  • $\begingroup$ Bailleul continues his sentence: "... so we provide a proof in an appendix" (p.4). If you look in the appendix, you find the independence assumption right at the beginning (p.29). Also in Theorem 1.2. $\endgroup$ Oct 22, 2020 at 8:25

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.

  • 2
    $\begingroup$ This is incorrect; the theorem depends on the linear relations between the $\theta_j$, because this defines the subtorus! $\endgroup$ Apr 9, 2014 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. $\endgroup$ Apr 9, 2014 at 19:44
  • 1
    $\begingroup$ Sounds right to me. I don't understand how Peter and Greg think their comments are responsive. $\endgroup$ Apr 9, 2014 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. $\endgroup$ Apr 9, 2014 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. $\endgroup$ Apr 10, 2014 at 23:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.