1 Introduction

If a measure-theoretic dynamical system \((X,T,\mu )\), with \(\mu \) being a probability invariant by T, is invertible and ergodic, then for arbitrary \(\varepsilon > 0\) and \(h \in {\mathbb {N}}\), it is always possible to find a measurable set B such that \(B,TB,\ldots ,TB^{h-1}\) are disjoint sets and have joint measure greater than \(1-\varepsilon \). This result is known as the famous Rokhlin lemma. Its proof is constructive, and hence it allows for finding an explicit set B satisfying the mentioned properties, see Sect. 2 for more details. The finite sequence of sets \((T^kB)_{k=0}^{h-1}\) is called a Rokhlin tower and B is its basis. In this paper, we exclusively consider a special class of transformations, namely rotations \(f_\alpha : x \mapsto x + \alpha \mod 1\) by an irrational angle \(\alpha \) of the unit torus \({\mathbb {T}}_1 = [0,1)\) with ends of the interval glued, and aim to contribute to a better understanding of Rokhlin towers when an extra condition on the topological structure of the basis B is imposed. The map \(f_\alpha \) yields a uniquely ergodic measure-theoretic dynamical system on the torus where \(\mu \) is the Lebesgue-measure. The properties of \(f_\alpha \) and also of the structure of its Rokhlin towers have been comprehensively studied, see e.g. [1, 3, 6,7,8,9] to name only a few references.

Here, we ask how to find nice sets B for \(f_\alpha \) given \(\varepsilon > 0, h \in {\mathbb {N}}\). In [3], a method based on the three gap theorem (Theorem 2.1) was introduced how to calculate the minimal admissible value \(\varepsilon \) for \(f_\alpha \) if we choose B as an interval, see Theorem 1.5. Necessary and sufficient conditions on \(\alpha \) have been described when \(\varepsilon = 0\) can be achieved for arbitrary high h. In this paper, we address basis sets consisting of a finite union of intervals and analyze differences and similarities in comparison to the case of only one interval.

Before we come to the presentation of our results, we mention at first that the dynamical system on \({\mathbb {T}}_1\) defined by \(f_\alpha \) is not only uniquely ergodic (for the Lebesgue measure) but even a so-called system of rank one, see [7]. That means that it does not only satisfy the properties stated in the Rokhlin lemma, but approximates every partition arbitrarily well by Rokhlin towers, which means that the diameter of B can be chosen arbitrarily small. As we have already indicated, this still does not guarantee that B can always be chosen as an interval. For the formal definition of a system of rank one, we recall at first that the distance between two partitions \(P=\left\{ P_1,P_2,\ldots ,P_r\right\} , P'=\left\{ P_1',P_2',\ldots ,P_r'\right\} \) of X is defined by \(|P-P'| = \sum \mu (P_i \Delta P_i')\).

Definition 1.1

A measure-theoretic dynamical system \((X,T,\mu )\) is of rank one if for any partition P of X and any \(\varepsilon > 0\), there exists a subset \(B \subset X\), a positive height h, and a partition \(P'\) of X such that

  • \(B, TB, \ldots , T^{h-1}B\) are disjoint,

  • \(|P-P'| < \varepsilon \),

  • \(P'\) is refined by the partition made of the sets \(B,TB,\ldots ,T^{h-1}B\) and \(X {\setminus } \cup _{k=0}^{h-1} T^kB\).

In fact, the map \(f_\alpha \) has rank one because it is an ergodic transformation of the compact group \({{\mathbb {R}}}/ {{\mathbb {Z}}}\), compare [14]. The notion of covering numbers (see [3, 10]) is closely linked to the definition of systems of rank one, and gives a quantitative measure of how close \((X,T,\mu )\) is to being of rank one.

Definition 1.2

The covering number \(F^*(T)\) of \((X,T,\mu )\) is the supremum of all \(z \in {\mathbb {R}}\) such that for every partition \(P = \left\{ P_1,\ldots ,P_r \right\} \) of X and for every \(\varepsilon > 0\) and every \(h_0 \in {\mathbb {N}}\), there exists a subset \(B \subset X\), an integer \(h \ge h_0\), and a partition \(P' = \left\{ P_1',\ldots ,P_r' \right\} \) of X such that we have

  1. (i)

    \(B, TB, \ldots , T^{h-1}B\) are disjoint,

  2. (ii)

    \(\mu (\cup _{k=0}^{h-1}T^kB) \ge z\),

  3. (iii)

    \(\sum _{i=1}^r \mu ((P_i \Delta P_i') \cap (\cup _{k=0}^{h-1}T^kB)) < \varepsilon \),

  4. (iv)

    each \(P_i'\cap (\cup _{i=0}^{h-1}T^iB)\) is a union of sets \(T^jB\) for some \(0 \le j \le h-1\).

The fact that \(f_\alpha \) is a system of rank one for any \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) is thus equivalent to \(F^*(f_\alpha ) = 1\). We call an arc of the torus which is closed on the left and open on the right an interval. If we can approximate any partition, in the sense of Definition 1.1, by a tower whose basis B is an interval, we say that \((X,T,\mu )\) is of rank one by intervals. This idea can be transferred to Definition 1.2 by restricting covering number to unions of \(n_B \in {\mathbb {N}}\) disjoint intervals and considering \(n_B = 1\).

Definition 1.3

Let \(F_n(T)\) be the supremum of all \(z \in {\mathbb {R}}\) such that for every \(h_0 \in {\mathbb {N}}\), there exists \(h \ge h_0\) and a set B consisting of \(n_B = n\) disjoint intervals with

  1. (i)

    \(B, TB, \ldots , T^{h-1}B\) are disjoint,

  2. (ii)

    \(\mu (\cup _{i=0}^{h-1}T^iB) \ge z\).

Note that \(F_1(T) = 1\) means for a transformation T that it is rank one by intervals. For \(f_\alpha \), the property \(F_1(f_\alpha )=1\) depends on the continued fraction of \(\alpha \). Therefore, we briefly fix notation and summarize some of the important properties of continued fractions. For more details, we refer the reader e.g. to [2, 11]. Let \([a_0;a_1;a_2;\ldots ]\) with partial quotients \(a_i \in {\mathbb {N}}_0\) be the continued fraction expansion of \(\alpha \in {\mathbb {R}}\) and denote the corresponding sequence of convergents by \((p_n/q_n)_{n \in {\mathbb {N}}_0}\). Recall that

$$\begin{aligned} p_{-2}&= 0, p_{-1} = 1, p_n = a_np_{n-1} + p_{{n-2}}, n \ge 0, \\ q_{-2}&= 1, q_{-1} = 0, q_n = a_nq_{n-1} + q_{{n-2}}, n \ge 0, \end{aligned}$$

Theorem 1.4

Let \(\alpha \in {{\mathbb {R}}}{\setminus } {{\mathbb {Q}}}\). Then \(F_1^*(f_\alpha ) = 1\) if and only if \(\alpha \) has unbounded partial quotients.

The result was already mentioned in [14]. A complete proof using the three gap theorem is given in [3]. For \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) with continued fraction expansion \([a_0;a_1;a_2;\ldots ]\) and convergents \(p_n,q_n\), a basis of the Rokhlin towers (depending on \(\varepsilon \)) is given by

$$\begin{aligned} B_n = \left[ q_n \left| \frac{p_n}{q_n} - \alpha \right| , \frac{1}{q_n} - q_n \left| \frac{p_n}{q_n} - \alpha \right| \right) , \end{aligned}$$

compare [7, Theorem 6]. Note that \(B_n\) is well-defined for all n if and only if \(\alpha \) has unbounded partial quotients. In general, the covering number of a base set consisting of \(n_B=1\) interval can be precisely calculated by the following theorem.

Theorem 1.5

(Checkhova, [3]). Let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) have continued fraction expansion \([a_0;a_1;a_2;\ldots ]\) with convergents \((p_n,q_n)\) and define \(v_n = [0;a_n;a_{n-1}; \ldots ;a_1]\) and \(t_n = [0;a_{n+1};a_{n+2};\ldots ]\). Then

$$\begin{aligned} F_1(f_\alpha ) = \lim _{n \rightarrow \infty } q_{n+1} |q_n \alpha - p_n| = \limsup _{n \rightarrow \infty } \frac{1}{1+t_nv_n}. \end{aligned}$$

Hence, \(F_1(f_\alpha ) = 1\) if and only if \(\alpha \) has unbounded partial quotients and \(F_1(f_\alpha ) \ge \frac{1}{5 + \sqrt{5}}{10}\) for all \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\). The minimal value of \(F_1(f_\alpha )\) is attained for any \(\alpha \in [0,1) \cap {\mathbb {Z}}\varphi + {\mathbb {Z}}\), where \(\varphi = \frac{1+\sqrt{5}}{2}\) is the golden mean.

If the basis does not only consist of one interval but is a union of several, we may without loss of generality assume that B is (after rotation) of the form

$$\begin{aligned} B = \underbrace{[c_1=0,\beta _1)}_{=:B_1} \cup \underbrace{[c_2, c_2 + \beta _2)}_{=:B_2} \cup \cdots \cup \underbrace{[c_{n_B}, c_{n_B} + \beta _{n_B})}_{=:B_{n_B}} \end{aligned}$$
(1)

with \(1 \ge c_i \ge c_{i-1} + \beta _{i-1}\) for all \(2 \le i \le n_B\), and \(c_i, \beta _i > 0\). As an analogue of Theorem 1.5, we can precisely calculate the covering number if \(n_B = 2\).

Theorem 1.6

Let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) with continued fraction expansion \([a_0;a_1;a_2;\ldots ]\).

  1. (i)

    If \(a_i = 1\) eventually, then \(F_2(f_\alpha ) = \frac{4\varphi +3}{10} = \frac{2\sqrt{5}+5}{10} > F_1(f_\alpha )\).

  2. (ii)

    If \(a_i > 1\) eventually, then we have

    $$\begin{aligned} F_2(f_\alpha ) = F_1(f_\alpha ). \end{aligned}$$

In contrast to \(F_1(f_\alpha )\), the covering number by two intervals \(F_2(f_\alpha )\) does thus not attain its minimum for \(\alpha = \varphi = \frac{1+\sqrt{5}}{2}\) because \(F_2(f_\varphi ) = \frac{2\sqrt{5}+5}{10}\) and for example for \(\beta = 1 + \sqrt{2}\), we have \(F_2(f_\beta ) = F_1(f_\beta ) = 1/(4-2\sqrt{2}) < F_2(f_\varphi )\) by Theorem 1.5. The proof of Theorem 1.6 relies on a more deep going application of the three gap theorem (Theorem 2.1) than in [3]. The reason why this is possible is that the combinatorics behind the orbits of the endpoints of the base intervals can still be controlled if \(n_B = 2\). As the number of base intervals \(n_B\) increases, the situation gets more and more complicated but we can nonetheless give a lower bound when \(F_k(f_\alpha ) > F_1(f_\alpha )\) holds for a given \(\alpha \).

Theorem 1.7

We have

$$\begin{aligned} F_k(f_\alpha ) > F_1(f_\alpha ) \end{aligned}$$

for all \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) with \(\limsup _{i \rightarrow \infty } a_i \le k-1\).

From Theorem 1.6 and numerical experiments, we conjecture that \(F_k(f_\alpha ) > F_1(f_\alpha )\) actually holds if and only if the partial quotients satisfy \(\limsup _{i \rightarrow \infty } a_i \le k-1\). From a quantitative point of view, Theorem 1.7 is nonetheless not very satisfactory because the lower bound therein is constant given \(\alpha \). For this reason, we also describe the behaviour of \(F_n(f_\alpha )\) as \(n \rightarrow \infty \). A lower bound in n can even be made explicit if the partial quotients of \(\alpha \) are constant. Our proof of this result makes use of the explicit construction of base sets B in the (standard) proof of the Rokhlin lemma, see Theorem 2.3 and explanations thereafter.

Theorem 1.8

Let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) have continued fraction expansion \([a_0;a_1;a_2;\ldots ]\) and assume that \(a_i \le s\) for all \(i \in {\mathbb {N}}\). Then \(F_n(f_\alpha ) \rightarrow 1\) for \(n \rightarrow \infty \). Moreover, if \(\alpha _i = s\) for all \(i \in {\mathbb {N}}\), then \(F_n(f_\alpha )\) can for \(n \ge \alpha ^{j-1}(\alpha +1)\) be bounded from below by

$$\begin{aligned} F_n(f_\alpha ) \ge F_1(f_\alpha ) \frac{1}{\alpha ^{j+2}}\left( \lfloor \alpha ^{j+1} \rfloor \lfloor \alpha \rfloor + \lfloor \alpha ^j \rfloor \right) . \end{aligned}$$
(2)

Note that assuming \(a_i \le s\) is not a real restriction because in the unbounded case we have \(F_1(f_\alpha ) = 1\) by Theorem 1.4 and thus \(F_n(f_\alpha ) = 1\) for all \(n \in {\mathbb {N}}\).

Remark 1.9

If the continued fraction expansion of \(\alpha \in {\mathbb {R}}\) has constant partial quotients, i.e., \(\alpha = [s;s;s;\ldots ]\), then \(\alpha = \frac{s+\sqrt{s^2+4}}{2}\), and

$$\begin{aligned} F_1(f_\alpha ) = \frac{1+\alpha ^2}{\alpha ^2} \end{aligned}$$

follows from Theorem 1.5. This implies that the explicit expression for the lower bound of \(F_n(f_\alpha )\) in (2) converges to 1 for \(n \rightarrow \infty \).

2 Proofs of main results

Three gap theorem. For \(\alpha \in {\mathbb {R}}\), the Kronecker sequence is defined as \((\left\{ n\alpha \right\} )_{n \in {\mathbb {N}}}\), where \(\left\{ x \right\} := x - \lfloor x \rfloor \) denotes the fractional part of \(x \in {\mathbb {R}}\). The Kronecker sequence may either be interpreted as a sequence on the unit interval [0, 1) or, after gluing its endpoints, on the torus. If the latter viewpoint is applied, the norm of a point \(x \in {{\mathbb {R}}}\) defined by \(\left\Vert x\right\Vert := \min (x-\lfloor x \rfloor , 1-(x-\lfloor x \rfloor ))\) is used. It measures the distance of \(x \in {\mathbb {R}}\) from the next integer.

Most of the results of this paper are based on the so-called three gap theorem which links the gap structure, i.e., the (Euclidean) distance of neighbouring elements of the Kronecker sequence on the torus, to the continued fraction expansion of \(\alpha \). Its first proof goes back to Sós in [13] but it turns out to be useful for us to formulate it here in terms of the Ostrowski expansion, as it is implicitly done in [12], see also [15] for a slightly different version. For that purpose, let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) have continued fraction expansion \([a_0;a_1;a_2;\ldots ]\) with convergents \(p_n/q_n\). Then for \(N \in {\mathbb {N}}\), the Ostrowski expansion of N is uniquely given by

$$\begin{aligned} N = \sum _{n=1}^N b_n q_n \end{aligned}$$

with \(0 \le b_n \le a_n\) and \(b_{n-1} = 0\) if \(b_n = a_n\).

Theorem 2.1

(Three gap theorem). Let \((\left\{ n\alpha \right\} )_{n \in {\mathbb {N}}}\) be the Kronecker sequence of \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\). Let \(N \in {\mathbb {N}}\) have the Ostrowski expansion

$$\begin{aligned} N = \sum _{n=1}^N b_n q_n. \end{aligned}$$

Then the gaps that can appear have lengths

$$\begin{aligned} L_1&= \left\Vert \left\{ q_n z \right\} \right\Vert ,\\ L_2&= \left\Vert \left\{ q_{n-1} z \right\} \right\Vert - (b_n-1)L_1 - \min (b_{n-1},1)L_1,\\ L_3&= L_1 + L_2, \end{aligned}$$

and their multiplicities are

$$\begin{aligned} N_1&= N - q_n,\\ N_2&= (b_{n-1}-1)q_{n-1} + \sum _{n=1}^{N-2} b_n q_n,\\ N_3&= N - N_1 - N_2. \end{aligned}$$

In fact, it is necessary to understand the dynamical behaviour of the Kronecker sequence in order to derive information about the covering numbers of unions of \(n_B\) disjoint intervals from it. This can be regarded as an extension of the three gap theorem. Our description here is inspired by the presentation in [3] and [15]: If \(N = q_i\) is increased to \(N'=q_i + q_{i-1} \le q_{i+1}\), then the former small gaps (for N) become large (for \(N'\)) and every former large gap gets split up into a large and a small gap (both for \(N'\)). If \(a_{i+1} > 2\), every large gap for \(N' = q_i + q_{i-1}\) gets split up into a small gap (same length as for \(N'\)) and a gap of length smaller than the large gap length (for \(N'\)) but greater than the small gap length (for \(N'\)) until we reach \(N'' = 2q_i + q_{i-1}\). This process is repeated until \(N''' = (a_{i+1}-1)q_i + q_{i-1}\). Afterwards, every large gap length (for \(N'''\)) gets split up into a small gap smaller than for \(N'''\) and a now large gap for \({\tilde{N}}\) (which is equal to the small gap length for \(N'\)), and this process ends when \({\tilde{N}} = q_{i+1}\) and then starts all over again. In total, a small gap for \(N=q_i\) is split up into \(a_{i+1}-1\) small gaps for \({\tilde{N}}=q_{i+1}\) and 1 large gap for \({\tilde{N}}=q_{i+1}\) when passing from \(q_i\) to \(q_{i+1}\). Similarly, a large gap for \(N=q_{i}\) is split up into \(a_{i+1}\) small gaps for \({\tilde{N}}=q_{i+1}\) and 1 large gap for \({\tilde{N}}=q_{i+1}\).

These observations imply four fundamental but important facts for disjoint orbits of sets of the form described in (1) with \(n_B = 2\).

Lemma 2.2

Let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) with continued fraction expansion \([a_0;a_1;a_2;\ldots ]\) and convergents \(p_n,q_n\). Let B be a set of the form (1) and let \(q_n < N \in {\mathbb {N}} \le q_{n+1}\) be arbitrary. If \(B, f_\alpha B, \ldots , f_\alpha ^{N-1}B\) are disjoint, then

  1. (i)

    \(\beta _i \le \left\Vert q_n \alpha \right\Vert \) holds for all \(\beta _i\).

  2. (ii)

    If \(n_B = 2\) and \(a_{n+1} > 1\), then B can be chosen such that \(\beta _1 + \beta _2 = 2\left\Vert q_n \alpha \right\Vert \) if and only if \(q_n < N \le \frac{1}{2}q_{n+1}\).

  3. (iii)

    If \(n_B = 2\) and \(a_{n+1} > 1\) and \(\left\Vert q_n \alpha \right\Vert< \beta _1 + \beta _2 < 2\left\Vert q_n \alpha \right\Vert \), then \(\beta _1 + \beta _2 \le \left\Vert q_{n-1}\alpha \right\Vert - (a_{n+1}-1)\left\Vert q_n\alpha \right\Vert \) and \(q_{n+1} < 2N \le q_{n+1}+q_n\). If \(q_{n+1} < 2N \le q_{n+1} + q_n\), then there exists a B with \(\beta _1+\beta _2 = \left\Vert q_{n-1}\alpha \right\Vert - (a_{n+1}-1)\left\Vert q_n\alpha \right\Vert \).

  4. (iv)

    If \(n_B=2\) and \(a_{n+1} = 1\) and \(\left\Vert q_n \alpha \right\Vert < \beta _1 + \beta _2\), then \(\beta _1 + \beta _2 \le \left\Vert q_{n+1}\alpha \right\Vert + \left\Vert q_{n+2}\alpha \right\Vert = \left\Vert q_n\alpha \right\Vert \) and \(q_{n} < N \le q_{n} + q_{n-1}/2\). If \(q_{n} < N \le q_{n} + q_{n-1}/2\), then there exists a B with \(\beta _1+\beta _2 = \left\Vert q_n\alpha \right\Vert \).

Observe that if \(a_{n+1} = 1\), then the situation described in part (ii) cannot occur because the long gap length is less than twice the small one in this case.

Proof

Assertion (i) is immediately clear by the three gap theorem since the orbits of the intervals \(B_i\) have to be disjoint.

For the second assertion, the following line of argument leads to the desired result: let us assume first that \(a_{n+1} > 2\) is even. If \(N = q_{n} + q_{n-1}\), then there are \(q_{n-1}\) small gaps of size \(\left\Vert q_n \alpha \right\Vert \) and \(q_n\) large gaps of size \(\left\Vert q_{n-1}\alpha \right\Vert \). At most \(a_{n+1}\) small gaps fit into a large gap. In order to have \(\beta _1 = \beta _2 = \left\Vert q_n \alpha \right\Vert \), there needs to be an extra small space on the right of each block of small gaps. However, for \(N = a_{n+1}/2 \cdot q_n\), the gap structure of the sequence is as follows: it consists solely of blocks with \((a_{n+1}/2-1)\) small gaps followed either by a medium or a large sized gaps. In each of the following j steps, one of the large gaps gets, as usual, split up into a medium and a small size gap, i.e., there are \(q_{n-1}-j\) large gaps while we get j additional small gaps. Therefore, we must have \(2j \le q_{n-1}\). This condition implies \(2N \le q_{n+1}\) and is necessary for \(\beta _1+\beta _2 = 2 \left\Vert q_n \alpha \right\Vert \). If \(a_{n+1} > 1\) is odd, then for \(N = \lfloor a_{n+1}/2 \rfloor \cdot q_n\), the gap structure of the sequence is as follows: it consists solely of blocks with \((\lfloor a_{n+1}/2 \rfloor -1)\) small gaps followed either by a medium or a large sized gaps and a similar argument as in the even case can be applied. Now let N satisfy the assumptions of (ii) and choose \(\beta _1 = \beta _2 = \left\Vert q_n \alpha \right\Vert \) and \(c_1 = \left\Vert N\alpha \right\Vert \). The orbit of the left endpoints of B thus equals the Kronecker sequence \(\alpha ,\ldots ,N\alpha ,(N+1)\alpha ,\ldots ,2N\alpha \). As \(2N \le q_{n+1}\), the joint sequence has minimal gap length \(\left\Vert q_n \alpha \right\Vert \) and thus the orbit of B is disjoint. Finally, in the case \(a_{n+1} = 2\) and \(N = q_n\), there are \(q_{n-1}\) large gaps and \(q_n-q_{n-1}\) small gaps and the gap sizes satisfy \(\left\Vert q_{n-1}\alpha \right\Vert< \left\Vert q_{n-2}\alpha \right\Vert < 2\left\Vert q_{n-1}\alpha \right\Vert \). Therefore \(N \le q_n + q_{n-1}/2 = \frac{1}{2}q_{n+1}\) must hold if \(\beta _1 = \beta _2 = \left\Vert q_n \alpha \right\Vert \). Again, the upper bound on \(\beta _1 + \beta _2\) can be realized by a rotation with angle, i.e., \(c_1 = \left\Vert N\alpha \right\Vert \)

If \(N \le \frac{1}{2}q_{n+1}\), then we are in the situation of (ii). So we may restrict to the case \(N > \frac{1}{2}q_{n+1}\) when we want to prove (iii). The maximal possible value for \(\beta _1 + \beta _2\) is indeed \(\left\Vert q_{n-1}\alpha \right\Vert - (a_{n+1}-1)\left\Vert q_n\alpha \right\Vert \) and this is the only possibility because \(\left\Vert q_{n-1}\alpha \right\Vert - a_{n+1} \left\Vert q_n\alpha \right\Vert < \left\Vert q_n\alpha \right\Vert \). Note that in the \(\lfloor q_{n}/2 \rfloor \) steps between the upper and the lower bound for N every medium and large gap gets diminished by \(\left\Vert q_n \alpha \right\Vert \). Thus it follows that \(\beta _1 + \beta _2 = \left\Vert q_{n-1}\alpha \right\Vert - (a_{n+1}-1)\left\Vert q_n\alpha \right\Vert \) implies \(2N \le q_{n+1} + q_n\). A rotation with angle, i.e., \(c_1=\left\Vert N\alpha \right\Vert \), yields the claim as for (ii).

In order to prove (iv), we only note that \(\beta _1+\beta _2\) can be realized by a rotation with angle \(c_1 = \left\Vert N\alpha \right\Vert \) because \(2N \le 2q_n + q_{n-1} = q_{n+1} + q_n\) and the three gap theorem. Counting the number of large gaps yields the upper bound on N. \(\square \)

Theorem 1.6 can be deduced by using Lemma 2.2.

Proof of Theorem 1.6

Let us at first take a look at the general setting for any set

$$\begin{aligned} B = [0,\beta _1) \cup [c_2, c_2 + \beta _2) \end{aligned}$$

consisting of two intervals. By Lemma 2.2 (iii) and (iv), it follows, for \(2N > q_n + q_{n-1}\), that

$$\begin{aligned} \mathrm{{vol}}(B \cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) \le q_{n+1} \left\Vert q_n \alpha \right\Vert \le F_1(f_\alpha ). \end{aligned}$$

If \(2N < q_{n+1}\), Lemma 2.2 (ii) yields

$$\begin{aligned} \mathrm{{vol}}(B \cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) = 2 \left\Vert q_n \alpha \right\Vert N < q_{n+1} \left\Vert q_n \alpha \right\Vert \le F_1(f_\alpha ). \end{aligned}$$

The cases which remain to be checked are the maximal values of N in Lemma 2.2 (iii) and (iv).

(i) If \(a_n=1\) eventually and \(N=q_{n+1} + \lfloor \tfrac{1}{2}q_n \rfloor \), then

$$\begin{aligned} \mathrm{{vol}}(B&\cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) = \left\Vert q_n \alpha \right\Vert N \\&= (q_{n+1}+\lfloor \tfrac{1}{2}q_n\rfloor ) \left\Vert q_n \alpha \right\Vert \end{aligned}$$

follows by Lemma 2.2 (iv). The right hand side of the equality converges to

$$\begin{aligned} \frac{1}{1+\frac{1}{\theta }\frac{1}{\theta }} + \frac{1}{2} \frac{1}{\theta } \frac{1}{1+\frac{1}{\theta }\frac{1}{\theta }} = \frac{4\theta +3}{10} = \frac{2\sqrt{5}+5}{10}. \end{aligned}$$

(ii) For \(a_{n+1} \ge 3\) and \(N=\lfloor \tfrac{1}{2} (q_n+q_{n+1}) \rfloor \), we get

$$\begin{aligned} \mathrm{{vol}}(B&\cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) = (\left\Vert q_n \alpha \right\Vert +\left\Vert q_{n+1}\alpha \right\Vert ) N \\&\le \frac{1}{2}(q_{n+1}+q_n) (\left\Vert q_n \alpha \right\Vert + \left\Vert q_{n+1}\alpha \right\Vert )\\&\le \frac{1}{2}F_1(f_\alpha ) + \frac{1}{6}F_1(f_\alpha ) + \frac{1}{6}(f_\alpha ) + \frac{1}{9}F_1(f_\alpha ) < F_1(f_\alpha ). \end{aligned}$$

For \(a_{n+1} =2\) and \(N=\lfloor \tfrac{1}{2} (q_n+q_{n+1}) \rfloor \), Lemma 2.2 (iii) implies

$$\begin{aligned} \mathrm{{vol}}(B&\cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) = (\left\Vert q_n \alpha \right\Vert +\left\Vert q_{n+1}\alpha \right\Vert ) N \\&= \frac{1}{2} (q_{n+1} + q_n) (\left\Vert q_n \alpha \right\Vert +\left\Vert q_{n+1}\alpha \right\Vert ). \\&= \frac{1}{2} q_{n+1} \left\Vert q_n\alpha \right\Vert \left( 1 + \frac{q_n}{q_{n+1}} + \frac{\left\Vert q_{n+1}\alpha \right\Vert }{\left\Vert q_n\alpha \right\Vert } + \frac{q_n}{q_{n+1}} \frac{\left\Vert q_{n+1}\alpha \right\Vert }{\left\Vert q_n\alpha \right\Vert }\right) . \end{aligned}$$

If \(a_{n+1} = 2\) eventually, then \(q_{n}/q_{n+1} \rightarrow \frac{1}{2+\sqrt{2}} = \sqrt{2}-1\) and \(\left\Vert q_{n+1}\alpha \right\Vert /\left\Vert q_n\alpha \right\Vert \rightarrow \frac{1}{2+\sqrt{2}}\). If \(a_{n+1} = 2\) and \(a_n > 2\), then \(q_n/q_{n+1} < \frac{1}{2+\sqrt{2}}\). As \((1 +\frac{1}{2+\sqrt{2}})^2 = 2\), we have in any case \(\lim _{N \rightarrow \infty } \mathrm{{vol}}(B \cup f_\alpha B \cup \cdots \cup f_\alpha ^{N-1}B) \le F_1(f_\alpha )\). In conclusion, we always get \(F_2(f_\alpha ) \le F_1(f_\alpha )\). \(\square \)

From the observation in the proof of Lemma 2.2 that the choice \(c_1 = N\alpha \) suffices to attain the maximum in the case of two intervals, it is possible to also derive a lower bound on \(F_k\) in the case \(a_i < k\) for all but finitely many \(i\in {\mathbb {N}}\). This is the content of Theorem 1.7.

Proof of Theorem 1.7

Let \(k^* = \limsup _{i \rightarrow \infty } a_i\). By considering the corresponding subsequence, we may without loss of generality assume \(a_i = k^*\) for all \(i \in {\mathbb {N}}\). Partitioning the complete Kronecker sequence \(z = \left\{ j\alpha \right\} _{j=1}^{q_{n+1}}\) into \(k^*\) blocks of elements of size \(\lfloor q_{n+1}/k^* \rfloor \) yields

$$\begin{aligned} F_{k^*+1}(f_\alpha ) \ge (k^*-1) \lfloor \frac{q_{n+1}}{k^*} \rfloor \left\Vert q_n \alpha \right\Vert + \lfloor \frac{q_{n+1}}{k^*} \rfloor (\left\Vert q_{n}\alpha \right\Vert + \left\Vert q_{n+1}\alpha \right\Vert ), \end{aligned}$$

where the length \(\left\Vert q_{n}\alpha \right\Vert + \left\Vert q_{n+1}\alpha \right\Vert \) stems from the last block. \(\square \)

Explicit Rokhlin towers. In order to improve the lower bound for the area covered by \(B,f_\alpha B,\ldots ,f_\alpha ^{h-1}B\) for circle rotations by \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\), we will make use of the construction used in the standard proof of the Rokhlin lemma, as it is presented e.g. in [5, Lemma 2.45].

Theorem 2.3

(Rokhlin lemma). Let \((X,T,\mu )\) be an invertible ergodic measure-theoretic dynamical system with non-atomic \(\mu \). Then for any \(h \ge 1\) and \(\varepsilon > 0\), there is a measurable set B such that \(B,TB,\ldots ,T^{h-1}B\) are disjoint and

$$\begin{aligned} \mu (B\cup TB \cup \cdots \cup T^{h-1}B) > 1 - \varepsilon . \end{aligned}$$

We therefore shortly recall the construction in the proof of Theorem 2.3 at first. Choose a measurable set A with \(0<\mu (A)<\varepsilon /h\) and define the sets

$$\begin{aligned} A_1&= A \cap T^{-1}A,\\ A_2&= A \cap T^{-2}A {\setminus } A_1,\\&\vdots \\ A_n&= A \cap T^{-n}A {\setminus } \bigcup _{i<n} A_i. \end{aligned}$$

The sets \(A_i\) are disjoint and so are the unions

$$\begin{aligned} A_k \cup TA_k \cup \cdots \cup T^{k-1}A_k \end{aligned}$$

for all \(k \ge 1\). Then the base set B defined by

$$\begin{aligned} B = \bigcup _{k \ge n} \bigcup _{j=0}^{\lfloor k/n \rfloor -1} T^{jn}A_k \end{aligned}$$

satisfies the properties mentioned in Lemma 2.3.

Now let \(\alpha \in {\mathbb {R}} {\setminus } {\mathbb {Q}}\) and let \(p_n,q_n\) be the sequence of the convergents. According to classical continued fraction theory, \(\left\Vert q_n \alpha \right\Vert \rightarrow 0\) for \(n \rightarrow \infty \) holds. For \(\varepsilon > 0\) and \(n \in {\mathbb {N}}\) arbitrary, set \(\varepsilon ' := \varepsilon / n\). In order to apply the following algorithm, we furthermore need to impose the condition that \(\varepsilon ' < 1\) (although this might seem to be a trivial remark, it will indeed have some importance in the remainder of the proof). Assume that \(\left\Vert q_k\alpha \right\Vert < \varepsilon ' \le \left\Vert q_{k-1}\alpha \right\Vert \). Given \(a_{k+1}>1\), the value \(\varepsilon '\) lies in one of the following intervals

$$\begin{aligned}&\underbrace{(\left\Vert (q_k+q_{k-1})\alpha \right\Vert ,\left\Vert q_{k-1}\alpha \right\Vert ]}_{:=I_0}, \quad \underbrace{(\left\Vert (2q_k+q_{k-1})\alpha \right\Vert ,\left\Vert (q_k+q_{k-1})\alpha \right\Vert ]}_{:=I_1}, \ldots , \\&\underbrace{(\left\Vert q_{k}\alpha \right\Vert ),\left\Vert ((a_{k+1}-1)q_k+q_{k-1})\alpha \right\Vert ]}_{:=I_{a_{k+1}}}. \end{aligned}$$

If \(\left\{ q_k\alpha \right\} > 0.5\), then \(\left\{ (iq_k + q_{k-1})\alpha \right\} < 0.5\) for \(1 \le i \le a_{k+1}\). If \(\varepsilon ' \in I_{j-1}\), let \(l_0(k) := q_k, l_1(k) := jq_k + q_{k-1}\), and \(l_2(k) := (j+1)q_k + q_{k-1}\) and \(\alpha _{k,0} := \left\{ l_0(k)\alpha \right\} , \alpha _{k,1} = \left\{ l_1(k)\alpha \right\} \), and \(\alpha _{k,2} = \left\{ l_2(k)\alpha \right\} \). Define the sets \(A_i\) as in the proof of the Rokhlin lemma. These intervals are explicitly given by

$$\begin{aligned} A_1&= A \cap f_\alpha ^{-1}A = \emptyset ,\\&\vdots \\ A_{l_0(k)-1}&= A \cap f_\alpha ^{-(l_0(k)-1)}A = \emptyset ,\\ A_{l_0(k)}&= A \cap f_\alpha ^{-l_0(k)}A = [\alpha _{k,0},\varepsilon '), \\ A_{l_0(k)+1}&= A \cap f_\alpha ^{-(l_0(k)+1)}A = \emptyset ,\\&\vdots \\ A_{l_1(k)-1}&= A \cap f_\alpha ^{-(l_1(k)-1)}A = \emptyset ,\\ A_{l_1(k)}&= A \cap f_\alpha ^{-l_1(k)}A = [0,\varepsilon '-\alpha _{k,1}), \\ A_{l_1(k)+1}&= A \cap f_\alpha ^{-(l_1(k)+1)}A = \emptyset ,\\&\vdots \\ A_{l_2(k)-1}&= A \cap f_\alpha ^{-(l_2(k)-1)}A = \emptyset ,\\ A_{l_2(k)}&= A \cap f_\alpha ^{-l_2(k)}A = [\varepsilon '-\alpha _{k,1},\alpha _{k,0}), \end{aligned}$$

and \(A_i = \emptyset \) for all \(i > l_2(k)\). Note that \(A_{l_1(k)} = \emptyset \) might happen (if the left and the right endpoint of the interval are equal). More precisely, we observe that only two intervals occur excatly if \(\varepsilon '\) is a right endpoint of one of the \(I_j\). For the cases \(\left\{ q_k\alpha \right\} < 0.5\) and \(a_{i+1} = 1\), the calculation of the \(A_i\) can be treated similarly. Thus, we may without loss of generality restrict our analysis to the case \(\left\{ q_k\alpha \right\} > 0.5\), which was described above explicitly.

If \(n=q_{k+1}\) and \(\varepsilon ' = \left\Vert q_k\alpha \right\Vert \), then it can be easily checked that \(l_0(k) = q_{k+1}, l_1(k) = q_{k+1} + q_k\) and these are the only non-empty sets \(A_i\). Since \(\lfloor l_1(k) / n \rfloor = 0\), it follows that B consists of one interval only. Moreover, the Rokhlin tower built by the base B satisfies \(\limsup _{n \rightarrow \infty } \mathrm{{vol}}(B \cup f_\alpha B \cup \cdots \cup f_\alpha ^{n-1}B) = F_1(f_\alpha )\). Thus, we see that for \(\varepsilon = q_{k+1} \left\Vert q_k\alpha \right\Vert < 1\), the construction from the Rokhlin lemma yields a basis consisting of one interval only which covers the maximal possible area.

Writing \(\varepsilon := n \varepsilon '\), our idea is now to keep n constant while alternating \(\varepsilon '\). Due to our preparatory work, the actual proof of Theorem 1.8 can be kept relatively short now.

Proof of Theorem 1.8

Let \(\varepsilon ' = \left\Vert q_{k+j}\alpha \right\Vert \) for some \(j \in {\mathbb {N}}_0\). The intervals \(A_{l_0(k+j)}\) and \(A_{l_1(k+j)}\) are adjacent by construction. Therefore the basis B consists of \(\lfloor l_1(k+j)/n \rfloor = \lfloor (q_{k+j+1} + q_{k+j})/n \rfloor \) distinct intervals. Hence \(\lfloor l_1(k+j)/n \rfloor \le \alpha _s^{j-1}(\alpha _s+1)\) for \(j \gg 1\) big enough. The area covered by \(B,f_\alpha B,\ldots ,f_\alpha ^{n-1}B\) is equal to

$$\begin{aligned} q_{k+1} \lfloor q_{k+j+1} / q_{k+1} \rfloor \left\Vert q_{k+j} \alpha \right\Vert + q_{k+1} \lfloor q_{k+j} / q_{k+1} \rfloor \left\Vert q_{k+j+1}\alpha \right\Vert . \end{aligned}$$
(3)

From the three gap Theorem 2.1, it follows that \(q_{k+j+1}\left\Vert q_{k+j}\alpha \right\Vert + q_{k+j}\left\Vert q_{k+j+1}\alpha \right\Vert =1\). Therefore, expression (3) converges to 1 as \(j \rightarrow \infty \) which implies \(F_n(\alpha ) \rightarrow 1\) as \(n \rightarrow \infty \). If \(a_i = s\) for all \(i \in {\mathbb {N}}\), then (3) can be bounded from below by

$$\begin{aligned} q_{k+1} \cdot&\left( \lfloor \alpha ^{j+1} \rfloor \left\Vert q_{k+j}\alpha \right\Vert + \lfloor \alpha ^j \rfloor \left\Vert q_{k+j+1}\alpha \right\Vert \right) \ge q_{k+1} \cdot \left\Vert q_{k+j+1}\alpha \right\Vert \left( \lfloor \alpha ^{j+1} \rfloor \lfloor \alpha \rfloor + \lfloor \alpha ^j \rfloor \right) \\&\ge q_{k+1} \cdot \left\Vert q_k\alpha \right\Vert \frac{1}{\alpha ^{j+2}} \left( \lfloor \alpha ^{j+1} \rfloor \lfloor \alpha \rfloor + \lfloor \alpha ^j \rfloor \right) . \end{aligned}$$

Applying \(\limsup \) with respect to k implies the desired result. \(\square \)

3 Constructive geometric definition of systems of rank one

At the end of this paper, we would like to comment on an alternative definition of systems of rank one, namely the constructive geometric one, and thereby round off the presentation of the topic. An advantage of Definition 3.1 in comparison to Definition 1.1 is that the bullet points in Definition 3.1 give an explicit possibility how to define a system of rank one and thereby yield an infinite class of examples. A proof for the equivalence of the two definitions can be found e.g. in [6, Lemme 16].

Definition 3.1

A dynamical system \((X,T,\mu )\) is of rank one if there exists a sequence of positive integers \(q_n, n \in {{\mathbb {N}}}\) and \(a_{n,i}, n \in {{\mathbb {N}}}, 1 \le i \le q_n -1\), such that if \(h_n\) is defined by

$$\begin{aligned} h_0 = 1, \qquad h_{n+1} = q_nh_n + \sum _{i=1}^{n-1}a_{n,i}, \end{aligned}$$

then

$$\begin{aligned} \sum _{n=0}^\infty \frac{h_{n+1}-q_nh_n}{h_{n+1}} < \infty , \end{aligned}$$

and subsets \(B_n \subset X, n \in {{\mathbb {N}}}, B_{n,i}, n \in {{\mathbb {N}}}, 1 \le i \le q_n\), and \(C_{n,i,j}, n \in {{\mathbb {N}}}, 1 \le i \le q_{n}-1, 1 \le j \le a_{n,i}\), such that for all n,

  • the \(B_{n,i}, 1 \le i \le q_n\), form a partition of \(F_n\),

  • the \(T^kB_n, 1 \le k \le h_{n}-1\), are disjoint,

  • \(T^{h_n}B_{n,i} = C_{n,i,1}\) if \(a_{n,i} \ne 0\) and \(i < q_n\),

  • \(T^{h_n}B_{n,i} = B_{n,i+1}\) if \(a_{n,i} = 0\) and \(i < q_n\),

  • \(TC_{n,i,j} = C_{n,i,j+1}\) if \(j < a_{n,i}\),

  • \(TC_{n,i,a_{n,i}} = B_{n,i+1}\) if \(j < a_{n,i}\),

  • \(B_{n+1} = B_{n,1}\),

and the partitions \(\left\{ B_n, TB_n, \ldots , T^{h_n-1}B_n, X {\setminus } \cup _{k=0}^{h_n-1}T^k B_n \right\} \) converge to the Lebesgue \(\sigma \)-algebra of X.

We call a rank one map (CG) rank one by intervals if the sets \(B_n\) can be chosen as intervals. A detailed description of the action of T for the case that X is the unit torus can be found e.g. in [8].

If, given a system of rank one in the sense of Definition 1.1, we were able to find the sets and calculate the coefficients in Definition 3.1, one of the main outcomes would be that the star-discrepancy (see [11]) of T-orbits could be easily calculated, see [4, Théorème 2.6] or [8, Theorem 86]. On the downside of Definition 3.1, this task seems to be very hard also for dynamical systems for which it is comparably easy to prove that they are of rank one in the sense of Definition 1.1 (despite that the proof of the equivalence of the two definitions is even partially though not completely constructive, compare [6, Lemme 16]). At the end of this article, we give a theoretical justification what makes this task particularly hard for rotations. The reason is that the sets \(B_n\) can never be chosen as intervals.

Theorem 3.2

For \(\alpha \in {{\mathbb {R}}}{\setminus } {{\mathbb {Q}}}\) arbitrary, the rotation map \(f_\alpha : x \mapsto x + \alpha \) is not (CG) rank one by intervals.

Proof

Assume that the claim is false and let \(B_1\) be an interval. By rotating (if necessary), we may without loss of generality assume that \(B_1 = [0,x)\) for some \(x > 0\). By definition, \(f_\alpha ^i(B_2)\) would then be of the form [0, x/k) with \(k \in {{\mathbb {N}}}\) for some \(i \in {{\mathbb {N}}}\). Since the finite sequence \((f_\alpha ^l(B_2))\) needs to build a partition of \(B_1\), we must have

$$\begin{aligned} \frac{\left\{ n \alpha \right\} }{\left\{ m \alpha \right\} } = k \end{aligned}$$

for some \(m,n \in {\mathbb {N}}\) which can only hold for \(n = mk\). This contradicts the fact that x is in the forward-orbit of x/k. Thus, the rotation cannot be (CG) rank one by intervals. \(\square \)