Double Rotations

The METR technical reports are published as a means to ensure timely dissemination of scholarly and technical work on a non-commercial basis. Copyright and all rights therein are maintained by the authors or by other copyright holders , notwithstanding that they have offered their works here electronically. It is understood that all persons copying this information will adhere to the terms and constraints invoked by each author's copyright. These works may not be reposted without the explicit permission of the copyright holder. Abstract. We consider a map called a double rotation, which is composed of two rotations on a circle. Specifically, a double rotation is a map on the interval [0, 1) that maps x ∈ [0, c) to {x + α}, and x ∈ [c, 1) to {x + β}. Although double rotations are discontinuous and non-invertible in general, we show that almost every double rotation can be reduced to a simple rotation, and the set of parameters such that the double rotation is irreducible to a rotation has a fractal structure. We also examine a characteristic number of double rotations that is called a discharge number. The discharge number as a function of c reflects the fractal structure, and is very complicated.


Introduction
In this paper, we consider the family of double rotations f (α,β,c) : [0, 1) → [0, 1) defined by 1) for (α, β, c) ∈ [0, 1) × [0, 1) × [0, 1]. A typical graph of a double rotation is shown in Figure 1(a). Although the map is apparently discontinuous and non-invertible in general, we show that, for almost every parameter (α, β, c), the double rotation f (α,β,c) can be reduced to a rotation. We also investigate a characteristic number of double rotations that is called a discharge number. For f (α,β,c) and x ∈ [0, 1), we consider the elements of the orbit starting from x, and define the discharge number as the ratio of the elements that fall within the interval [c, 1). Figure 1(b) shows a graph of the discharge number as a function of c for fixed α and β. Despite the simple definition of the double rotation, the graph is complicated and like a devil's staircase. The complex appearance indicates a fractal structure in the parameter space.
In the context of studies on piecewise isometries, the class of double rotations can be considered as a special form of interval translation mappings (ITMs) introduced by Boshernitzan and Kornfeld [1]. An ITM is defined as a transformation on an interval such that there exists a partition of the interval into finitely many intervals and, on each of the intervals, it is a translation. Then every double rotation is an ITM, because a partition consisting of at most four such intervals always exists. In fact, the example ITM given by them is a double rotation with a special parameter. The class of ITMs recently investigated by Bruin and Troubetzkoy [2] is also a subclass of double rotations such that c = 1 − α. In both studies, induced transformations (first return maps) of the maps are considered, and self-similarity in the maps are reported. We also investigate such self-similarity in double rotations by considering induced transformations in the present paper. Piecewise isometries such as two-parameter piecewise rotations [3] are also closely related with double rotations.
Double rotations appear in the field of electrical engineering as well. A simple three capacitance equivalent circuit model of partial discharge phenomena [4] can be reduced to a double rotation [5]. The parameter c of the double rotation corresponds to the amplitude of the voltage applied to the model, and the discharge number corresponds to the average discharge rate of the model. Although the three capacitance model is an old model, it is still important, for most partial discharge models are based on it.

Partition of the parameter space
In this section, we define a partition of the parameter space and a transformation on it. The meanings of the partition and the transformation will be given in the next section.
Let  T (e,j) : D e,j → D by Then, we define a transformation T : D * ,1 ∪ D * ,3 → D simply as With the transformation T , for a given initial parameter (α, β, c), the sequence of parameters T (α, β, c), T 2 (α, β, c), . . . can be considered, as far as T is defined for each parameter. Once the parameter is mapped into the set D * ,2 ∪ B, this process terminates. When we consider such sequences, throughout this paper, we

Induced transformation of double rotation
Before starting investigation on characteristics of double rotations, it should be noted that the family of double rotations naturally has two symmetries as shown in the following proposition. (i) f (α,β,c) is isomorphic to f H (α,β,c) , where H is the transformation given by Proof. (i) Let Φ θ denote a rotation given by Φ θ (x) = x + θ (mod 1). Then, for arbitrary x ∈ [0, 1), we have (ii) Let h be the transformation given by h( Note that both 0 and 1 − c are the discontinuity points of Because of these two symmetries, f (α,β,c) is also isomorphic to f ({1−α},{1−β},c) except for the discontinuity points. By the first symmetry, there is a one-to-one correspondence of elements between D 0,1 and D 1,3 , between D 0,2 and D 1,2 , and between D 0,3 and D 1,1 . By the second symmetry, domains D e,1 and D e,3 are symmetric to each other for e = 0 and 1. Furthermore, it should be noted that the definition of T is also symmetric in the sense that H • T • H(α, β, c) = T (α, β, c) for arbitrary (α, β, c) ∈ D * ,1 ∪ D * , 3 .
In the case when (α, β, c) ∈ B, the double rotation f (α,β,c) is considered to be trivial in the sense that it is a simple rotation or has an identity map on [0, c) or [c, 1). In the following, for a non-trivial double rotation with a parameter (α, β, c) ∈ D * , we consider the induced transformation of f (α,β,c) on the set I (α,β,c) , which is defined as Then, we show that the induced transformation is isomorphic to another double rotation or a simple rotation, as shown in Figure 3. The definition of I (α,β,c) is also symmetric in the sense that I (α,β,c) = Φ c (I H(α,β,c) ) for (α, β, c) ∈ D * . Proof. From the assumption (α, β, c) ∈ D 0,1 , we have α < β and c ≤ 1 − β. Therefore f (α,β,c) (x) can be expressed as Then, with h(x) = (1 − β)x, the induced transformation is isomorphic to the map Proof. From the assumption (α, β, c) ∈ D 0,2 , we have 1 − β < c < 1 − α. Therefore f (α,β,c) (x) can be expressed as . It should be noted that every x ∈ [0, 1) is mapped into I (α,β,c) by f (α,β,c) in a finite time.
In this way, reductions of double rotations correspond to the transformation T . The example ITM given by Boshernitzan and Kornfeld [1] can be considered as a double rotation with the parameter (α, α 2 , 1 − α) ∈ D, where α ≈ 0.311108 is the unique root of the equation x 3 − x 2 − 3x + 1 = 0 such that α ∈ [0, 1). In fact, (α, α 2 , 1−α) is a periodic point of T with period three, because (α, α 2 , 1−α) ∈ D 1,3 and Thus f (α,α 2 ,1−α) can be reduced to itself, as shown in [1]. The class of ITMs investigated by Bruin and Troubetzkoy [2] also corresponds to the plane in D specified by c = 1 − α. In the region D 0 , this plane is located at the boundary between D 0,2 and D 0,3 . Therefore, f (α,β,1−α) can be reduced to a rotation. In the region D 1 , although the behaviour of T is complicated as shown in [2], the plane is mapped into itself by T . In this sense, the reductions are closed within the class.

Cantor structure
As shown in the previous section, reductions of double rotations can be described by the transformation T on the parameter space. In this section, to investigate the induced transformations successively reduced from a double rotation, we will investigate the dynamics of T .
Firstly, in the following proposition, we consider the orbits starting from two distinct initial parameters (α, β, c) and (α, β, c ) that share same α and β.
Let C = [0, 1], because C can be considered as C j 1 ,...,j n for n = 0. To consider the case n = 1, let us assume that (α, β) is in the set S * defined by Then, from the definitions of the regions D e,j , the sets C 1 , C 2 and C 3 turn out to be the intervals given by If (α, β) ∈ S * , the sets C 1 , C 2 and C 3 are not well-defined (empty sets), because (α, β, c) is in the boundary B for any c. We also define σ j (α, β) for j ∈ {1, 2, 3} and (α, β) ∈ S * as the length of C j . Specifically, σ j (α, β) is given by In the following proposition, we consider the structure of the sets C j 1 ,...,j n in the parameter space C = [0, 1] for fixed α and β.
Then the following statements are true.
In the proposition, we assumed (α i , β i ) ∈ S * for 0 ≤ i ≤ n. In fact, it will be proven in Proposition 6.1 that, if α and β are linearly independent over Q, then Under the assumption, we can consider the structure of the sets C j1,...,jn without taking account of the boundary. Parameters outside of the setŜ will be discussed in Section 6.
. If 0 ≤ i < n and j 1 = · · · = j i = 1, then the lengths of the intervals C j1,...,ji,ji+1 are given by |C j1,...,ji, Proof. The statement (ii) is symmetric to (i). Therefore, we prove (i) by induction on i. For i = 0, the equations are nothing but the definitions of σ j (α, β). Let us assume the lemma is true for i and verify it for i + 1 < n. Since the lengths of C j1,...,ji,1 and C j1,...,ji are given by the lemma for i and i − 1, we obtain Then Therefore, we obtain Applying this result to Proposition 4.2(iv), we obtain Therefore, the lemma is also true for i + 1.
In the following, we consider the set Γ defined by In fact, the previous proposition implies that Γ is a Cantor set as shown later in Theorem 4.1. For the investigation of the Lebesgue measure of Γ, we define Γ n for n ≥ 0 as Then the sets Γ and Γ n can be expressed by the intervals C j1,...,jn as Since C j1,...,jn are always closed for j 1 , . . . , j n ∈ {1, 3}, the sets Γ n and Γ are also closed. Let us consider a subset V of S * defined by Then note that the condition of (α, β) ∈ V is equivalent to each of the following two inequalities: Intuitively, if (α, β) ∈ V , the measure of Γ 1 becomes relatively large and near to one, because σ 2 (α, β) is relatively small. Therefore, the set V plays an important role in evaluating the measure of Γ n . The following lemma claims that it is impossible for any sequence (α i , β i ) to stay in the set V forever.  Let (α, β) ∈Ŝ. There exists an integer n > 0 such that, for any sequence j 1 , . . . , j n ∈ {1, 3} of length n, a non-negative integer i < n such that (α i , β i ) ∈ V can be found.
Proof. Let us assume that there exist a parameter (α, β) ∈Ŝ and an infinite sequence of j i ∈ {1, 3} such that (α n , β n ) ∈ V for all n ≥ 0. It follows from the definition of T that Taking into account that σ 2 (α, β) can be both β − α and α − β, we obtain because of Inequality (1). Therefore, in both cases, Applying this inequality to Proposition 4.2(iv), for arbitrary n > 0, we obtain On the other hand, according to Proposition 4.3, the limit of |C j 1 ,...,j n | as n goes to infinity must be zero. This is a contradiction and thus the first assumption turns out to be false.
Thus, it is impossible for any sequence (α i , β i ) to stay in the set V forever. Furthermore, if (α, β) ∈ V , by taking adequate n, µ(Γ n ) becomes relatively small, where µ denotes the Lebesgue measure. Therefore, now we can show the following lemma concerning the measure of Γ n .
Proof. Let us consider the set of sequences Since n ≥ k + [E(α k , β k )], it follows from Lemma 4.1 that j k+1 ,...,j n ∈{1,3} where [E(α k , β k )] −1 ≤ 1 is applied to the last inequality. Therefore, for l = 1, the set of sequences W l satisfies the following conditions.  1,j k+1 ,...,jn + C j1,...,j k−1 ,3,j k+1 ,...,jn Thus W l+1 satsifies the conditions. The number of elements in W l+1 decreased by one from W l . By induction, there exists a finite integer l such that W l contains only one element and still satisfies the conditions. The only element must be an empty string, because of the condition (ii). Then the condition (i) for the empty string, namely is nothing but the statement of the lemma. Now we are ready to prove the first theorem. Proof. Γ is compact, because it is closed and bounded within [0, 1]. The connected components in Γ n are the intervals C j 1 ,...,j n for j 1 , . . . , j n ∈ {1, 3}, because they are separated by the intervals C j 1 ,...,j n−1 ,2 . The length of each connected component as n goes to infinity is zero because of Proposition 4.3. Therefore, every connected component in Γ is a single point, which means Γ is totally disconnected. Γ is also a perfect set, because, for arbitrary c ∈ Γ, we can construct a sequence γ i ∈ Γ that never equals to c but converges to c in the following way. For a positive integer i, there exists a sequence j 1 , . . . , j i ∈ {1, 3} such that c ∈ C j 1 ,...,j i . Then one of the intervals C j 1 ,...,j i ,1 and C j 1 ,...,j i ,3 does not contain c. Let γ i ∈ Γ be one of the endpoints of the interval not containing c. Then γ i = c. Since |c − γ i | ≤ |C j 1 ,...,j i |, it follows from Proposition 4.3 that lim i→∞ |c − γ i | = 0. Thus Γ is compact, totally disconnected and perfect. Therefore it is a Cantor set. Let n be a positive integer. Then, by applying Lemma 4.4 for (α n , β n ), we can find an integer n > n such that jn+1,...,j n C j 1 ,...,j n < 5 6 |C j 1 ,...,j n | for any sequence j 1 , . . . , j n ∈ {1, 3}. Taking summation over all the sequences, we obtain µ(Γ n ) = j 1 ,...,j n C j 1 ,...,j n < 5 6 j 1 ,...,j n |C j 1 ,...,j n | = 5 6 µ(Γ n ).
Therefore, for arbitrary > 0, we can find finite n such that the µ(Γ n ) < by repeating this process recursively. Then we obtain µ(Γ) = 0, because Γ ⊂ Γ n for all n > 0.
Almost every (α, β) ∈ S satisfies the assumption (α, β) ∈Ŝ, and then almost every c ∈ [0, 1] is not in the set Γ. Therefore, we can conclude that almost every double rotation is reducible to a rotation. Besides, since f (α n ,β n ,c n ) becomes a rotation if c n ∈ {0, 1}, it should be noted that f (α,β,c) can also be reduced to a rotation if c is an endpoint of an interval C j 1 ,...,j n for a sequence j 1 , . . . , j n ∈ {1, 3}, even if c ∈ Γ.

Discharge number
In this section, we consider a characteristic number of double rotations, which is called the discharge number. For a parameter (α, β, c) ∈ D and an initial state x ∈ [0, 1), the discharge number of f (α,β,c) for x is defined as if the limit exists, where χ [c,1) denotes the characteristic function of [c, 1). Figure 1(b) shows a graph of q (α,β,c) (x) as a function of c, where α and β are fixed. In the graph, q (α,β,c) (x) is independent of x, so that the graph takes only a single value for each c. At first glance, we notice that the graph is like a devil's staircase. It is monotonically non-increasing and seems closely related to the Cantor structure we discussed in the previous section. Each flat step in the graph seems likely to correspond to each interval C j1,...,jn,2 .
We will assume (α, β, c) ∈D =Ŝ × [0, 1] and investigate such properties of the discharge number in this section.
For the first step, in the following lemma, we show that the discharge number is well-defined, independent of x, and constant on the interval C 2 . It should be noted that, from the symmetry, q (α,β,c) (x) = 1 − q H(α,β,c) (Φ −1 c (x)).
For further investigation of discharge numbers, we define two other characteristic numbers p (α,β,c) and r (α,β,c) for the sets [0, c) and U , respectively, in a similar way that we have defined q (α,β,c) for the interval [c, 1). We also define P n (x), Q n (x) and R n (x) for a positive integer n as the number of integers i ∈ {0, . . . , n − 1} such that f i (x) is in [0, c), [c, 1) and U , respectively. Then, we have It should be noted that, using these numbers, we can express f n (α,β,c) (x) as . Similarly, the values p (α,β,c) (x), q (α,β,c) (x) and r (α,β,c) (x) are closely related to each other, as shown in the following lemma.
Lemma 5.1. If p (α,β,c) (x), q (α,β,c) (x) and r (α,β,c) (x) have respective limits, then they satisfy the equations Proof. Since [0, c) ∪ [c, 1) is a disjoint union and equals to the whole state space [0, 1), it is clear that P n (x) + Q n (x) = n. Therefore, we obtain the first equation We also obtain the second equation, by dividing Equation (3) by n and taking the limit as n goes to infinity, where note that both x and f n (x) are bounded within the interval [0, 1).
This lemma claims that, if one of the values p (α,β,c) (x), q (α,β,c) (x) and r (α,β,c) (x) is obtained, other two can be derived from the equations. In this sense, these three values can be considered as equivalent. Now we consider dependencies between the discharge numbers of f (α,β,c) and f T (α,β,c) , for (α, β, c) ∈ D * ,1 ∪ D * , 3 .
For an initial state x ∈ [0, 1), let us consider the orbit . Then the subsequence that consists of the elements such that x i ∈ I (α,β,c) is nothing but an orbit of the induced transformation f (α,β,c) I (α,β,c) . Let x k be the first element of the orbit. By the isomorphism between f (α,β,c) I (α,β,c) and f T (α,β,c) , the element x k ∈ I (α,β,c) has a corresponding point x in [0, 1). Let us call x as the initial state induced from x. In the following lemma, we will evaluate q (α,β,c) in the form where ∼ means that (p (α,β,c) (x), q (α,β,c) (x)) and (u, v) are equivalent to each other in the sense that one is a non-zero real multiple of the other. Then the value of q (α,β,c) (x) is obtained by v/(u + v).
Lemma 5.2. Let (α, β, c) ∈ D 0,1 and x ∈ [0, 1). Suppose the discharge number q T (α,β,c) (x ) is well-defined, where x is the initial state induced from x. Then, the discharge number q (α,β,c) (x) is also well-defined and given by Proof. Firstly, consider the first n elements of the orbit starting from x. Let n be the number of the elements of the orbit that is in I (α,β,c) = [0, 1 − β). Then we have n = n − R n (x), because U = [1 − β, 1). As we have defined before, P n (x) and Q n (x) are the number of elements in the intervals [0, c) and [c, 1), respectively. As for the induced transformation, let P n (x ) and Q n (x ) be the number of elements in the intervals [0, c) and [c, 1 − β), respectively. Then we obtain P n (x) = P n (x ) and Q n (x) = Q n (x ) + R n (x). Applying these results to Equation (3), we obtain where note that |x − f n (x)| < 1. Then, because n also goes to infinity as n goes to infinity. From Lemma 5.1, we obtain which is equivalent to the equation in the statement.
x is the initial state induced from x. Then, the discharge number q (α,β,c) (x) is also well-defined and given by Proof. Similarly to the proof of Lemma 5.2, consider the first n elements of the orbit starting from x and let n bet the number of the elements in I (α,β,c) = [1 − α, 1). In this case we have n = R n (x), because U = I (α,β,c) . Let P n (x ) and Q n (x ) be the number of elements in the intervals [1−α, c) and [c, 1), respectively. Then we obtain P n (x) = P n (x )+n−R n (x) and Q n (x) = Q n (x ). Since R n (x) = P n (x )+Q n (x ), from Equation (3), we obtain where |f n (x) − x| < 1. Then because n also goes to infinity as n goes to infinity. From Lemma 5.1, we obtain which is equivalent to the equation in the statement.
Lemma 5.2 and 5.3 can be generalized as the following proposition.
Proof. (i) If α < β, the statement is equivalent to Lemma 5.2 and 5.3. Suppose β > α. By the symmetry, we obtain where j ∈ {1, 3} is the one that is different from j. Then, since M j and M j satisfies the equation the statement is also true even if β > α.
(ii) Let us define a matrix by Thus we can reduce the calculation of the discharge number of a double rotation to the calculation for its induced transformation. Accordingly, we can show the following proposition and theorem. Proposition 5.3. Let (α, β) ∈ S * and c ∈ C j1,...,jn,2 . Then the discharge number q (α,β,c) (x) is independent of x and given by Proof. By applying Proposition 5.1 to (α n , β n , c n ) ∈ D * ,2 , we obtain for arbitrary x. Then, applying Proposition 5.2 recursively, we obtain the equation in the statement. Proof. Since c ∈ [0, 1] \ Γ, there exists a sequence j 1 , . . . , j n ∈ {1, 3} such that c ∈ C j 1 ,...,j n ,2 . From Proposition 5.3, it is shown that q (α,β,c) (x) is well-defined for all x ∈ [0, 1) and independent of x. Every connected component in [0, 1]\Γ equals to an open interval C j 1 ,...,j n ,2 for a certain sequence j 1 , . . . , j n ∈ {1, 3}, because the two endpoints of C j 1 ,...,j n ,2 belongs to Γ. Therefore, q (α,β,c) (x) is constant on the component.
For (α, β, c) ∈ D, let us define the rotation rank of the double rotation f (α,β,c) as the dimension of the vector space over Q spanned by α, β and 1. The rotation rank differs from the rank of ITMs defined in [1], for the rank of f (α,β,c) as an ITM is the dimension of the space spanned by α, β, c and 1. For each possible rotation rank d = 1, 2 and 3, let Ω d denote the set of parameters (α, β, c) ∈ D such that the rotation rank of f (α,β,c) is d. Then the three classes can be denoted by (i) We have investigated the class Ω 3 in the previous sections, and we will consider the classes Ω 1 and Ω 2 in the following.
Firstly, we assume the parameter (α, β, c) is in the second class Ω 2 . In other words, we assume that α and β are linearly dependent over Q and that at least one of α and β is irrational. Then there exists (s, t) ∈ L that satisfies sα + tβ ∈ Z, and the pair (s, t) is unique except for its rational multiples. Moreover, if (α, β, c) ∈ D * ,1 ∪ D * ,3 , the pair (s 1 , t 1 ) obtained from Equation (4) is the unique solution such that s 1 α 1 + t 1 β 1 ∈ Z. Recursively, for n ≥ 0, (s n , t n ) that satisfies s n α n + t n β n ∈ Z can also be obtained uniquely by Let us divide L into two disjoint regions L + = {(s, t) ∈ L | st ≥ 0} and L − = {(s, t) ∈ L | st < 0}. Then the following lemma shows that, intuitively speaking, (s i , t i ) flows from L + to L − and toward the origin (0, 0) as i increases. 3 . Suppose (s, t) and (s 1 , t 1 ) ∈ L satisfies Equation (4).
Proof. By reversing the sign of (s, t) if necessary, we only have to consider the case that s is positive if s is non-zero, and that t is positive if s is zero. Furthermore, we can assume (α, β, c) ∈ D * ,1 without any loss of generality, because all the statements are symmetric. Then, (s 1 , t 1 ) is given by s 1 = s and t 1 = −sσ 3 (α, β) + tσ 1 (α, β).
By using these results on the behaviour of (s i , t i ), we can examine the behaviour of (α i , β i , c i ) in the following proposition. Proposition 6.2. Let (α, β, c) ∈ Ω 2 . There exists a finite integer n such that (α n , β n , c n ) ∈ D * ,2 ∪ B or c n ∈ {0, 1}.

Concluding remarks
We have shown that, for every non-trivial double rotation with the parameter (α, β, c) ∈ D * , there exists an interval such that the induced transformation on it becomes a simple rotation or another double rotation. Although this approach does not seem directly applicable to general ITMs, it is a convenient tool at least for double rotations. In the sense that this approach is effective, double rotations seem to be an important class of ITMs.
By considering the induced transformations, we have investigated the fractal structure in the parameter space, and shown that almost every double rotation can be reduced to a rotation. The discharge number as a function of c for fixed α and β turned out to reflect the fractal structure. Continuity of the discharge number as a function of c for fixed (α, β) ∈Ŝ is an interesting open problem.
Besides, there may be some intriguing relation between double rotations and injective discontinuous piecewise-linear functions such as the function reduced from Caianiello's equation [6], for similar fractal structures in the parameter spaces are observed in both maps.