# The Existence and Structure of Rotational Systems in the Circle

Authors: Jayakumar Ramanathan
Journal: Abstract and Applied Analysis
 (2018)

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## Abstract

By a rotational system, we mean a closed subsetX of the circle, T=R/Z, together with a continuous transformation f:X→X with the requirements that the dynamical system (X,f) be minimal and that f respect the standard orientation of T. We show that infinite rotational systems (X,f), with the property that map f has finite preimages, are extensions of irrational rotations of the circle. Such systems have been studied when they arise as invariant subsets of certain specific mappings, F:T→T. Because our main result makes no explicit mention of a global transformation on T, we show that such a structure theorem holds for rotational systems that arise as invariant sets of any continuous transformation F:T→T with finite preimages. In particular, there are no explicit conditions on the degree of F. We then give a development of known results in the case where Fθ=d·θmod⁡1 for an integer d>1. The paper concludes with a construction of infinite rotational sets for mappings of the unit circle of degree larger than one whose lift to the universal cover is monotonic.


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## Body

## 1. Introduction

In what follows,T=R/Z denotes the unit circle with the standard orientation.SupposeX is a compact metric space and f:X→X is a continuous transformation. The dynamical system (X,f) isminimal if, for every θ∈X, the orbit(1)Oθ=fnθ:n∈Nis dense in X.Definition 1.
LetX⊆T. A continuous transformation f:X→Xpreserves cyclic order if, for any A,B,C∈X with distinct images, the arcs ABC and f(A)f(B)f(C) have the same orientation.Definition 2.
Arotational system is a subset X⊆T and a continuous transformation f:X→X, with the properties that(i)
the dynamical system(X,f) is minimal,(ii)
the transformationf:X→X preserves cyclic order.
In this situation, we will simply say that(X,f) is rotational.We need to recall one more definition, before stating the main theorem.Definition 3.
Letfi:Xi→Xi be continuous transformations of compact metric spaces for i=1,2. The dynamical system X1,f1 is anextension of (X2,f2) if there is a continuous, surjective function ϕ:X1→X2 such that ϕ∘f1=f2∘ϕ.Our main result is as follows.Theorem 4.
Let(X,f) be a rotational system such that X is an infinite, proper subset of T. In addition, suppose that the continuous mapping f:X→X has finite preimages, that is, T-1x<∞ for each x∈X. Then,(i)
the dynamical system(X,f) is an extension of an irrational rotation of the circle via a map ϕ:X→T that is compatible with the cyclic ordering on both X and T;(ii)
the functionϕ has preimages of cardinality one except at countably many points of T. The preimages of these exceptional points have cardinality two and are the endpoints of gaps of the set X in T;(iii)
(X,f) has a unique ergodic measure μ and ϕ∗μ is the standard Lebesgue measure on T.The angle of the rotation ofT in the preceding theorem is called therotation number of (X,f).Such systems are of particular interest when they arise as invariant subsets of a continuous mapping on the whole circle,F:T→T. Recall that a closed subset X of T isinvariant with respect to F if F(X)⊆X. In this situation, we may set f=F|X and consider the dynamical system (X,f). Such a system, (X,f), is asubsystem of (T,F).Our main theorem has the following obvious corollary.Corollary 5.
LetF:T→T be a continuous mapping with finite preimages. Suppose, moreover, that X is a closed, infinite, proper subset of T that is invariant with respect to F. If (X,f) is rotational, then all three conclusions of Theorem 4 hold.In the case whereF:T→T is given by Fθ=d·θ  mod⁡1 for an integer d>1, Corollary 5 has an extensive history. Ideas related to the d=2 case were studied by Morse and Hedlund [1] in their work on Sturmian trajectories. The problem was taken up later by several authors, including Gambaudo et al. [2], Veerman [3, 4], Goldberg [5], Goldberg and Milnor [6], and Bullett and Sentenac [7]. The d>2 case was studied by Goldberg and Tresser [8], Blokh et al. [9], and Bowman et al. [10]. In sum, these works provide a complete characterization of the rotational subsystems, with rational and irrational rotation number, for the uniform cover of T with positive degree.In this paper, we point out that parts of the analysis of rotational systems with irrational rotation number can be done without explicit reference to an ambient transformation on the unit circle. This leads to a structure result for rotational subsystems of a wide class of continuous transformationsF:T→T, those with finite preimages.The proof of Theorem4 will be accomplished over the next two sections. An important step is to solve the functional equation found in Proposition 16. This equation is mentioned in the appendix in [8] where an analytical approach to the uniform cover case is sketched. (In particular, for the direction we are interested in, a solution to the functional equation is claimed but not given.) We provide a solution to this problem in our more general setting using the existence of an invariant measure together with the Mean Ergodic Theorem. Sections 4 and 5 then revisit the known d-fold cover case (see [3, 4, 8, 9]) from our point of view.For a given continuousF:T→T, Theorem 4 and its corollary shed no light on how to determine which irrational numbers can be realized by rotational subsystems of (T,F). When deg⁡F=1, it is well known that there can be at most one such rotation number. In the uniform cover case with d>1, every irrational number in 0,1 can be achieved. In the last section, we consider degree d>1 mappings of T that are monotonic with respect to the usual orientation. Mappings of this type can have intervals of constancy or an arbitrary number of fixed points (many of which may be attractive). In particular, they are not conjugate to the uniform cover case via a homeomorphism. We show that, with at most countable exceptions, every irrational rotation number can be realized when the degree is two. When the degree is larger than 2, there are examples for every irrational rotation number.


## 2. Structure of Rotational Subsets

For this section and the next, we will work under the assumptions of Theorem4:(i)
(X,f) is a rotational system.(ii)
X is an infinite, proper subset of T.(iii)
f:X→X has finite preimages.Proposition 6.
The setX is a Cantor set; that is, it is a compact set that is perfect and has empty interior.Proof.
Supposeθ0∈X is an isolated point of such a dynamical system. Minimality implies that the orbit, Oθ0={fn(θ0):n>0}, is dense in X. As a consequence, we have that fn(θ0)=θ0 for some positive integer n. The set Oθ0 must then be finite as well as dense. Therefore, X=fnθ0:n>0. This contradiction implies that X cannot have any isolated points. Thus, X is a perfect subset of T; that is, it is closed and has no isolated points.
Now, consider the possibility thatX contains a closed interval of positive length. Since X is closed and a proper subset of T, we may choose a closed interval I⊆X of maximal length. The set I cannot be left invariant by any power of f. If n were such a power, then fn|I would have a fixed point. The orbit of this fixed point would be a finite invariant subset of X, which is impossible. Next, fix θ0∈I. Since the orbit of this point is dense and I has a nonempty interior, there must be a positive integer n such that fn(θ0)∈I. On the other hand, there must a point θ1∈I with fn(θ1)∉I. Let θt, 0≤t≤1, be the linear path in I that connects θ0 to θ1. Then fn(θt) connects fn(θ0)∈I with fn(θ1)∉I. But the range of fn is contained in X. An argument using the intermediate value theorem shows that this contradicts the maximality of I. We have shown that X has empty interior.Remark 7.
The properties of the dynamical system(X,f) used in the above result are that (X,f) is minimal and that X is an infinite, closed, proper subset of T.SinceX is a proper subset of T, we may select a point θ0∈T with θ0∉X. Parameterize T by the map χ:[0,1[→T defined by(2)χt=θ0+tmod⁡1. This continuous bijection is orientation preserving, provided we orient [0,1[ in the standard way. χ is also a homeomorphism from ]0,1[ to the open set T\θ0. Set Ξ=χ-1(X) and g=χ-1∘f∘χ (see Figure 1). Thus, Ξ is homeomorphic to X and (X,f) and (Ξ,g) are isomorphic dynamical systems. In particular, Ξ is an infinite, perfect compact subset of the open interval ]0,1[ and (Ξ,g) is a minimal dynamical system.Figure 1
The dynamical systems(X,f) and (Ξ,g) are equivalent.There is also a useful reformulation of the condition thatf preserves cyclic order. Let A,B and C be three distinct points in X⊂T with distinct images under f. Then, the points a=χ-1(A), b=χ-1(B), and c=χ-1(C) have distinct images under g. The action of g on this triple can be conveniently encoded by an arrow diagram as in Figure 2. That the arcs ABC and f(A)f(B)f(C) have the same orientation in T is equivalent to saying that the corresponding arrow diagram has an even number (zero or two) of crossings. This elementary fact is easy to check and will be convenient in the proof of Proposition 9.Figure 2
f preserving cyclic order is equivalent to all arrow diagrams for g having 0 or 2 crossings.Lemma 8.
Supposex,y∈Ξ and g(x)=y. Then, there are sequences an and bn in Ξ such that(i)
an and bn are strictly monotone sequences,(ii)
g(an)=bn for all n∈N,(iii)
limn→∞an=x and limn→∞bn=y.Proof.
The proof uses the following elementary fact: suppose one is given a convergent sequence of real numbers whose terms are distinct from the limit. Then, there is a subsequence which is strictly monotonic.
Now, sinceΞ is perfect, one can construct a sequence an∈Ξ such that an≠x for all n∈N and an→x as n→∞. By passing to a subsequence, one can further arrange an to be strictly monotonic. Moreover, by the continuity of g, g(an)→g(x)=y. Set bn=g(an). Since g has finite preimages, g(an)≠y for all but finitely many n. By once again passing to a subsequence, if necessary, we may arrange that bn is strictly monotone.Proposition 9.
The preimages ofg:Ξ→Ξ have cardinality at most two.Proof.
Suppose, to the contrary, thatx0, x1, x2∈Ξ are three distinct points with the same image under g. So g(xi)=y for i=0,1 and 2. We may assume that x0, x1, and x2 are arranged in increasing order in [0,1[. Let an(i),i=0,1,2 and bn(i),  i=0,1,2 be the sequences in Ξ obtained by applying Lemma 8 to x0, x1, and x2, respectively.
The proof proceeds according to howbn(0) and bn(2) approach y.
Case b n ( 0 ) and b n ( 2 ) Approach y from the Same Side. We give the argument when both sequences approach y from below; the other possibility is dealt with similarly. In this case, we can use the properties of sequence bn(i) to find indices k and l so that(3)bk0<bl2<yand ak(0), x1 and al(2) are in increasing order. This would mean that the arrow diagram for the triple ak(0),x1,al(2) has an odd number (one) of crossings, which is not possible.
Case b n ( 0 ) ↘ y and b n ( 2 ) ↗ y . Choose n sufficiently large so that(4)an0<x1<an2.But g(an(2))=bn(2)<y<bn(0)=g(an(0)). Hence, the arrow diagram for an(0),x,an(2) has an odd number (three) of crossings.
Case b n ( 0 ) ↗ y and b n ( 2 ) ↘ y . By invoking Lemma 8 again, we construct an x′ that is sufficiently close to x1 so that x0<x′<x2 and g(x′)≠y. We treat the case where g(x′)<y=g(x)—the other case is handled similarly. In this situation, there is a sufficiently large n with the property that x′<an(2). So,(5)x0<x′<an2,gx′<y=gx0<bn2=gan2.
Thus, the arrow diagram forx0,x′,an(2) has an odd number (one) of crossings.
Hence, in all three cases, we have a contradiction.Setα=inf⁡Ξ and β=sup⁡Ξ. Since Ξ is a compact, subset of the open interval ]0,1[ we have(6)0<α<β<1.Since (Ξ,g) is minimal, g is surjective. Consequently, we may put α′=sup⁡x∈Ξ:gx=β and β′=inf⁡{x∈Ξ:g(x)=α}.Proposition 10.
The inequality(7)α<α′<β′<βholds. Moreover, [α′,β′] is a gap for Ξ, that is, X∩]α′,β′[=∅.Proof.
Suppose thatα′>β′. Minimality ensures that β′>α. By using Lemma 8, we can find x near α such that α<g(x)<β. However, this would lead to an arrow diagram for the triple x, β′, α′ with an odd number of crossings.
Now, considerx∈Ξ with α′<x<β′. By the definition of α′ and β′, g(x) must be distinct from α or β. Hence g(x) must be strictly between α and β. This is a contradiction due to an impossible arrow diagram—this time for the triple α′, x, β′.
Finally, sinceΞ has no isolated points, α′≠α and β′≠β.Lemma 11.
Letx,y∈Ξ. If x<y and g(x)<g(y), then, for any z∈Ξ with x<z<y, we must have g(x)≤g(z)≤g(y).Proof.
Failure of the conclusion clearly leads to an impossible arrow diagram for the triplex, z, y.Proposition 12.
α < g ( β ) ≤ g ( α ) < β.Proof.
Minimality rules out the possibilities thatg(α)=α and g(β)=β.
Ifg(β)=α, then by Lemma 8 there is x∈Ξ near β such that g(α)>g(x)>α. This means that the arrow diagram for α, x, β is not allowed. In a similar way, g(α)=β can also be ruled out.
Only the middle inequality remains. Suppose, to the contrary, thatg(β)>g(α). The previous proposition implies that g(α)≤g(x)≤g(β), for any x∈Ξ that is strictly between α and β. Thus, Im g is a proper subset of X. This violates the minimality assumption.For anyx0,x1∈[0,1), set(8)Ξx0,x1=Ξ∩x0,x1.Proposition 13.
g is monotone increasing on the sets Ξα,α′ and Ξβ′,β. Moreover,(9)gx>x∀x∈Ξα,α′,gx<x∀x∈Ξβ,β′.Proof.
Letx,y∈Ξ∩[α,α′] with x<y. Suppose g(y)<g(x). Since y<β′, the definition of β′ forces α=g(β′)<g(y). Thus, the arrow diagram for the triple x, y, β′ is impossible. This contradiction shows that g is monotonic increasing on [α,α′]. A similar argument applies to Ξ∩[β′,β].
Next, letx∈Ξ∩[α,α′] and suppose g(x)≤x. Since g has no fixed points, g(x)<x. For the same reason, α<g(α). Appealing to the monotonicity property proved in the previous paragraph, we have α<g(α)≤g(x)<x. Lemma 11 then implies that Ξ∩[α,x] is a nonempty, proper, closed invariant subset of Ξ—a contradiction. Thus g(x)>x for x∈Ξ∩[α,α′]. The last inequality can be verified similarly. See Figure 3 for a diagram of the structure of g.Figure 3
Schematic for the action ofg. Ξ is a proper subset of the two intervals marked in blue.


## 3. Coding by Irrational Rotations of the Circle

By the Krylov-Bogoliubov theorem (see, e.g., [11, pp. 98]), there is a Borel probability measure on Ξ that is invariant under g. Fix one such, μ. Regard μ as a measure on [0,1) and write γ~ for its cumulative distribution function:(10)γ~x=∫0xμt.Since every point of the infinite set Ξ has dense orbit, the invariant probability measure μ cannot include any point masses. Thus γ~ is continuous. Hence, the restriction γ=γ~|Ξ is also a continuous, monotone increasing map from Ξ to [0,1]. Finally, because(11)γ~α=0,γ~β=1and the fact that γ~ is locally constant on the complement of Ξ, γ:X→[0,1] is surjective.Proposition 14.
(i) Ifx0,x1∈Ξ and Ξ∩(x0,x1)=∅, then γ(x0)=γ(x1).
(ii) On the other hand, ifΞ∩(x0,x1)≠∅, then γ(x0)>γ(x1).Proof.
The first statement follows directly from the facts that sptμ⊆Ξ and μ contains no point masses.
To prove the second statement, first putU=Ξ∩(x0,x1) and(12)νnU,x=k:k=0,…,n-1,gkx∈U.Finally, write P:L2(Ξ,μ)→L2(Ξ,μ) for the projection onto the subspace of functions left invariant by g. By the Mean Ergodic Theorem (see, e.g., [11, pp. 32]), the averages(13)Anx=νnU,xnconverge in L2(Ξ,μ) to the projection P[1U](x). On the other hand, since U is a nonempty open set and (Ξ,g) is a minimal dynamical system, there is an ϵ>0 such that(14)liminfn→∞νnU,xn>ϵ∀x∈Ξ.See, e.g., Proposition  4.7 in [12]. Hence, keeping Fatou’s theorem in mind,(15)γx1-γx0=μU=1U2≥P1U2=limn→∞⁡An2≥∫liminfn→∞Anx2dμx≥ϵ2>0.The above result can be restated as follows: for any distinct pairx, x′∈Ξ, γ(x)=γ(x′) if and only if x and x′ are endpoints of a gap of Ξ⊂]0,1[. A further useful corollary of this is that any nonempty open subset of Ξ has positive μ-measure.Lemma 15.
One has(16)γ∗μ=L,where L denotes Lebesgue measure on [0,1[.Proof.
It is enough to show that(17)γ∗μ0,t=L0,t∀t∈0,1.The left hand side is just μ(γ-1[0,t]). Since γ is monotone increasing, continuous, and surjective, γ-1[0,t]=[α,x] for some x with γ(x)=t. But this just means μ(Ξ[α,x])=t=L([0,t]).Proposition 16.
There is anω0∈(0,1) such that(18)γgx=ω0+γxmod⁡1for all x∈Ξ.Proof.
Putω0=μβ′,β. If x∈Ξ∩[α,α′], then the g invariance of the measure μ implies that (19)γgx=μα,gx=μα,gβ+μgα,gx=μgβ′,gβ+μgα,gx=ω0+μα,x=ω0+γx.If x∈X∩[β′,β], then(20)γgx=μα,gx=μgβ′,gx=μβ′,x=γx-μα,α′=γx+μβ′,β-1=γx+ω0mod⁡1.Remark 17.
In the appendix to the article by Goldberg and Tresser [8], an analytic argument for characterizing rotational subsystems for the map Fθ=d·θ  mod⁡1 (where d>1) is sketched. The preceding proposition is claimed, but no argument is given. In addition, the authors then start with data equivalent to what is given in Section 5 and solve this functional equation to produce rational systems with a given rotation number.Define the mapsΓ:Ξ→T and r:T→T by(21)Γx=γxmod⁡1,rθ=θ+ω0mod⁡1.Proposition 16 proved that(22)Γ∘g=r∘Γ.In other words, Γ is a surjective, continuous mapping of the dynamical system (Ξ,g) to the dynamical system (T,r). In view of Proposition 14, if Γ(x)=Γ(x′) for distinct x and x′, then either x and x′ are endpoints of a gap for Ξ or x=α and x′=β. Finally, from Lemma 15, it is clear that Γ∗μ is Lebesgue measure on T.Proposition 18.
ω 0 is irrational.Proof.
Ifω0 were rational, the set(23)rnω0:n∈N=nω0mod⁡1:n∈Nwould be finite. Recalling that Γ has preimages of cardinality no greater that 2, this means that(24)x∈Ξ:Γx=nω0mod⁡1for  somen∈Nis finite. It is easy to check that it is also invariant under g. This contradicts the minimality of (Ξ,g).Lemma 19.
LetZ0 be those points x∈Ξ with the property that for every δ>0, the intervals (x,x+δ) and (x,x-δ) contain infinitely many points of Ξ. Then Z0 is a dense, uncountable subset of Ξ.Proof.
A point is in the complement ofZ0 in Ξ precisely when it is the endpoint of a gap, that is, a maximal open interval contained in [α,β]\Ξ. Since there are at most countably many such open intervals, we conclude that Ξ\Z0 is countable. Since the perfect set Ξ is uncountable, Z0 is uncountable as well.
The density follows from a small modification of the standard argument that a perfect subset of a complete metric space is uncountable. SupposeU⊆Ξ is a nonempty open set with U∩Z0=∅. Then, U is countable. Let U={x1,x2,…}. Write Ui=Ξ\{xi} and note that Ui is dense and open in Ξ for each i. Baire’s theorem yields that ⋂i=1∞Ui is dense in Ξ. But, by construction, U∩(⋂i=1∞Ui)=∅—a contradiction. So, Z0 is dense.Theorem 20.
The dynamical system(Ξ,g) is uniquely ergodic.Proof.
Letx0∈Z0. By Proposition 14 and Lemma 19,(i)
x∈Ξ and x<x0⇒γ(x)<γ(x0),(ii)
x∈Ξ and x>x0⇒γ(x)>γ(x0).
Therefore, for anyy∈Ξ and n≥0,(25)gny≤x0⟺γy+nθ0≤γx0.(Here the braces denote fractional part.) Consequently,(26)limn→∞k:gky≤x0,0≤k<nn=limn→∞k:γy+nθ0≤γx0,0≤k<nn=γx0=μΞ∩α,x0.The second limit has been evaluated by invoking Weyl’s equidistribution theorem (see [13, 14]). Since such x0 are dense in Ξ, the cumulative distribution of the measure μ is uniquely determined by (Ξ,g). In other words, (Ξ,g) is uniquely ergodic.With this last result, we have completed the proof of Theorem4. Indeed, the analysis of this section transfers over to the dynamical system (X,f) by means of the isomorphism, χ-1, of dynamical systems. In particular, one sets ϕ=Γ∘χ-1.


## 4. The Uniformd-Fold Cover

In this section and the next we are concerned with the mapF:T→T defined by(27)Fx=d·xmod⁡1for some fixed integer d≥2. We will analyse the structure of rotational systems (X,f) with f=F|X using the findings of the last two sections.Since0  mod⁡1 is a fixed point of F, it is not in X. So, choosing this point for θ0, we have that the parameterization χ:[0,1[→T is given by(28)χt=tmod⁡1and that G=χ-1∘F∘χ satisfies(29)Gx=d·x.Define Ξ and g as before, and note that g=G|Ξ. We will find it convenient to work with (Ξ,g), [0,1[ and G, noting that findings in this setting convert readily to statements about (X,f), T and F.The inverse image of0 under G consists of the d points:(30)ξk=kdk=0,…,d-1.In addition, G has d-1 fixed points:(31)ηk=kd-1k=0,…,d-2.We also set ξd=ηd-1=1.SetIi=[ξi,ξi+1[ for i=0,…,d-1 and note that the interior of Ii are precisely those points in T with a canonical d-adic expansion that starts with the digit k. (Recall that every real number has a canonical d-adic expansion that does not end with an infinite string of d-1’s.) Note also that GIk=[0,1], since G is just the shift map on the d-adic expansion. Moreover, G is monotonic increasing on the interior of each Ii. Each closed interval Ii contains a unique fixed point ηi, for 0≤i<d-1. The behavior of G at these fixed points can be readily determined. In particular, one checks that(32)Gx<xforξi<x<ηi,Gx>xforηi<x<ξi+1.(See Figure4 for the d=5 case.) Under our operating assumptions, Ξ is infinite and minimal. Hence, none of ξi and ηi can lie in Ξ. In particular, any point in Ξ must lie in the interior of precisely one of the intervals I0,…,Id-1.Figure 4
Plot ofG(x) when d=5.WriteA={0,1,…,d-1} and AN0 for the set of functions from the nonnegative integers to A. We use the usual product topology on AN0 and let S denote the usual continuous shift on AN0. Define a map E:Ξ→AN0 by(33)Ex=a0a1a2…,where ai are defined by the requirement that gix∈[ξai,ξai+1). E(x) is just the canonical d-adic expansion of x. Since each x∈[0,1) has a unique expansion of this type, E is injective. It is also obvious that S∘E=E∘G.Proposition 21.
The mapE is a homeomorphism from Ξ to Im(E).Proof.
SinceΞ is compact, it is enough to show that E is continuous on Ξ. Suppose xi,x∈Ξ and xi→x as i→∞. For any n, we must have that gnx lies in an open interval of the form I°j=(ξj,ξj+1). Since g is continuous on [0,1)\{ξ0,ξ1,…,ξd-1},(34)gnxi⟶gnxasi⟶∞for each n∈N0. This implies that the nth digit of xi coincides with that of x for all sufficiently large i. Hence E is continuous.LetΞ1,…,Ξl be the nonempty sets in the list(35)Ξ∩I0,Ξ∩I1,…,Ξ∩Id-1.The indexing can be arranged so Ξi⊂Iki for some 0≤ki<d with(36)k1<k2<⋯<kl.Set(37)αi=inf⁡Ξi,βi=sup⁡Ξi.Since Ξ is perfect, αi<βi for i=1,…,l and Ξ∩(αi,βi) is nonempty. Consequently, γ(Ξi) is the closed interval [γ(αi),γ(βi)] and has positive length (see Proposition 14). Because μ has no mass on the open interval (βi,αi+1), γ(βi)=γ(αi+1). Set t0=0 and ti=γ(βi) for i=1,…,l. Then,(38)0=t0<t1<⋯<tl=1,γΞi=ti-1,tifori=1,…,l.Recall a consequence of our previous analysis: every point x∈Ξ with g(x)>x must lie to the left of every point y∈Ξ with g(y)<y. Thus, each Ξi must lie completely to one side of the sole fixed point within Iki. Moreover, if supΞi<ηki and infΞj>ηkj then i>j and ki>kj. Let m be the last index, j between 1 and l with infΞj>ηkj. Clearly, m<l and sup⁡Ξm=α′ and inf⁡Ξm+1=β′. Hence, tm=γ(αm+1)=γ(βm)=1-ω0.We next seek to understand the preimages ofγ. Define ρ:[0,1[→[0,1[ by(39)ρt=t+ω0.(See Figure 5.) Set(40)D0=t∈0,1:ρnt=t+nω0≠tifor  anyn≥0,i=1,…,l.(Except for a countable subset, every t∈[0,1[ is in D0.) For any such t, each member of the sequence of points ρn(t)={t+nω0} with n≥0 lies in exactly one of the intervals of the form [ti,ti+1[. Write in for the index with this property:(41)ρnt∈tin,tin+1.Then, for any x in γ-1(t),(42)gnx∈Ξ∩ξkin,ξkin+1∀n≥0.This just says the string ki0ki1ki2⋯ is the canonical d-adic expansion of x. Thus, there is a unique point in the preimage γ-1(t).Figure 5
Plot ofρ.In summary, this discussion shows how any rotational subsystem(X,g) of the uniform d-fold cover of the unit circle must arise from the symbolic flow of an irrational rotation of T relative to a suitable partition. The next section shows that this process can be reversed.


## 5. The Inverse Process

In this section, we begin with an irrational numberω0∈[0,1[ and a partition of [0,1] into l≤d subintervals with the requirement that one of the interior nodes is 1-ω0:(43)0=t0<t1<⋯<tm<tm+1<⋯<tl=1and tm=1-ω0. Set Jk=]tk,tk+1[ for k=0,…,l-1. Next, select a coding that maps {0,…,l-1} to the set of digits {0,…,d-1}. More precisely, choose integers k1,…,kl that satisfy(44)0≤k1<k2<⋯<kl<d.We will show that this data determines a rotational subset of T that inverts the process described in previous section.Letρ:[0,1[→[0,1[ be defined by ρ(t)={t+ω0}. Write r:T→T for the corresponding rotation of T and note that χ∘ρ=r∘χ:(45)Remark 22.
It is well known that(T,r) is a minimal dynamical system. Hence, for any non-empty open V⊆T there is an N>0 with the following property: for every θ∈T,(46)rkθ∈Vfor at least one k with 0≤k≤N. (See, e.g., Proposition  4.7 in [12].)
It follows easily that the analogous statement holds for([0,1[,ρ) and any non-empty open U⊂]0,1[. (Simply apply the previous observation to V=χ(U).)As before, letD0 be the set of t∈[0,1[ satisfying ρn(t)={t+nω0}≠ti for all n∈N0 and i=0,…,l. In other words, D0 consists of those points of [0,1[\{t0,t1,t2,…,tl-1} whose forward orbit does not contain any of the nodes ti. D0 is invariant under the map ρ.Proposition 23.
(i) The complement ofD0 in [0,1[ is countable.
(ii) For everyt∈[0,1[, there is an integer n≥0 with the property that ρn(t)∈D0.Proof.
The mapρ is invertible. The complement of D0 is just the countable set:(47)ρ-kti:i=1,…,l,k≥0.
For the proof of the second claim, fixt∈[0,1[. If the orbit of t hits the {ti:i=0,…,l} infinitely often, then there must be an index i such that(48)t+nω0=tifor infinitely many n∈N. This means that there are two distinct, positive integers n0, n1 with the property that(49)n1-n0ω0=0mod⁡1.But this contradicts the condition that ω0 is irrational. This argument proves that the forward orbit of such a t must eventually lie completely in D0.The trajectory of any pointt∈D0 can be encoded by an infinite string:(50)a0a1a2…,where an=ki precisely when ω+nω0 is in Ji. The Kronecker approximation theorem implies that each of the digits ki, i=0,…,l-1 occurs infinitely often. Thus,(51)Eω=0.a0a1a2…may be interpreted as the canonical d-adic expansion of a real number in the open unit interval ]0,1[. So, E:D0→[0,1). Note that Eω+ω0 is the shift 0.a1a2….Proposition 24.
The mapE:D0→[0,1[ is a continuous, injective, strictly monotonic increasing and(52)Eρt=GEt.Proof.
Kronecker’s theorem also implies injectivity ofE: Let t and t′ be distinct points in D0 with t′>t. Set Δ=t′-t and note that there must be an s with the property that s and s′=s+Δ lie in the interior of different Ji’s. Note that if |ρnt-s| is sufficiently small, then |ρnt-s|=|ρnt′-s′|. In particular, by choosing n so that |ρnt-s| is sufficiently small, we may insure that ρnt and ρnt′ will be in different Ji’s. Hence the d-adic encodings of t and t′ are different. Since neither of these encodings can end with an infinite string of d-1’s, E(t)≠E(t′).
Equation (52) follows directly from (51) and the ensuing discussion.
Lett∈D0 and tn be a sequence in D0 that converges to t. Then, since ρkt is in the interior of one of Ji, ρktn must eventually have this property as well. Thus, E must be continuous.
It only remains to prove thatE is monotone increasing. Let t, t′∈D0 with t<t′. Write E(t)=a0a1a2… and E(t′)=a0′a1′a2′…. Because E is injective, there is a first index i≥0 for which ai≠ai′. If i=0, t<t′ implies a0<a0′. This in turn yields E(t)<E(t′). If i>0, then ai-1=ai-1′ entails that both ρi-1t and ρi-1t′ are in interior of the same Jk. Since ρ is increasing on the interior of any Jn, we must have ai<ai′. As a consequence, E(t)<E(t′) in this case as well.WriteΞ0 for the image of D0 under the map E and let Ξ be its closure in [0,1[. It is straightforward to check that Ξ0 is invariant under G. Establishing this for its closure is complicated by the possibility that G might not be continuous on Ξ. Most of the effort in the proof of the next result is to rule this out.Proposition 25.
Ξ is a compact subset of ]0,1[ and is invariant under G.Proof.
By Remark22, there is an N with the property that for each t∈[0,1[, the set(53)ρit:0≤i≤Nhas nonempty intersection with the interiors of each of the intervals J0,…,Jl-1. Now consider the corresponding set of words. In particular, let W denote the set of all words in the alphabet {k1,…,kl} of length N for which each ki occurs at least once. Let w and w′ be the smallest and largest word, respectively, in the finite set W. Let k be the first digit of w′ that is not d-1. Write v for the word obtained by replacing k with k+1 in w′. The words w and v are lexicographically smaller and larger, respectively, than every element of Ξ0. Write m and M for the numbers in [0,1[ corresponding to w and v, noting that 0<m<M<1. It is clear that Ξ0⊆[m,M]. Hence Ξ is compact in the open unit interval.
By examiningG explicitly, it is easy to check that G-1[m,M] is a compact subset of [0,1[ that does not contain any of the points ξ0,…,ξd-1. Since these are precisely the points of discontinuity of G, G is continuous on G-1[m,M]. Because Ξ0 is invariant, GΞ0⊆Ξ0⊂[m,M]. Consequently, Ξ=Ξ0¯⊂G-1[m,M] and G is continuous on Ξ.
Letx∈Ξ. Then xn→x for a sequence xn∈Ξ0. By the observations in the previous paragraph, G(xn)→G(x). Since, Ξ0 is G invariant, this proves that G(x)∈Ξ0¯=Ξ.As before, defineg:Ξ→Ξ by g=G|Ξ.Theorem 26.
The dynamical system(Ξ,g) is rotational.Proof.
We first prove the minimality of(Ξ,g).
Consider a pointx0∈Ξ0. An y∈Ξ can be approximated to within an arbitrary accuracy by an element of y0∈Ξ0. Write t0, s0 for the elements of D0 with E(t0)=x0 and E(s0)=y0. By Kronecker’s theorem, there is a sequence ni of indices such that ρni(t0)→s0. The continuity of E together with Proposition 24 yields(54)limi⁡Gnix0=y0.Hence, y can be approximated to arbitrary accuracy by an element of the orbit, under g, of x0. Thus, for any x0 in Ξ0, the orbit of x0 is dense in Ξ.
Now letx∈Ξ\Ξ0. It suffices to prove that an element in the forward orbit of x is in Ξ0. We may write x as a limit of a sequence xi in Ξ0. Write ti for the elements of [0,1[ with E(ti)=xi. By passing to a subsequence, if needed, we may assume that χ(ti) converges to an element of T, say χ(t). By Proposition 23, there is m such that ρm(t)∈D0. By continuity of r, we have that χ(ρm(ti))=rm(χ(ti))→rm(χ(t))=χ(ρm(t)). Since ρm(t)∈D0, χ(ρm(t))∈T\{0mod⁡1}. Since χ-1 is continuous on the latter set, ρm(ti)→ρm(t). We may apply E to both sides of this last limit to get(55)Eρmt=limi⁡Eρmti=limi⁡GmEti=limi⁡Gmxi=Gmx.It follows that Gm(x)∈Ξ0. Since the orbit of this point is dense in Ξ, so is the orbit of x.
The only point left is to prove thatg respects cyclic order on Ξ. To this end, let x0, x1, and x2 be an increasing triple of points in Ξ with distinct images under g. We may approximate each of these points by elements of Ξ0 without changing the number of crossings in the arrow diagram. Hence, we may assume that x0, x1, and x2 are in Ξ0. Let t0, t1, and t2 be the preimages under E of x0, x1, and x2, respectively. The ti’s are in increasing order due to Proposition 24. This monotonicity property of E together with the identity in that proposition, also force the arrow diagram for t0, t1, and t2 under the map ρ to have the same number of crossings as that for x0, x1, and x2 under g. But it is easy to verify that ρ respects cyclic order. Hence, so does g.


## 6. Examples for a Class of Continuous Maps

As the last two sections show, every irrationalω0∈[0.1) arises as the rotation number of some rotational system with respect to a standard cover of degree d>1. (This is true for rational ω0∈[0,1) as well.) It is tempting to ask if this is true for any continuous transformation of degree d>1.In this section, we look at mappings of the oriented unit circle,K:T→T, whose lifts to the universal cover are monotonic increasing (though not strictly). The degree of K will be assumed to be at least 2. It is well known that such maps must have at least one fixed point. Conjugating by the appropriate rotation, we may arrange that 0 mod⁡1 is a fixed point. As before, we parameterize by the unit interval: H=χ-1∘K∘χ. There is a partition(56)0=s0<s1<⋯<sd=1with the property that, for each i=0,…,d-1,(i)
H is monotone increasing function on each half-open interval [sk,sk+1),(ii)
H(sk)=0 and limx→sk+1-H(x)=1.Applying a piecewise linear change of coordinates, we may assume thatsi=i/d for i=0,…,d.Next, we set up a standard symbolic encoding forH in terms of infinite strings of the alphabet A={0,1,…,d-1}. In particular, for each finite (nonempty) word w=i1i2⋯in, let(57)Aw=x∈0,1:∀k=1,…,n,Hk-1x∈sik,sik+1.As in Section4, denote by AN0 the set of infinite words in the alphabet A with the usual product topology. The shift on S:AN0→AN0 defined by(58)Si0i1i2…=i1i2….is continuous with respect to this topology.Definition 27.
A finite wordv is aprefix of a word w if v⋆u where ⋆ means concatenation of words.We collect some useful remarks that are easy to check.Remark 28.
(i)Aw are half-open intervals that are closed on the left.
(ii) Letw and v be finite words. Suppose that w is a prefix for the word v. Then Av⊆Aw. In addition, if vk≠d-1 for some k>j, then Av¯⊂Aw.
(iii) For any fixedw and n≥|w|, the collection(59)Av:v=nandwisaprefixforvis a partition of Aw.
(iv) Ifw=i1i2…in and w′=i2…in, then(60)x∈Aw⟹Hx∈Aw′.
(v) Ifw and v are of the same length and w precedes v in lexicographical order, then(61)x∈Aw,y∈Av⟹x<y.Each pointx∈[0,1) determines a unique infinite word ι(x)∈AN0. Specifically, ι(x)=i0i1i2… if(62)Hnx∈sin,sin+1∀n∈N0.By construction,(63)ι∘H=S∘ι.Finally, for a given finite word w=i0i1…in, the half-interval Aw consists precisely of those x∈[0,1) with the property that wn is a prefix for ι(x).Remark 29.
The lexicographical ordering onAN0 and the usual order on [0,1[ are also compatible with the factor map ι. In particular, for any x,y∈[0,1[,(i)
x<y implies ι(x)≤ι(y),(ii)
ι(x)<ι(y) implies x<y.
Thus, for anyw∈AN0, the preimage ι-1{w} is either empty, a singleton, or an interval of positive length. In particular, the cardinality of ι-1{w} exceeds one for only countable many words w∈AN0.Lemma 30.
Letxn be a sequence in [0,1) with the property that ι(xk) converges in AN0 to w. Suppose, in addition, that the infinite word w does not end with an infinite sequence of d-1’s. Then any convergent subsequence xki with a limit, say x, has the property that x∈[0,1) with ι(x)=w.Proof.
Letw=i0i1i2… and set wn=i0i1…in for each n∈N. Since ι(xi)→w, we have that, for each m, xk∈Awm for all sufficiently large k. By Remark 28 and the hypothesis on w, we have that for any n there is an m>n with the property that(64)Awm¯⊂Awn.Thus, x must be in Awn⊂[0,1) for each n∈N0. This just means that ι(x)=w.Remark 31.
If the infinite wordw∈AN0 does not terminate with an infinite sequence of d-1’s, there is an x∈[0,1[ with the property that ι(x)=w. To see this, first construct a sequence xn by the requirement that xn∈Awn for each n∈N0. (Here wn is defined as in the previous proof.) By construction, ι(xn)→w as n→∞. Let x be any limit point of this sequence in R. By Lemma 30, x∈[0,1) and ι(x)=w.The mapι is not continuous everywhere on [0,1). However, one can characterize the points of continuity.Lemma 32.
Letx0∈[0,1) be a point with the property that ι(x0) does not terminate with an infinite sequence of 0’s. Then x0 is a point of continuity for the map ι:[0,1)→AN0.Proof.
Writew=i0i1i2… for ι(x0) and let wn=i0i1…in. The hypothesis entails that x0 does not lie on the left endpoint of any of the Awn. In other words, x0 is in the interior of all the Awn. Suppose xk→x0 as k→∞. For any n∈N, xk∈Awn for all sufficiently large k. For such k, ι(xk) will have wn as a prefix. Since n∈N0 is arbitrary, ι(xn)→w=ι(x0) as n→∞.Now choose an irrational numberω0∈[0,1) and a compatible partition as in (43) of the last section. Let (X,f) be the rotational system constructed from this data and write (Ξ,g) for the associated system on [0,1).LetZ⊂[0,1) consist of those points z with the property that the associated word ι(z) is in E(Ξ). The word w=E(x) associated with an x∈Ξ cannot end with an infinite sequence of d-1’s; this would contradict the minimality of the infinite system (Ξ,g), for instance. Hence, by Remark 31, for any x∈Ξ there is a z∈Z with ι(z)=E(x).We next claim thatZ is (sequentially) compact in [0,1). Indeed, recall that E(Ξ) is compact (see Proposition 21). Let zn be any sequence in Z. By passing to a subsequence, we may assume that zn converges in R and ι(zn) converges to a w∈E(Ξ). By Lemma 30, we have that z=limzi satisfies z∈[0,1) and ι(z)=w.Another feature is that noz∈Z can have the property that ι(z) ends with an infinite string of 0’s. (The same statement holds for any E(x) where x∈Ξ.) Consequently, by Lemma 32, ι is continuous on Z.Consider the compositionψ=E-1∘ι. The function ψ:Z→Ξ assigns to a z∈Z the element in Ξ with the same symbolic representation. In view of Proposition 21 and the previous paragraph, ψ is continuous. It is also clear that it is a map between the dynamical systems (Z,H|Z) and (Ξ,g), that is, ψ∘H=g∘ψ.Let(Y,H|Y) be a compact, minimal subsystem of (Z,H) and write h=H|Y. The factor map ψ restricts to Y in a natural way: ψ∘h=g∘ψ. Since (Ξ,g) is minimal as well, ψ maps Y to Ξ surjectively. In particular, Y is infinite.The only point left open is whether(Y,h) preserves cyclic order. We do not know if this is true in general, but it is true generically as will now be shown.Proposition 33.
In thed=2 case, (Y,h) is a rotational system for all irrational ω0∈0,1 except for a countable number of possible exceptions.Proof.
In this case, the choice ofω0 determines the partition in (43). We will write (Ξω0,hω0) for the associated rotational system. Note that, for distinct choices ω0 and ω0′, the corresponding sets, Ξω0 and Ξω0′ must be disjoint. Indeed, because of minimality, the orbit of any common point would have to be dense in both sets. This would imply that Ξω0=Ξω0′, which yields ω0=ω0′.
Thus, the family of sets,Yω0=ι-1(EΞω0), is an uncountable collection of disjoint sets. By Remark 29, ι is injective on all but countably many of Yω0. For such ω0, h=hω is a continuous bijection between compact sets. So h is a homeomorphism. Since h respects the ordering of [0,1), (Y,h) preserves cyclic order.Thed>2 case differs in that, for each ω0, there are uncountably many choices for the partition, P, described in (43). Let (Ξω0,P,g) denote the rotational system uniquely associated with ω0 and P, according to the method of Section 5. For distinct partitions P and P′ compatible with θ0, Ξω0,P and Ξω0,P′ are disjoint. Thus, ι will be injective on ι-1EΞω0,P for all but countably many of the partitions P compatible with ω0. This yields the following.Proposition 34.
Ifd>2 and ω0 is an irrational number in 0,1, there are uncountable many rotational systems (Y,h) with rotation number ω0.



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