The Existence and Structure of Rotational Systems in the Circle

and Applied Analysis 3


Introduction
In what follows, T = R/Z denotes the unit circle with the standard orientation.
Suppose  is a compact metric space and  :  →  is a continuous transformation.The dynamical system (, ) is minimal if, for every  ∈ , the orbit is dense in .
Definition 1.Let  ⊆ T. A continuous transformation  :  →  preserves cyclic order if, for any , ,  ∈  with distinct images, the arcs  and ()()() have the same orientation.
Definition 2. A rotational system is a subset  ⊆ T and a continuous transformation  :  → , with the properties that (i) the dynamical system (, ) is minimal, (ii) the transformation  :  →  preserves cyclic order.
In this situation, we will simply say that (, ) is rotational.
We need to recall one more definition, before stating the main theorem.Definition 3. Let   :   →   be continuous transformations of compact metric spaces for  = 1, 2. The dynamical system ( 1 ,  1 ) is an extension of ( 2 ,  2 ) if there is a continuous, surjective function  :  1 →  2 such that  ∘  1 =  2 ∘ .
Our main result is as follows.
Theorem 4. Let (, ) be a rotational system such that  is an infinite, proper subset of T. In addition, suppose that the continuous mapping  :  →  has finite preimages, that is, | −1 | < ∞ for each  ∈ .Then, (i) the dynamical system (, ) is an extension of an irrational rotation of the circle via a map  :  → T that is compatible with the cyclic ordering on both  and ; (ii) the function  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  in T; The angle of the rotation of T in the preceding theorem is called the rotation number of (, ).
Such systems are of particular interest when they arise as invariant subsets of a continuous mapping on the whole circle,  : T → T. Recall that a closed subset  of T is invariant with respect to  if () ⊆ .In this situation, we may set  = |  and consider the dynamical system (, ).Such a system, (, ), is a subsystem of (T, ).
Our main theorem has the following obvious corollary.
Corollary 5. Let  : T → T be a continuous mapping with finite preimages.Suppose, moreover, that  is a closed, infinite, proper subset of T that is invariant with respect to .If (, ) is rotational, then all three conclusions of Theorem 4 hold.
In the case where  : T → T is given by () =  ⋅  mod 1 for an integer  > 1, Corollary 5 has an extensive history.Ideas related to the  = 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  > 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 transformations  : T → T, those with finite preimages.
The proof of Theorem 4 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 -fold cover case (see [3,4,8,9]) from our point of view.
For a given continuous  : 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, ).When deg  = 1, it is well known that there can be at most one such rotation number.In the uniform cover case with  > 1, every irrational number in [0, 1) can be achieved.In the last section, we consider degree  > 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.

Structure of Rotational Subsets
For this section and the next, we will work under the assumptions of Theorem 4: (i) (, ) is a rotational system.
(ii)  is an infinite, proper subset of T.
(iii)  :  →  has finite preimages.Proposition 6.The set  is a Cantor set; that is, it is a compact set that is perfect and has empty interior.
Proof.Suppose  0 ∈  is an isolated point of such a dynamical system.Minimality implies that the orbit, O  0 = {  ( 0 ) :  > 0}, is dense in .As a consequence, we have that   ( 0 ) =  0 for some positive integer .The set O  0 must then be finite as well as dense.Therefore,  = {  ( 0 ) :  > 0}.This contradiction implies that  cannot have any isolated points.Thus,  is a perfect subset of T; that is, it is closed and has no isolated points.Now, consider the possibility that  contains a closed interval of positive length.Since  is closed and a proper subset of T, we may choose a closed interval  ⊆  of maximal length.The set  cannot be left invariant by any power of .If  were such a power, then   |  would have a fixed point.The orbit of this fixed point would be a finite invariant subset of , which is impossible.Next, fix  0 ∈ .Since the orbit of this point is dense and  has a nonempty interior, there must be a positive integer  such that   ( 0 ) ∈ .On the other hand, there must a point  1 ∈  with   ( 1 ) ∉ .Let   , 0 ≤  ≤ 1, be the linear path in  that connects  0 to  1 .Then   (  ) connects   ( 0 ) ∈  with   ( 1 ) ∉ .But the range of   is contained in .An argument using the intermediate value theorem shows that this contradicts the maximality of .We have shown that  has empty interior.Remark 7. The properties of the dynamical system (, ) used in the above result are that (, ) is minimal and that  is an infinite, closed, proper subset of T.
Since  is a proper subset of T, we may select a point  0 ∈ T with  0 ∉ .Parameterize T by the map  : [0, 1[ → T defined by  () =  0 +  mod 1. ( This continuous bijection is orientation preserving, provided we orient [0, 1[ in the standard way. is also a homeomorphism from ]0, 1[ to the open set T\{ 0 }.Set Ξ =  −1 () and  =  −1 ∘  ∘  (see Figure 1).Thus, Ξ is homeomorphic to  and (, ) and (Ξ, ) are isomorphic dynamical systems.In particular, Ξ is an infinite, perfect compact subset of the open interval ]0, 1[ and (Ξ, ) is a minimal dynamical system.There is also a useful reformulation of the condition that  preserves cyclic order.Let ,  and  be three distinct points in  ⊂ T with distinct images under .Then, the points  =  −1 (),  =  −1 (), and  =  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, since Ξ is perfect, one can construct a sequence   ∈ Ξ such that   ̸ =  for all  ∈ N and   →  as  → ∞.By passing to a subsequence, one can further arrange   to be strictly monotonic.Moreover, by the continuity of , (  ) → () = .Set   = (  ).Since  has finite preimages, (  ) ̸ =  for all but finitely many .By once again passing to a subsequence, if necessary, we may arrange that   is strictly monotone.Proposition 9.The preimages of  : Ξ → Ξ have cardinality at most two.
Case  (0)  and  (2)   Approach  from the Same Side.We give the argument when both sequences approach  from below; the other possibility is dealt with similarly.In this case, we can use the properties of sequence  ()   to find indices  and  so that and  (0)  ,  1 and  (2)   are in increasing order.This would mean that the arrow diagram for the triple  (0)  ,  1 ,  (2)   has an odd number (one) of crossings, which is not possible.
Case  (0)  ↗  and  (2)    ↘ .By invoking Lemma 8 again, we construct an   that is sufficiently close to  1 so that  0 <   <  2 and (  ) ̸ = .We treat the case where (  ) <  = ()-the other case is handled similarly.In this situation, there is a sufficiently large  with the property that   <  (2)   .So, Thus, the arrow diagram for  0 ,   ,  (2)    has an odd number (one) of crossings.
Hence, in all three cases, we have a contradiction.
By using Lemma 8, we can find  near  such that  < () < .However, this would lead to an arrow diagram for the triple ,   ,   with an odd number of crossings.Now, consider  ∈ Ξ with   <  <   .By the definition of   and   , () must be distinct from  or .Hence () must be strictly between  and .This is a contradiction due to an impossible arrow diagram-this time for the triple   , ,   .
Proof.Failure of the conclusion clearly leads to an impossible arrow diagram for the triple , , .
If () = , then by Lemma 8 there is  ∈ Ξ near  such that () > () > .This means that the arrow diagram for , ,  is not allowed.In a similar way, () =  can also be ruled out.
Only the middle inequality remains.Suppose, to the contrary, that () > ().The previous proposition implies that () ≤ () ≤ (), for any  ∈ Ξ that is strictly between  and .Thus, Im  is a proper subset of .This violates the minimality assumption.

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 .Fix one such, .Regard  as a measure on [0, 1) and write γ for its cumulative distribution function: Since every point of the infinite set Ξ has dense orbit, the invariant probability measure  cannot include any point masses.Thus γ is continuous.Hence, the restriction  = γ| Ξ is also a continuous, monotone increasing map from Ξ to [0, 1].Finally, because and the fact that γ is locally constant on the complement of Ξ,  :  → [0, 1] is surjective.
Proof.The first statement follows directly from the facts that spt  ⊆ Ξ and  contains no point masses.
To prove the second statement, first put  = Ξ ∩ ( 0 ,  Finally, write P :  2 (Ξ, ) →  2 (Ξ, ) for the projection onto the subspace of functions left invariant by .By the Mean Ergodic Theorem (see, e.g., [11, pp. 32]), the averages converge in  2 (Ξ, ) to the projection P[1  ]().On the other hand, since  is a nonempty open set and (Ξ, ) is a minimal dynamical system, there is an  > 0 such that See, e.g., Proposition 4.7 in [12].Hence, keeping Fatou's theorem in mind, The above result can be restated as follows: for any distinct pair ,   ∈ Ξ, () = (  ) if and only if  and   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 measure.
The left hand side is just ( −1 [0, ]).Since  is monotone increasing, continuous, and surjective,  ( Remark 17.In the appendix to the article by Goldberg and Tresser [8], an analytic argument for characterizing rotational subsystems for the map () =  ⋅  mod 1 (where  > 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 maps Γ : Ξ → T and  : T → T by Proposition 16 proved that In other words, Γ is a surjective, continuous mapping of the dynamical system (Ξ, ) to the dynamical system (T, ).In view of Proposition  (26) The second limit has been evaluated by invoking Weyl's equidistribution theorem (see [13,14]).Since such  0 are dense in Ξ, the cumulative distribution of the measure  is uniquely determined by (Ξ, ).In other words, (Ξ, ) is uniquely ergodic.
With this last result, we have completed the proof of Theorem 4. Indeed, the analysis of this section transfers over to the dynamical system (, ) by means of the isomorphism,  −1 , of dynamical systems.In particular, one sets  = Γ ∘  −1 .

The Uniform 𝑑-Fold Cover
In this section and the next we are concerned with the map  : T → T defined by for some fixed integer  ≥ 2. We will analyse the structure of rotational systems (, ) with  = |  using the findings of the last two sections.Since 0 mod 1 is a fixed point of , it is not in .So, choosing this point for  0 , we have that the parameterization  : [0, 1[ → T is given by and that  =  −1 ∘  ∘  satisfies Define Ξ and  as before, and note that  = | Ξ .We will find it convenient to work with (Ξ, ), [0, 1[ and , noting that findings in this setting convert readily to statements about (, ), T and .
The inverse image of 0 under  consists of the  points: In addition,  has  − 1 fixed points: We also set   =  −1 = 1.
Set   = [  ,  +1 [ for  = 0, . . .,  − 1 and note that the interior of   are precisely those points in T with a canonical -adic expansion that starts with the digit .(Recall that every real number has a canonical -adic expansion that does not end with an infinite string of  − 1's.)Note also that   = [0, 1], since  is just the shift map on the -adic expansion.Moreover,  is monotonic increasing on the interior of each   .Each closed interval   contains a unique fixed point   , for 0 ≤  <  − 1.The behavior of  at these fixed points can be readily determined.In particular, one checks that (See Figure 4 for the  = 5 case.)Under our operating assumptions, Ξ is infinite and minimal.Hence, none of   and   can lie in Ξ.In particular, any point in Ξ must lie in the interior of precisely one of the intervals  0 , . . .,  −1 .
Write A = {0, 1, . . ., −1} and A N 0 for the set of functions from the nonnegative integers to A. We use the usual product topology on A N 0 and let  denote the usual continuous shift on A N 0 .Define a map E : Ξ → A N 0 by where   are defined by the requirement that    ∈ [   ,    +1 ).E() is just the canonical -adic expansion of .Since each  ∈ [0, 1) has a unique expansion of this type, E is injective.
It is also obvious that  ∘ E = E ∘ .
1 Proposition 21.The map E is a homeomorphism from Ξ to Im(E).
Proof.Since Ξ is compact, it is enough to show that E is continuous on Ξ. Suppose   ,  ∈ Ξ and   →  as  → ∞.For any , we must have that    lies in an open interval of the form for each  ∈ N 0 .This implies that the th digit of   coincides with that of  for all sufficiently large .Hence E is continuous.
Because  has no mass on the open interval (  ,  +1 ), (  ) = ( +1 ).Set  0 = 0 and   = (  ) for  = 1, . . ., ℓ.Then, Recall a consequence of our previous analysis: every point  ∈ Ξ with () >  must lie to the left of every point  ∈ Ξ with () < .Thus, each Ξ  must lie completely to one side of the sole fixed point within    .Moreover, if sup Ξ  <    and inf Ξ  >    then  >  and   >   .Let  be the last index,  between 1 and ℓ with inf Ξ  >    .Clearly,  < ℓ and sup Ξ  =   and inf Ξ +1 =   .Hence, We next seek to understand the preimages of .Define (See Figure 5.) Set (Except for a countable subset, every  ∈ [0, 1[ is in D 0 .)For any such , each member of the sequence of points   () = { +  0 } with  ≥ 0 lies in exactly one of the intervals of the form [  ,  +1 [.Write   for the index with this property: Then, for any  in  −1 (), This just says the string   0   1   2 ⋅ ⋅ ⋅ is the canonical -adic expansion of .Thus, there is a unique point in the preimage  −1 ().In summary, this discussion shows how any rotational subsystem (, ) of the uniform -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.

The Inverse Process
In this section, we begin with an irrational number  0 ∈ [0, 1[ and a partition of [0, 1] into ℓ ≤  subintervals with the requirement that one of the interior nodes is 1 −  0 : and Next, select a coding that maps {0, . . ., ℓ−1} to the set of digits {0, . . .,  − 1}.More precisely, choose integers  1 , . . .,  ℓ that satisfy We will show that this data determines a rotational subset of T that inverts the process described in previous section.
Proof.The map  is invertible.The complement of D 0 is just the countable set: For the proof of the second claim, fix  ∈ [0, 1[.If the orbit of  hits the {  :  = 0, . . ., ℓ} infinitely often, then there must be an index  such that for infinitely many  ∈ N.This means that there are two distinct, positive integers  0 ,  1 with the property that But this contradicts the condition that  0 is irrational.This argument proves that the forward orbit of such a  must eventually lie completely in D 0 .
The trajectory of any point  ∈ D 0 can be encoded by an infinite string: where   =   precisely when {+ 0 } is in   .The Kronecker approximation theorem implies that each of the digits   ,  = 0, . . ., ℓ − 1 occurs infinitely often.Thus, Let  ∈ D 0 and   be a sequence in D 0 that converges to .Then, since    is in the interior of one of   ,     must eventually have this property as well.Thus,  must be continuous.
Write Ξ 0 for the image of D 0 under the map  and let Ξ be its closure in [0, 1[.It is straightforward to check that Ξ 0 is invariant under .Establishing this for its closure is complicated by the possibility that  might not be continuous on Ξ.Most of the effort in the proof of the next result is to rule this out.
Consider a point  0 ∈ Ξ 0 .An  ∈ Ξ can be approximated to within an arbitrary accuracy by an element of  0 ∈ Ξ 0 .
Write  0 ,  0 for the elements of D 0 with ( 0 ) =  0 and ( 0 ) =  0 .By Kronecker's theorem, there is a sequence   of indices such that    ( 0 ) →  0 .The continuity of  together with Proposition 24 yields lim     ( 0 ) =  0 . (54) Hence,  can be approximated to arbitrary accuracy by an element of the orbit, under , of  0 .Thus, for any  0 in Ξ 0 , the orbit of  0 is dense in Ξ.Now let  ∈ Ξ\Ξ 0 .It suffices to prove that an element in the forward orbit of  is in Ξ 0 .We may write  as a limit of a sequence   in Ξ 0 .Write   for the elements of [0, 1[ with (  ) =   .By passing to a subsequence, if needed, we may assume that (  ) converges to an element of T, say ().By Proposition 23, there is  such that   () ∈ D 0 .By continuity of , we have that (  (  )) =   ((  )) →   (()) = (  ()).Since   () ∈ D 0 , (  ()) ∈ T\{0 mod 1}.Since  −1 is continuous on the latter set,   (  ) →   ().We may apply  to both sides of this last limit to get (55) It follows that   () ∈ Ξ 0 .Since the orbit of this point is dense in Ξ, so is the orbit of .
The only point left is to prove that  respects cyclic order on Ξ.To this end, let  0 ,  1 , and  2 be an increasing triple of points in Ξ with distinct images under .We may approximate each of these points by elements of Ξ 0 without changing the number of crossings in the arrow diagram.Hence, we may assume that  0 ,  1 , and  2 are in Ξ 0 .Let  0 ,  1 , and  2 be the preimages under  of  0 ,  1 , and  2 , respectively.The   's are in increasing order due to Proposition 24.This monotonicity property of  together with the identity in that proposition, also force the arrow diagram for  0 ,  1 , and  2 under the map  to have the same number of crossings as that for  0 ,  1 , and  2 under .But it is easy to verify that  respects cyclic order.Hence, so does .

Examples for a Class of Continuous Maps
As the last two sections show, every irrational  0 ∈ [0.1) arises as the rotation number of some rotational system with respect to a standard cover of degree  > 1. (This is true for rational  0 ∈ [0, 1) as well.)It is tempting to ask if this is true for any continuous transformation of degree  > 1.
In this section, we look at mappings of the oriented unit circle,  : T → T, whose lifts to the universal cover are monotonic increasing (though not strictly).The degree of  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:  =  −1 ∘  ∘ .There is a partition with the property that, for each  = 0, . . .,  − 1,

Figure 3 :
Figure 3: Schematic for the action of .Ξ is a proper subset of the two intervals marked in blue.
14, if Γ() = Γ(  ) for distinct  and   , then either  and   are endpoints of a gap for Ξ or  =  and   = .Finally, from Lemma 15, it is clear that Γ Recalling that Γ has preimages of cardinality no greater that 2, this means that { ∈ Ξ : Γ () =  0 mod 1 for some  ∈ N} (24) is finite.It is easy to check that it is also invariant under .This contradicts the minimality of (Ξ, ).Proof.A pointis in the complement of  0 in Ξ precisely when it is the endpoint of a gap, that is, a maximal open interval contained in [, ]\Ξ.Since there are at most countably many such open intervals, we conclude that Ξ\ 0 is countable.Since the perfect set Ξ is uncountable,  0 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.Suppose  ⊆ Ξ is a nonempty open set with  ∩  0 = 0.Then,  is countable.Let  = { 1 ,  2 , . ..}. Write   = Ξ\{  } and note that   is dense and open in Ξ for each .Baire's theorem yields that ⋂ ∞ *  is Lebesgue measure on T. Proposition 18.  0 is irrational.Proof.If  0 were rational, the set {  ( 0 ) :  ∈ N} = { 0 mod 1 :  ∈ N} (23) would be finite.=1   is dense in Ξ.But, by construction,  ∩ (⋂ ∞ =1   ) = 0-a contradiction.So,  0 is dense.
) may be interpreted as the canonical -adic expansion of a real number in the open unit interval ]0, 1[.So,  : D 0 → [0, 1).Note that ({ +  0 }) is the shift 0. 1  2 . ... Proof.Kronecker's theorem also implies injectivity of : Let  and   be distinct points in D 0 with   > .Set Δ =   −  and note that there must be an  with the property that  and   = +Δ lie in the interior of different   's.Note that if |  −| is sufficiently small, then |   − | = |    −   |.In particular, by choosing  so that |   − | is sufficiently small, we may insure that    and     will be in different   's.Hence the adic encodings of  and   are different.Since neither of these encodings can end with an infinite string of  − 1's, (t) ̸ = (  ).
Proposition 25.Ξ is a compact subset of ]0, 1[ and is invariant under .Proof.By Remark 22, there is an  with the property that for each  ∈ [0, 1[, the set{  () : 0 ≤  ≤ }(53)has nonempty intersection with the interiors of each of the intervals  0 , ...,  ℓ−1 .Now consider the corresponding set of words.In particular, let W denote the set of all words in the alphabet { 1 , ...,  ℓ } of length  for which each   occurs at least once.Let  and   be the smallest and largest word, respectively, in the finite set W. Let  be the first digit of   that is not  − 1. Write V for the word obtained by replacing  with  + 1 in   .The words  and V are lexicographically smaller and larger, respectively, than every element of Ξ 0 .Write  and  for the numbers in [0, 1[ corresponding to  and V, noting that 0 <  <  < 1.It is clear that Ξ 0 ⊆ [, ].Hence Ξ is compact in the open unit interval.By examining  explicitly, it is easy to check that  −1 [, ] is a compact subset of [0, 1[ that does not contain any of the points  0 , . . .,  −1 .Since these are precisely the points of discontinuity of ,  is continuous on