Abstract
By a rotational system, we mean a closed subset of the circle, , together with a continuous transformation with the requirements that the dynamical system be minimal and that respect the standard orientation of . We show that infinite rotational systems , with the property that map 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, . Because our main result makes no explicit mention of a global transformation on , we show that such a structure theorem holds for rotational systems that arise as invariant sets of any continuous transformation with finite preimages. In particular, there are no explicit conditions on the degree of . We then give a development of known results in the case where for an integer . 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.
1. Introduction
In what follows, 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 orbitis dense in .
Definition 1. Let . 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 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 . The dynamical system is an extension of if there is a continuous, surjective function such that .
Our main result is as follows.
Theorem 4. Let be a rotational system such that is an infinite, proper subset of . In addition, suppose that the continuous mapping has finite preimages, that is, for each . Then,(i)the dynamical system is an extension of an irrational rotation of the circle via a map that is compatible with the cyclic ordering on both and ;(ii)the function has preimages of cardinality one except at countably many points of . The preimages of these exceptional points have cardinality two and are the endpoints of gaps of the set in ;(iii) has a unique ergodic measure and is the standard Lebesgue measure on .
The angle of the rotation of 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, . Recall that a closed subset of is invariant with respect to if . In this situation, we may set and consider the dynamical system . Such a system, , is a subsystem of .
Our main theorem has the following obvious corollary.
Corollary 5. Let be a continuous mapping with finite preimages. Suppose, moreover, that is a closed, infinite, proper subset of that is invariant with respect to . If is rotational, then all three conclusions of Theorem 4 hold.
In the case where is given by for an integer , Corollary 5 has an extensive history. Ideas related to the 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 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 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 , 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 , Theorem 4 and its corollary shed no light on how to determine which irrational numbers can be realized by rotational subsystems of . When , it is well known that there can be at most one such rotation number. In the uniform cover case with , every irrational number in can be achieved. In the last section, we consider degree mappings of 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 , 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 Theorem 4:(i) is a rotational system.(ii) is an infinite, proper subset of .(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 is an isolated point of such a dynamical system. Minimality implies that the orbit, , is dense in . As a consequence, we have that for some positive integer . The set must then be finite as well as dense. Therefore, . This contradiction implies that cannot have any isolated points. Thus, is a perfect subset of ; 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 , 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 . Since the orbit of this point is dense and has a nonempty interior, there must be a positive integer such that . On the other hand, there must a point with . Let , , be the linear path in that connects to . Then connects with . 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 .
Since is a proper subset of , we may select a point with . Parameterize by the map defined by This continuous bijection is orientation preserving, provided we orient in the standard way. is also a homeomorphism from to the open set . Set and (see Figure 1). Thus, is homeomorphic to and and are isomorphic dynamical systems. In particular, is an infinite, perfect compact subset of the open interval and 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 with distinct images under . Then, the points , , and have distinct images under . The action of on this triple can be conveniently encoded by an arrow diagram as in Figure 2. That the arcs and have the same orientation in 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.
Lemma 8. Suppose and . Then, there are sequences and in such that(i) and are strictly monotone sequences,(ii) for all ,(iii) 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 is perfect, one can construct a sequence such that for all 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 have cardinality at most two.
Proof. Suppose, to the contrary, that , , are three distinct points with the same image under . So for and . We may assume that , , and are arranged in increasing order in . Let and , be the sequences in obtained by applying Lemma 8 to , , and , respectively.
The proof proceeds according to how and approach .
Case and 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 thatand , and are in increasing order. This would mean that the arrow diagram for the triple has an odd number (one) of crossings, which is not possible.
Case and . Choose sufficiently large so thatBut . Hence, the arrow diagram for has an odd number (three) of crossings.
Case and . By invoking Lemma 8 again, we construct an that is sufficiently close to so that and . We treat the case where —the other case is handled similarly. In this situation, there is a sufficiently large with the property that . So,Thus, the arrow diagram for has an odd number (one) of crossings.
Hence, in all three cases, we have a contradiction.
Set and . Since is a compact, subset of the open interval we haveSince is minimal, is surjective. Consequently, we may put and .
Proposition 10. The inequalityholds. Moreover, is a gap for , that is, .
Proof. Suppose that . Minimality ensures that . 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 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 , , .
Finally, since has no isolated points, and .
Lemma 11. Let . If and , then, for any with , we must have .
Proof. Failure of the conclusion clearly leads to an impossible arrow diagram for the triple , , .
Proposition 12. .
Proof. Minimality rules out the possibilities that and .
If , then by Lemma 8 there is 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 that is strictly between and . Thus, Im is a proper subset of . This violates the minimality assumption.
For any , set
Proposition 13. is monotone increasing on the sets and . Moreover,
Proof. Let with . Suppose . Since , the definition of forces . Thus, the arrow diagram for the triple , , is impossible. This contradiction shows that is monotonic increasing on . A similar argument applies to .
Next, let and suppose . Since has no fixed points, . For the same reason, . Appealing to the monotonicity property proved in the previous paragraph, we have . Lemma 11 then implies that is a nonempty, proper, closed invariant subset of —a contradiction. Thus for . The last inequality can be verified similarly. See Figure 3 for a diagram of the structure of .
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 that is invariant under . Fix one such, . Regard as a measure on and write for its cumulative distribution function:Since every point of the infinite set has dense orbit, the invariant probability measure cannot include any point masses. Thus is continuous. Hence, the restriction is also a continuous, monotone increasing map from to . Finally, becauseand the fact that is locally constant on the complement of , is surjective.
Proposition 14. (i) If and , then .
(ii) On the other hand, if , then .
Proof. The first statement follows directly from the facts that spt and contains no point masses.
To prove the second statement, first put andFinally, write for the projection onto the subspace of functions left invariant by . By the Mean Ergodic Theorem (see, e.g., [11, pp. 32]), the averagesconverge in to the projection . On the other hand, since is a nonempty open set and is a minimal dynamical system, there is an such thatSee, 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 , , if and only if and are endpoints of a gap of . A further useful corollary of this is that any nonempty open subset of has positive -measure.
Lemma 15. One haswhere denotes Lebesgue measure on .
Proof. It is enough to show thatThe left hand side is just . Since is monotone increasing, continuous, and surjective, for some with . But this just means .
Proposition 16. There is an such thatfor all .
Proof. Put . If , then the invariance of the measure implies that If , then
Remark 17. In the appendix to the article by Goldberg and Tresser [8], an analytic argument for characterizing rotational subsystems for the map (where ) 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 and byProposition 16 proved thatIn other words, is a surjective, continuous mapping of the dynamical system to the dynamical system . In view of Proposition 14, if for distinct and , then either and are endpoints of a gap for or and . Finally, from Lemma 15, it is clear that is Lebesgue measure on .
Proposition 18. is irrational.
Proof. If were rational, the setwould be finite. Recalling that has preimages of cardinality no greater that , this means thatis finite. It is easy to check that it is also invariant under . This contradicts the minimality of .
Lemma 19. Let be those points with the property that for every , the intervals and contain infinitely many points of . Then is a dense, uncountable subset of .
Proof. A point is in the complement of in precisely when it is the endpoint of a gap, that is, a maximal open interval contained in . Since there are at most countably many such open intervals, we conclude that is countable. Since the perfect set is uncountable, 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 is a nonempty open set with . Then, is countable. Let . Write and note that is dense and open in for each . Baire’s theorem yields that is dense in . But, by construction, —a contradiction. So, is dense.
Theorem 20. The dynamical system is uniquely ergodic.
Proof. Let . By Proposition 14 and Lemma 19,(i) and ,(ii) and . Therefore, for any and ,(Here the braces denote fractional part.) Consequently,The second limit has been evaluated by invoking Weyl’s equidistribution theorem (see [13, 14]). Since such are dense in , the cumulative distribution of the measure is uniquely determined by . In other words, 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, , of dynamical systems. In particular, one sets .
4. The Uniform -Fold Cover
In this section and the next we are concerned with the map defined byfor some fixed integer . We will analyse the structure of rotational systems with using the findings of the last two sections.
Since is a fixed point of , it is not in . So, choosing this point for , we have that the parameterization is given byand that satisfiesDefine and as before, and note that . We will find it convenient to work with , and , noting that findings in this setting convert readily to statements about , and .
The inverse image of under consists of the points:In addition, has fixed points:We also set .
Set for and note that the interior of are precisely those points in 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 ’s.) Note also that , 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 . The behavior of at these fixed points can be readily determined. In particular, one checks that
(See Figure 4 for the case.) Under our operating assumptions, is infinite and minimal. Hence, none of and can lie in . In particular, any point in must lie in the interior of precisely one of the intervals .
Write and for the set of functions from the nonnegative integers to . We use the usual product topology on and let denote the usual continuous shift on . Define a map bywhere are defined by the requirement that . is just the canonical -adic expansion of . Since each has a unique expansion of this type, is injective. It is also obvious that .
Proposition 21. The map is a homeomorphism from to .
Proof. Since is compact, it is enough to show that is continuous on . Suppose and as . For any , we must have that lies in an open interval of the form . Since is continuous on ,for each . This implies that the th digit of coincides with that of for all sufficiently large . Hence is continuous.
Let be the nonempty sets in the listThe indexing can be arranged so for some withSetSince is perfect, for and is nonempty. Consequently, is the closed interval and has positive length (see Proposition 14). Because has no mass on the open interval , . Set and for . Then,Recall a consequence of our previous analysis: every point with must lie to the left of every point with . Thus, each must lie completely to one side of the sole fixed point within . Moreover, if and then and . Let be the last index, between and with . Clearly, and and . Hence, .
We next seek to understand the preimages of . Define by(See Figure 5.) Set(Except for a countable subset, every is in .) For any such , each member of the sequence of points with lies in exactly one of the intervals of the form . Write for the index with this property:Then, for any in ,This just says the string is the canonical -adic expansion of . Thus, there is a unique point in the preimage .
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 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 number and a partition of into subintervals with the requirement that one of the interior nodes is :and . Set for . Next, select a coding that maps to the set of digits . More precisely, choose integers that satisfyWe will show that this data determines a rotational subset of that inverts the process described in previous section.
Let be defined by . Write for the corresponding rotation of and note that :
Remark 22. It is well known that is a minimal dynamical system. Hence, for any non-empty open there is an with the following property: for every ,for at least one with . (See, e.g., Proposition 4.7 in [12].)
It follows easily that the analogous statement holds for and any non-empty open . (Simply apply the previous observation to .)
As before, let be the set of satisfying for all and . In other words, consists of those points of whose forward orbit does not contain any of the nodes . is invariant under the map .
Proposition 23. (i) The complement of in is countable.
(ii) For every , there is an integer with the property that .
Proof. The map is invertible. The complement of is just the countable set:For the proof of the second claim, fix . If the orbit of hits the infinitely often, then there must be an index such thatfor infinitely many . This means that there are two distinct, positive integers , with the property thatBut this contradicts the condition that is irrational. This argument proves that the forward orbit of such a must eventually lie completely in .
The trajectory of any point can be encoded by an infinite string:where precisely when is in . The Kronecker approximation theorem implies that each of the digits , occurs infinitely often. Thus,may be interpreted as the canonical -adic expansion of a real number in the open unit interval . So, . Note that is the shift .
Proposition 24. The map is a continuous, injective, strictly monotonic increasing and
Proof. Kronecker’s theorem also implies injectivity of : Let and be distinct points in with . Set and note that there must be an with the property that and 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 ’s, .
Equation (52) follows directly from (51) and the ensuing discussion.
Let and be a sequence in that converges to . Then, since is in the interior of one of , must eventually have this property as well. Thus, must be continuous.
It only remains to prove that is monotone increasing. Let , with . Write and . Because is injective, there is a first index for which . If , implies . This in turn yields . If , then entails that both and are in interior of the same . Since is increasing on the interior of any , we must have . As a consequence, in this case as well.
Write for the image of under the map and let be its closure in . It is straightforward to check that is invariant under . Establishing this for its closure is complicated by the possibility that might not be continuous on . Most of the effort in the proof of the next result is to rule this out.
Proposition 25. is a compact subset of and is invariant under .
Proof. By Remark 22, there is an with the property that for each , the sethas nonempty intersection with the interiors of each of the intervals . Now consider the corresponding set of words. In particular, let denote the set of all words in the alphabet of length for which each occurs at least once. Let and be the smallest and largest word, respectively, in the finite set . Let be the first digit of that is not . Write for the word obtained by replacing with in . The words and are lexicographically smaller and larger, respectively, than every element of . Write and for the numbers in corresponding to and , noting that . It is clear that . Hence is compact in the open unit interval.
By examining explicitly, it is easy to check that is a compact subset of that does not contain any of the points . Since these are precisely the points of discontinuity of , is continuous on . Because is invariant, . Consequently, and is continuous on .
Let . Then for a sequence . By the observations in the previous paragraph, . Since, is invariant, this proves that .
As before, define by .
Theorem 26. The dynamical system is rotational.
Proof. We first prove the minimality of .
Consider a point . An can be approximated to within an arbitrary accuracy by an element of . Write , for the elements of with and . By Kronecker’s theorem, there is a sequence of indices such that . The continuity of together with Proposition 24 yieldsHence, can be approximated to arbitrary accuracy by an element of the orbit, under , of . Thus, for any in , the orbit of is dense in .
Now let . It suffices to prove that an element in the forward orbit of is in . We may write as a limit of a sequence in . Write for the elements of with . By passing to a subsequence, if needed, we may assume that converges to an element of , say . By Proposition 23, there is such that . By continuity of , we have that . Since , . Since is continuous on the latter set, . We may apply to both sides of this last limit to getIt follows that . Since the orbit of this point is dense in , so is the orbit of .
The only point left is to prove that respects cyclic order on . To this end, let , , and be an increasing triple of points in with distinct images under . We may approximate each of these points by elements of without changing the number of crossings in the arrow diagram. Hence, we may assume that , , and are in . Let , , and be the preimages under of , , and , 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 , , and under the map to have the same number of crossings as that for , , and under . But it is easy to verify that respects cyclic order. Hence, so does .
6. Examples for a Class of Continuous Maps
As the last two sections show, every irrational arises as the rotation number of some rotational system with respect to a standard cover of degree . (This is true for rational as well.) It is tempting to ask if this is true for any continuous transformation of degree .
In this section, we look at mappings of the oriented unit circle, , whose lifts to the universal cover are monotonic increasing (though not strictly). The degree of will be assumed to be at least . It is well known that such maps must have at least one fixed point. Conjugating by the appropriate rotation, we may arrange that 0 is a fixed point. As before, we parameterize by the unit interval: . There is a partitionwith the property that, for each ,(i) is monotone increasing function on each half-open interval ,(ii) and .
Applying a piecewise linear change of coordinates, we may assume that for .
Next, we set up a standard symbolic encoding for in terms of infinite strings of the alphabet . In particular, for each finite (nonempty) word , let
As in Section 4, denote by the set of infinite words in the alphabet with the usual product topology. The shift on defined byis continuous with respect to this topology.
Definition 27. A finite word is a prefix of a word if where means concatenation of words.
We collect some useful remarks that are easy to check.
Remark 28. (i) are half-open intervals that are closed on the left.
(ii) Let and be finite words. Suppose that is a prefix for the word . Then . In addition, if for some , then .
(iii) For any fixed and , the collectionis a partition of .
(iv) If and , then(v) If and are of the same length and precedes in lexicographical order, then
Each point determines a unique infinite word . Specifically, ifBy construction,Finally, for a given finite word , the half-interval consists precisely of those with the property that is a prefix for .
Remark 29. The lexicographical ordering on and the usual order on are also compatible with the factor map . In particular, for any ,(i) implies ,(ii) implies .Thus, for any , the preimage is either empty, a singleton, or an interval of positive length. In particular, the cardinality of exceeds one for only countable many words .
Lemma 30. Let be a sequence in with the property that converges in to . Suppose, in addition, that the infinite word does not end with an infinite sequence of ’s. Then any convergent subsequence with a limit, say , has the property that with .
Proof. Let and set for each . Since , we have that, for each , for all sufficiently large . By Remark 28 and the hypothesis on , we have that for any there is an with the property thatThus, must be in for each . This just means that .
Remark 31. If the infinite word does not terminate with an infinite sequence of ’s, there is an with the property that . To see this, first construct a sequence by the requirement that for each . (Here is defined as in the previous proof.) By construction, as . Let be any limit point of this sequence in . By Lemma 30, and .
The map is not continuous everywhere on . However, one can characterize the points of continuity.
Lemma 32. Let be a point with the property that does not terminate with an infinite sequence of ’s. Then is a point of continuity for the map .
Proof. Write for and let . The hypothesis entails that does not lie on the left endpoint of any of the . In other words, is in the interior of all the . Suppose as . For any , for all sufficiently large . For such , will have as a prefix. Since is arbitrary, as .
Now choose an irrational number and a compatible partition as in (43) of the last section. Let be the rotational system constructed from this data and write for the associated system on .
Let consist of those points with the property that the associated word is in . The word associated with an cannot end with an infinite sequence of ’s; this would contradict the minimality of the infinite system , for instance. Hence, by Remark 31, for any there is a with .
We next claim that is (sequentially) compact in . Indeed, recall that is compact (see Proposition 21). Let be any sequence in . By passing to a subsequence, we may assume that converges in and converges to a . By Lemma 30, we have that satisfies and .
Another feature is that no can have the property that ends with an infinite string of ’s. (The same statement holds for any where .) Consequently, by Lemma 32, is continuous on .
Consider the composition . The function assigns to a the element in with the same symbolic representation. In view of Proposition 21 and the previous paragraph, is continuous. It is also clear that it is a map between the dynamical systems and , that is, .
Let be a compact, minimal subsystem of and write . The factor map restricts to in a natural way: . Since is minimal as well, maps to surjectively. In particular, is infinite.
The only point left open is whether 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 the case, is a rotational system for all irrational except for a countable number of possible exceptions.
Proof. In this case, the choice of determines the partition in (43). We will write for the associated rotational system. Note that, for distinct choices and , the corresponding sets, and 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 , which yields .
Thus, the family of sets, , is an uncountable collection of disjoint sets. By Remark 29, is injective on all but countably many of . For such , is a continuous bijection between compact sets. So is a homeomorphism. Since respects the ordering of , preserves cyclic order.
The case differs in that, for each , there are uncountably many choices for the partition, , described in (43). Let denote the rotational system uniquely associated with and , according to the method of Section 5. For distinct partitions and compatible with , and are disjoint. Thus, will be injective on for all but countably many of the partitions compatible with . This yields the following.
Proposition 34. If and is an irrational number in , there are uncountable many rotational systems with rotation number .
Conflicts of Interest
The author declares that he has no conflicts of interest regarding the publication of this paper.
Acknowledgments
This work was supported in part by a Faculty Research Fellowship from Eastern Michigan University (Fall 2016).