The Existence and Structure of Rotational Systems in the Circle

Definition 1. Let $$. A continuous transformation f : X → X preserves 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. A rotational system is a subset $$ and a continuous transformation f : X → X, with the properties that

• (i)

  the dynamical system (X, f) is minimal,

• (ii)

  the transformation f : X → X preserves cyclic order.

In this situation, we will simply say that (X, f) is rotational.

Definition 3. Let f_i : X_i → X_i be continuous transformations of compact metric spaces for i = 1,2. The dynamical system (X₁, f₁) is an extension of (X₂, f₂) if there is a continuous, surjective function ϕ : X₁ → X₂ such that ϕ∘f₁ = f₂∘ϕ.

Theorem 4. Let (X, f) be a rotational system such that X is an infinite, proper subset of $$. 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 $$ that is compatible with the cyclic ordering on both X and T;

• (ii)

  the function ϕ has preimages of cardinality one except at countably many points of $$. The preimages of these exceptional points have cardinality two and are the endpoints of gaps of the set X in $$;

• (iii)

  (X, f) has a unique ergodic measure μ and ϕ_∗μ is the standard Lebesgue measure on $$.

Corollary 5. Let $$ be a continuous mapping with finite preimages. Suppose, moreover, that X is a closed, infinite, proper subset of $$ that is invariant with respect to F. If (X, f) is rotational, then all three conclusions of Theorem 4 hold.

Proposition 6. The set X is a Cantor set; that is, it is a compact set that is perfect and has empty interior.

Proof. Suppose θ₀ ∈ X is an isolated point of such a dynamical system. Minimality implies that the orbit, $$, is dense in X. As a consequence, we have that fⁿ(θ₀) = θ₀ for some positive integer n. The set $$ must then be finite as well as dense. Therefore, X = {fⁿ(θ₀) : n > 0}. This contradiction implies that X cannot have any isolated points. Thus, X is a perfect subset of $$; that is, it is closed and has no isolated points.

Now, consider the possibility that X contains a closed interval of positive length. Since X is closed and a proper subset of $$, 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 θ_t, 0 ≤ t ≤ 1, be the linear path in I that connects θ₀ to θ₁. Then fⁿ(θ_t) 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 $$.

Lemma 8. Suppose x, y ∈ Ξ and g(x) = y. Then, there are sequences a_n and b_n in Ξ such that

• (i)

  a_n and b_n are strictly monotone sequences,

• (ii)

  g(a_n) = b_n for all $$,

• (iii)

  lim_n→∞a_n = x and lim_n→∞b_n = 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, since Ξ is perfect, one can construct a sequence a_n ∈ Ξ such that a_n ≠ x for all $$ and a_n → x as n → ∞. By passing to a subsequence, one can further arrange a_n to be strictly monotonic. Moreover, by the continuity of g, g(a_n) → g(x) = y. Set b_n = g(a_n). Since g has finite preimages, g(a_n) ≠ y for all but finitely many n. By once again passing to a subsequence, if necessary, we may arrange that b_n is strictly monotone.

Proposition 9. The preimages of g : Ξ → Ξ have cardinality at most two.

Proof. Suppose, to the contrary, that x₀, x₁, x₂ ∈ Ξ are three distinct points with the same image under g. So g(x_i) = y for i = 0,1 and 2. We may assume that x₀, x₁, and x₂ are arranged in increasing order in [0,1[. Let $$ and $$,  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 how $$ and $$ approach y.

Case $$ and $$ 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 $$ to find indices k and l so that

Case $$ and $$. Choose n sufficiently large so that

Case $$ and $$. 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 $$. So,

Thus, the arrow diagram for $$ has an odd number (one) of crossings.

Hence, in all three cases, we have a contradiction.

Proposition 10. The inequality

Proof. Suppose that α^′ > 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, consider x ∈ 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, since Ξ has no isolated points, α^′ ≠ α and β^′ ≠ β.

Lemma 11. Let x, 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 triple x, z, y.

Proposition 12. α < g(β) ≤ g(α) < β.

Proof. Minimality rules out the possibilities that g(α) = α and g(β) = β.

If g(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, that g(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.

Proposition 13. g is monotone increasing on the sets $$ and $$. Moreover,

Proof. Let x, 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, let x ∈ 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.

Proposition 14. (i) If x₀, x₁ ∈ Ξ and Ξ∩(x₀, x₁) = ∅, then γ(x₀) = γ(x₁).

(ii) On the other hand, if Ξ∩(x₀, x₁) ≠ ∅, then γ(x₀) > γ(x₁).

Proof. The first statement follows directly from the facts that spt μ⊆Ξ and μ contains no point masses.

To prove the second statement, first put U = Ξ∩(x₀, x₁) and

Lemma 15. One has

Proof. It is enough to show that

Proposition 16. There is an ω₀ ∈ (0,1) such that

Proof. Put ω₀ = μ([β^′, β]). If x ∈ Ξ∩[α, α^′], then the g invariance of the measure μ implies that

Remark 17. In the appendix to the article by Goldberg and Tresser [8], an analytic argument for characterizing rotational subsystems for the map F(θ) = 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.

Proposition 18. ω₀ is irrational.

Proof. If ω₀ were rational, the set

Lemma 19. Let Z₀ 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 of Z₀ 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. Suppose U⊆Ξ is a nonempty open set with U∩Z₀ = ∅. Then, U is countable. Let U = {x₁, x₂, …}. Write U_i = Ξ\{x_i} and note that U_i is dense and open in Ξ for each i. Baire’s theorem yields that $$ is dense in Ξ. But, by construction, $$—a contradiction. So, Z₀ is dense.

Theorem 20. The dynamical system (Ξ, g) is uniquely ergodic.

Proof. Let x₀ ∈ 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 any y ∈ Ξ and n ≥ 0,

Proposition 21. The map $$ is a homeomorphism from Ξ to $$.

Proof. Since Ξ is compact, it is enough to show that $$ is continuous on Ξ. Suppose x_i, x ∈ Ξ and x_i → x as i → ∞. For any n, we must have that gⁿx lies in an open interval of the form $$. Since g is continuous on [0,1)\{ξ₀, ξ₁, …, ξ_d−1},

Remark 22. It is well known that $$ is a minimal dynamical system. Hence, for any non-empty open $$ there is an N > 0 with the following property: for every $$,

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).)

Proposition 23. (i) The complement of $$ in [0,1[ is countable.

(ii) For every t ∈ [0,1[, there is an integer n ≥ 0 with the property that $$.

Proof. The map ρ is invertible. The complement of $$ is just the countable set:

For the proof of the second claim, fix t ∈ [0,1[. If the orbit of t hits the $$ infinitely often, then there must be an index i such that

Proposition 24. The map $$ is a continuous, injective, strictly monotonic increasing and

Proof. Kronecker’s theorem also implies injectivity of E: Let t and t^′ be distinct points in $$ 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 J_i’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 J_i’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.

Let $$ and t_n be a sequence in $$ that converges to t. Then, since ρ^kt is in the interior of one of J_i, ρ^kt_n must eventually have this property as well. Thus, E must be continuous.

It only remains to prove that E is monotone increasing. Let t, $$ with t < t^′. Write E(t) = a₀a₁a₂… and $$. Because E is injective, there is a first index i ≥ 0 for which $$. If i = 0, t < t^′ implies $$. This in turn yields E(t) < E(t^′). If i > 0, then $$ entails that both ρⁱ⁻¹t and ρⁱ⁻¹t^′ are in interior of the same J_k. Since ρ is increasing on the interior of any J_n, we must have $$. As a consequence, E(t) < E(t^′) in this case as well.

Proposition 25. Ξ is a compact subset of ]0,1[ and is invariant under G.

Proof. By Remark 22, there is an N with the property that for each t ∈ [0,1[, the set

By examining G 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, …, ξ_d−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, $$ and G is continuous on Ξ.

Let x ∈ Ξ. Then x_n → x for a sequence x_n ∈ Ξ₀. By the observations in the previous paragraph, G(x_n) → G(x). Since, Ξ₀ is G invariant, this proves that $$.

Theorem 26. The dynamical system (Ξ, g) is rotational.

Proof. We first prove the minimality of (Ξ, g).

Consider a point x₀ ∈ 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 $$ with E(t₀) = x₀ and E(s₀) = y₀. By Kronecker’s theorem, there is a sequence n_i of indices such that $$. The continuity of E together with Proposition 24 yields

Now let x ∈ 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 x_i in Ξ₀. Write t_i for the elements of [0,1[ with E(t_i) = x_i. By passing to a subsequence, if needed, we may assume that χ(t_i) converges to an element of $$, say χ(t). By Proposition 23, there is m such that $$. By continuity of r, we have that χ(ρ^m(t_i)) = r^m(χ(t_i)) → r^m(χ(t)) = χ(ρ^m(t)). Since $$, $$. Since χ⁻¹ is continuous on the latter set, ρ^m(t_i) → ρ^m(t). We may apply E to both sides of this last limit to get

The only point left is to prove that g 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 t_i’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.

Definition 27. A finite word v is a prefix of a word w if v⋆u where ⋆ means concatenation of words.

Remark 28. (i) A_w are half-open intervals that are closed on the left.

(ii) Let w and v be finite words. Suppose that w is a prefix for the word v. Then A_v⊆A_w. In addition, if v_k ≠ d − 1 for some k > j, then $$.

(iii) For any fixed w and n≥|w|, the collection

(iv) If w = i₁i₂ … i_n and w^′ = i₂ … i_n, then

(v) If w and v are of the same length and w precedes v in lexicographical order, then

Remark 29. The lexicographical ordering on $$ 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 any $$, the preimage ι⁻¹{w} is either empty, a singleton, or an interval of positive length. In particular, the cardinality of ι⁻¹{w} exceeds one for only countable many words $$.

Lemma 30. Let x_n be a sequence in [0,1) with the property that ι(x_k) converges in $$ to w. Suppose, in addition, that the infinite word w does not end with an infinite sequence of d − 1’s. Then any convergent subsequence $$ with a limit, say x, has the property that x ∈ [0,1) with ι(x) = w.

Proof. Let w = i₀i₁i₂… and set w_n = i₀i₁ … i_n for each $$. Since ι(x_i) → w, we have that, for each m, $$ 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

Remark 31. If the infinite word $$ does not terminate with an infinite sequence of d − 1’s, there is an x ∈ [0,1[ with the property that ι(x) = w. To see this, first construct a sequence x_n by the requirement that $$ for each $$. (Here w_n is defined as in the previous proof.) By construction, ι(x_n) → w as n → ∞. Let x be any limit point of this sequence in $$. By Lemma 30, x ∈ [0,1) and ι(x) = w.

Lemma 32. Let x₀ ∈ [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 $$.

Proof. Write w = i₀i₁i₂… for ι(x₀) and let w_n = i₀i₁ … i_n. The hypothesis entails that x₀ does not lie on the left endpoint of any of the $$. In other words, x₀ is in the interior of all the $$. Suppose x_k → x₀ as k → ∞. For any $$, $$ for all sufficiently large k. For such k, ι(x_k) will have w_n as a prefix. Since $$ is arbitrary, ι(x_n) → w = ι(x₀) as n → ∞.

Proposition 33. In the d = 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 of ω₀ determines the partition in (43). We will write $$ for the associated rotational system. Note that, for distinct choices ω₀ and $$, 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, $$, is an uncountable collection of disjoint sets. By Remark 29, ι is injective on all but countably many of $$. For such ω₀, h = h_ω 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.

Proposition 34. If d > 2 and ω₀ is an irrational number in [0,1, there are uncountable many rotational systems (Y, h) with rotation number ω₀.