On the structure of the totally ordered set of unimodal cycles

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ON THE STRUCTURE OF THE TOTALLY ORDERED SET

OF UNIMODAL CYCLES

IRENE MULVEY

(Received 26 April 1999)

Abstract.We continue the study of a class of unimodal cycles where each cycle in the class is forced by every unimodal cycle not in the class. For every order, we identify the cycle in the class of that order, which is maximal with respect to the forcing relation. 2000 Mathematics Subject Classiﬁcation. Primary 37E05, 37E15.

1. Introduction. In 1964, Šarkovs’ki˘ı deﬁned a linear order on the set of natural numbers:

3≺5≺7≺ ··· ≺2·3≺2·5··· ≺ ··· ≺22_·₃_≺₂2_·₅_{··· ≺ ··· ≺}₂3_≺₂2_≺₂_≺₁ (1.1) and proved the following theorem.

Theorem1.1(Šarkovs’ki˘ı [5]). Letf:R→Rbe a continuous map. The set of (least) periods off is a tail of the above order. Conversely, for every tail of the above order, there is a continuous mapf:R→Rhaving exactly those periods.

In 1987, Baldwin [2] considered not only the least period of a periodic point, but also the orbit type. He deﬁned the forcing relation on ﬁnite cyclic permutations (cycles), proved that the forcing relation is a partial order, and provided an algorithm to decide when one cycle forces another.

A cycle isunimodalif the canonical linear map it determines has exactly one turning point. Throughout this paper, we assume a unimodal cycle has exactly one turning point and it is a maximum. It is shown in [3] that the forcing relation is a total order on the set of unimodal cycles.

In [4], we described a class of unimodal cycles, where every cycle in the class is forced by every unimodal cycle not in the class. In this paper, for each ordern, we identify the maximal cycle of ordernin this class with respect to the forcing relation.

2. Deﬁnitions. Letf :I→I be a continuous map of a compact interval to itself. We deﬁnef0_(x)₌_x_{and for}_n_∈_N_,_n_≥_1,_fn_(x)₌_{f (f}n−1_{(x)). If there exists}_k_∈_N such thatfk_(x)₌_{x, then we say}_x_{is periodic for}_f_and_x_{has least period}_{s, where}

s is smallest element ofN such thatfs_(x)₌_{x. Fo r}_x_∈_{I, the orbit of}_x _{is the set} {fn_(x)_|_n_≥₀_}_{. If}_x_{is periodic with least period}_{s, then the orbit of}_x_{is the ﬁnite set}

X= {x,f (x),f2_(x),...,fs−1_(x)_}_.

A cycle of ordermis a bijectionη:{1,2,...,m} → {1,2,...,m}such thatηk₍₁₎_≠₁

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η(km)=k1. We assume, without loss of generality, thatk1=1. Write the elements of

a periodic orbitXin increasing order:x1< x2<···< xs. We sayXhas orbit typeηif

ηis a cycle of ordersand for eachi∈ {1,2,...,s}, f (xi)=xη(i). In fact, we say each

xi∈Xhas orbit typeη.

The forcing relation on cycles is deﬁned as follows:θforcesηif and only if every continuous map of the interval that has a periodic orbit of typeθhas a periodic orbit of type η. Baldwin [2] proved that this relation induces a partial order on the set of cycles.

Letθbe a cycle of ordern. The canonicalθ-linear mapLθis deﬁned byLθ:[1,n]→

[1,n], whereLθ=θon{1,2,...,n}andLθis linear on[i,i+1]for 1≤i≤n−1. It is

known thatθforcesηif and only ifLθhas a periodic orbit of orbit typeη[1]. A cycle

θis called unimodal ifLθhas exactly one turning point. Throughout this paper, every

unimodal cycle is assumed to have one turning point and it is a maximum. The forcing relation deﬁned by Baldwin induces a total order on the set of unimodal cycles [3].

Letθbe a cycle of ordern.

Deﬁnition 2.1. The RL-pattern for θ is the element S = S1S2···Sn ∈ {R,L}n

satisfying

Si=

  R ifθ

i_{(1) > θ}i−1_(1),

L ifθi_{(1) < θ}i−1_(1). (2.1) For example, theRL-pattern forθ=(12354)isRRRLL. Bothθ=(15234)andθ=

(13245)have theRL-patternRLRRL. EveryRL-pattern begins withRand ends with L. IfXis a periodic orbit of orbit typeθ, then theRL-pattern forXis theRL-pattern forθ.

3. Preliminaries. The next four results are proved in [4].

Lemma3.1. Ifθis unimodal, thenLθhas a unique ﬁxed point.

Lemma3.2. Ifθis unimodal, then theRL-pattern forθcannot contain two consec-utiveL’s.

Deﬁnition3.3. LetCdenote the class of unimodal cycles whoseRL-pattern does not contain two consecutiveR’s.

Lemma3.4. Letθ∈Cbe of ordern. Then (1)nis even andn≥4.

(2)θ(n−i)=i+1fori=0,1,2,...,k−1. (3)θ(1)=k+1.

Theorem3.5. Letη∈Cand letθbe a unimodal cycle not inC. Thenθforcesη. Deﬁnition3.6. For evenn≥4,

¯

θn=1,k+1,k,k+2,k−1,k+3,k−2,...,n−1,2,n. (3.1)

Remark3.7. In the main theorem, we prove that ¯θnis the maximal cycle of order

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Lemma3.8. Letθ∈Cbe of ordern. Thenθ=θ¯nif and only ifθ(2)=n.

Proof. The mapLθis a bijection which maps{3,4,...,k}to{k+2,k+3,...,n−1}.

Sinceθ(2)=nandθis unimodal,Lθis decreasing on{3,4,...,k}.

The next lemma restates Lemma 3.8 in the way we will often use it.

Lemma3.9. Letθ∈C be of order n. Then θ=θ¯n if and only if there existsi∈ {3,4,...,k}such thatθ(i)=n.

4. Main theorem

Lemma4.1. Letθ∈Cbe of ordern. Then the only periodic point forLθin(k,k+1) is the unique ﬁxed point forLθ.

Proof. Lθ(k+1)=k and Lθ(k)≥k+2. If x < p, then for j ≥0, L2θj(x) < p

andL2θj+1(x) > p. Let ¯kbe the unique preimage ofk+1 in(k,p). Ifx∈(k,k], then¯

Lθ(x)≥k+1 andx is not periodic since[1,k]∪[k+1,n] is invariant underLθ. If

x∈(¯k,p), thenL2

θ(x) < p. IfL2θ(x) <k, then¯ xis not periodic. IfL2θ(x)∈(k,p), then¯

(p−L2

θ(x))≥4(p−x)sinceLθ≤ −2 o n(k,k+1).

Remark4.2. In the next few results, we assumeθ∈Cis of ordern,θ=θ¯n, andθ

forces ¯θn. By Deﬁnition 3.6 this means thatLθ has a periodic orbitXof type ¯θn. As

standard notation, we write the elements ofXin increasing order and label them as follows:

X=x1< x2<···< xn, (4.1)

where

Lθxi=x¯θn(i). (4.2)

Lemma 4.3. Suppose that θ ∈C is of order n andθ = θ¯n. Ifθ forces θ¯n, then {x1,x2,...,xk} ⊆(1,k)and{xk+1,xk+2,...,xn} ⊆(k+1,n).

Proof. SinceX= {x1< x2<···< xn}is an orbit of type ¯θn, we have

x1,x2,...,xk⊆x|Lθ(x) > xandL2θ(x) < Lθ(x)⊆[1,p],

xk+1,xk+2,...,xn⊆x|Lθ(x) < xandL2θ(x) > Lθ(x)⊆[p,n].

(4.3)

But by Lemma 4.1, noxican be in(k,k+1).

Lemma4.4. Letθ∈C be of ordernandθ=θ¯n. Ifθ forcesθ¯n, then there are at most twoxiin(1,3).

Proof. If not, then (at least)x1< x2< x3are in(1,3). By Lemma 3.9,Lθis

increas-ing on(1,3). SoLθ(x3) > Lθ(x2)orxθ¯n(3)> x¯θn(2) orxn−1> xn. Hence, there are at most twoxiin(1,3).

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Deﬁnition4.5. Forθ∈Cof ordern, letPj= {i|θ(i)≥j}. Lemma4.6. Letθ∈Cbe of ordern. Then

(1) the order ofPjisn−(j−1),

(2) for1≤j≤n, either the largest element inPjor the smallest element inPjmaps tojunderLθ.

Proof. The proof of (1) is obvious.

(2) is proved inductively. We know thatP1= {1,2,...,n}andLθ(n)=1, so(2) is

true forj=1. Forj >1,Pj is a ﬁnite sequence of consecutive integers since each

Pj is derived fromPj−1 by removing the maximum or minimum ofPj−1 fromPj−1. Assume that (2) is true for j. Consider Pj+1 = {i|θ(i) ≥j+1}. Let a=minPj+1 andb=maxPj+1. If neitheranorbmaps toj+1 underLθ, thenLθ(a) > j+1 and

Lθ(b) > j+1. But there exists an integerc∈(a,b)such thatLθ(c)=j+1, and this

contradicts the fact thatθis unimodal.

Lemma4.7. Letθ∈Cbe of ordernand letk=n/2. Letj¯≥0be the largest integer such that2+j < k¯ −j¯. Then for allj=0,1,2,...,j¯,

Lθ((2+j,k−j))⊆(k+2+j,n]. (4.4)

Proof. Forj=0, consider, by Lemma 3.4,

Pk+2= {i|θ(i)≥k+2} = {2,3,...,k}. (4.5) Soifx∈(2,k), thenLθ(x) > k+2. That is,

Lθ((2,k))⊆(k+2,n]. (4.6)

Note thatPk+3properly contains{3,4,...,k−1}since by Lemma 4.6 exactly one of

{2,k}maps tok+2 underLθ.

Inductively, suppose thatjsatisﬁes 0≤j <j¯and

Lθ((2+j,k−j))⊆(k+2+j,n]. (4.7)

Consider

Pk+2+(j+1)=i|θ(i)≥k+2+(j+1)⊃2+(j+1),...,k−(j+1) (4.8)

(a proper containment). So ifx∈(2+(j+1),k−(j+1)), then

Lθ(x) > k+2+(j+1). (4.9)

That is,

Lθ2+(j+1),k−(j+1)⊆k+2+(j+1),n, (4.10)

as needed.

Lemma4.8. Letθ∈Cbe of ordern, and letk=n/2. Assume thatθ=θ¯nandθforces

¯

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Proof. We have θ ∈ C of order n, θ = θ¯n, and θ forces ¯θn. By Lemma 4.3, {x1,x2,...,xk} ⊆(1,k). Let ¯j be the largest integer such that 2+j < k−j. The proof¯ is inductive. First, forj=0, suppose that there are two or morexi∈(k−1,k).Then

(at least)xk,xk−1are in(k−1,k). It follows from Lemma 3.4 that

xi+1,xk+2∈(k+1,k+2) (4.11) sincexk+1andxk+2are preimages ofxkandxk−1and

L−1

θ

[k−1,k]=[k+1,k+2]. (4.12)

Now, we havexk∈(k−1,k)and

θxk=xk+2∈(k+1,k+2). (4.13) But

θxk∈Lθ(k−1,k)⊆Lθ(2,k)⊆(k+2,n] (4.14)

by Lemma 4.7, which is a contradiction. Therefore, there is at most onexi∈(k−1,k).

Next, suppose that ˜jsatisﬁes 0≤j <˜ j¯and for allj=0,1,2,...,j,˜there is at most onexi in(k−j−1,k−j). Suppose that there are two or morexi inJ=(k−j˜−2,

k−j˜−1). There are ˜j+2 ways for this to occur, since each of the ˜j+1 intervals

(k−1,k),(k−2,k−1),...,k−j˜−1,k−j˜ (4.15)

contains at most onexi. As in the case whenj=0, the preimages of any elements in

J=(k−j˜−2,k−j˜−1)must be inK=(k+j˜+2,k+j˜+3).

Case1(xk,xk−1∈J). Thenxk+1,xk+2∈K.

Case2(xk−1,xk−2∈J). Thenxk+2,xk+3∈K. and soon, until

Casej˜+2(xk−j˜−1,xk−j˜−2∈J). Thenxk+j˜+2,xk+j˜+3∈K. In every case there existsxr∈JandLθ(xr)∈K. Now,

Lθxr∈Lθ(J)=Lθk−j˜−2,k−j˜−1⊆Lθ2+j˜+1,k−j˜−1⊆k+2+j˜+1,n, (4.16)

by Lemma 4.7. SoLθ(xr) > k+j˜+3. But

Lθxr∈K=k+j˜+2,k+j˜+3, (4.17)

which is a contradiction. This proves the lemma. Note that 2+(j˜+1) < k−(j˜+1)by hypothesis, so 2+j˜+1≤k−j˜−2, as needed in (4.16).

Lemma4.9. Letθ∈Cbe of ordern, and letk=n/2. Assume thatθ=θ¯nandθforces

¯

θn. Letj¯be the largest integer such that2+j < k¯ −j¯. Ifθ(i)=n, then3≤i≤k−j¯−2. Proof. There are at most twoxithat are less thani, as in the proof of Lemma 4.4.

Sothere are at least(k−2) xi in(i,k). Ifi > k−j¯−2, then there are at most ¯j+

1 intervals and each interval contains at most one xi. Sincek−2>j¯+1,this is a

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Lemma4.10. Letθ∈C be of ordernand letk=n/2. Assume thatθ=θ¯nandθ forcesθ¯n. Letj¯be the largest integer such that2+j < k¯ −j¯. Then there is at most one

xiin each of the intervals

(3,4),(4,5),...,k−j¯−2,k−j¯−1. (4.18)

Proof. Let ¯ibe such thatθ(¯i)=n. Thenx1<¯i < x3. By Lemma 4.9, 3≤¯i≤k−j¯−2. Deﬁne ˜iby ¯i=k−j¯−i. Suppose that there are at least two˜ xi∈(¯i,i¯+1).

Either

x2,x3∈k−j¯+˜i,k−j¯+˜i+1, (4.19) in which case

xn−1,xn−2∈k+j¯+i˜,k+j¯+˜i+1, (4.20) or

x3,x4∈k−j¯+˜i,k−j¯+˜i+1, (4.21) in which case

xn−2,xn−3∈k+j¯+i˜,k+j¯+˜i+1. (4.22) In either case, there existsxr∈(k−(j¯+i),k˜ −(j¯+˜i)+1)such that

θxr∈k+j¯+˜i,k+j¯+i˜+1. (4.23)

This guarantees that

θk−j¯+˜i+1≤k+j¯+˜i. (4.24) But, by(2)in Lemma 3.4,

θk+j¯+˜i=θn−k+j¯+i˜=θn−k−j¯−˜i=k−j¯−˜i+1=k−j¯+˜i+1. (4.25)

Sincen≥4,

θk−j¯+˜i+1< k+j¯+˜i. (4.26) This leaves the set A = {2,3,...,¯i−1} tobe mapped bijectively by θ onto a set B⊇ {k+j¯+˜i,...,n−1}. But

|A| =¯i−2=k−j¯−˜i−2,

|B| ≥(n−1)−k+j¯+˜i+1=(n−k)−j¯−˜i=k−j¯−˜i. (4.27)

Hence, there is at most onexi∈(¯i,¯i+1).

The proof continues in this fashion. We have proved that there is at most one xi∈(¯i,¯i+1). Suppose that there are at least twoxi∈(¯i+1,¯i+2). There are three

ways for this to occur. In each of these three cases, there existsxr∈(¯i+1,¯i+2)such

that

θxr∈k+j¯+˜i−1,k+j¯+˜i. (4.28)

As in the previous case,

θ¯i+2≤k+j¯+˜i−1. (4.29) This leaves the setA= {2,3,...,i¯−1} ∪ {¯i+1}with |A| =¯i−1=k−j¯−˜i−1 to be mapped bijectively onto a set containing B ⊇ {k+j¯+i˜−1,...,n−1}, where|B| ≥

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Theorem4.11. Letθ∈Cof ordern. Thenθ¯nforcesθ.

Proof. Ifθ=θ¯n, the conclusion holds trivially. Assume thatθ=θ¯nand suppose

that ¯θndoes not forceθ. Since forcing induces a total order onC,θforces ¯θn, sothere

existx1< x2<···< xnin(1,n)such that for eachi,Lθ(xi)=x¯θn(i). By Lemma 4.3, {x1,x2,...,xk} ⊆(1,k), wherek=n/2. By Lemma 4.4, there are at most twoxi∈(1,3).

This leaves at leastk−2 o f thexiin the union of intervals(3,4)∪(4,5)∪···∪(k−1,k).

But each one of thesek−3 intervals contains at most onexiby Lemmas 4.8 and 4.10.

References

[1] L. Alsedà, J. Llibre, and M. Misiurewicz,Combinatorial Dynamics and Entropy in Dimension One, Advanced Series in Nonlinear Dynamics, vol. 5, World Scientiﬁc Publishing Co., Inc., River Edge, NJ, 1993. MR 95j:58042. Zbl 843.58034.

[2] S. Baldwin,Generalizations of a theorem of Sarkovskii on orbits of continuous real-valued functions, Discrete Math.67(1987), no. 2, 111–127. MR 89c:58057. Zbl 632.06005. [3] P. Collet and J.-P. Eckmann,Iterated Maps on the Interval as Dynamical Systems, Progress

in Physics, vol. 1, Birkhäuser, Boston, Mass., 1980. MR 82j:58078. Zbl 458.58002. [4] I. Mulvey,Symbolic representation for a class of unimodal cycles, preprint, 1998.

[5] O. M. Šarkovs’ki˘ı,Co-existence of cycles of a continuous mapping of the line into itself, Ukrain. Mat. ˘Z.16(1964), 61–71 (Russian). MR 28#3121.

Irene Mulvey: Mathematics and Computer Sciences, College of Arts and Sciences, Fairﬁeld University, North Benson Road, Fairﬁeld, Connecticut06430-5195, USA