ESTHER BARRAB ´ES AND DAVID JUHER
Received 5 May 2005 and in revised form 22 September 2005
We answer the following question: given anyn∈N, which is the minimum number of endpointsenof a tree admitting a zero-entropy map f with a periodic orbit of periodn? We prove thaten=s1s2···sk−
k
i=2sisi+1···sk, wheren=s1s2···sk is the
decomposi-tion ofninto a product of primes such thatsi≤si+1for 1≤i < k. As a corollary, we get
a criterion to decide whether a map f defined on a tree with eendpoints has positive entropy: iff has a periodic orbit of periodmwithem> e, then the topological entropy of
f is positive.
1. Introduction
In the last decades, many authors have studied the dynamical behaviour of continuous self-maps of one-dimensional spaces. There are also some books where most of the results are collected in (see, e.g., [3,7]). In particular, the study of the set of periods of continuous mapsf :X→X, whereXis a tree (a graph without circles), has been one of the problems that have centered the attention. The first and most famous result in this direction is ˇSarkovs’ki˘ı’s theorem [13], which gives a complete characterization of the set of periods of f whenXis a closed interval of the real line. Baldwin [5] extended this result to the case of ann-star (a tree consisting ofnedges attached to a unique central point). Recently, Alsed`a, Juher, and Mumbr ´u [2] have developed a characterization of the set of periods of f whenX is any generic tree, in terms of the topological structure ofX (number and arrangement of vertices, edges, and endpoints ofX).
One way to study the dynamical complexity of a continuous mapf :X→Xof a com-pact metric space is computing its topological entropy, a nonnegative constant which measures how the iterates of the map mix the points ofX(see [1]). It is known that an interval or line map with positive topological entropy ischaoticin the sense of Li and Yorke (see [11]). IfXis a general compact metric space, the same result has been recently obtained in [6]. It is also well known that the topological entropy of f is closely related with the sizes of the periodic orbits exhibited by f. Some results give upper or lower bounds of the set of periods of f depending on whether f has a positive entropy or not. Of course, these bounds depend strongly on the particular spaceXunder consideration.
Copyright©2005 Hindawi Publishing Corporation
WhenX is an interval, a map f :X→X has a positive entropy if and only if it has a periodic orbit of period different from a power of 2 (several authors have contributed to prove this result, see [3] for a historical survey). The zero-entropy self-maps of a generic treeXhave been characterized by Alsed`a and Ye [4] by using the notion ofdivisionof a pe-riodic orbit (seeSection 3for a definition), which is an important tool in the proof of the main result of this paper. The same authors give a maximal set of periods for zero-entropy maps in terms of the number of endpoints ofX. Blokh, in [8], proves the following result in the same spirit.
Theorem1.1 (Blokh). Let X be a tree and let f :X→X be continuous. The topological entropy of f is zero if and only if the period of each periodic orbit of f has the formk·2l, wherekis odd and not larger than the number of edges ofXand all the prime divisors ofk are not larger than the number of endpoints ofX.
Llibre and Misiurewicz, in [12], study the properties of the topological entropy of con-tinuous self-maps of graphs and its relationship with the set of periods. They introduce the notions of “godof a number” and “pantheonof a set of natural numbers”. Fork∈N, god(k) denotes the greatest odd divisor ofk. In other words, ifk=n·2lwithnodd and l≥0, then god(k)=n. ForS⊂N, the set{god(k) :k∈S}is called the pantheon ofS. Also, fork∈N, the following number is defined:
Γ(k)=
3≤p≤4k pprime
log(2k)
log (p/2)
+ 1, (1.1)
where [·] denotes the integer part. Llibre and Misiurewicz show that if f is defined on a graph withs edges and the pantheon of the set of periods of f has more thansΓ(s) elements, then the entropy of f is positive. As the authors remark, the estimatesΓ(s) is not the best possible one and they have not tried to optimize it. In the case of the interval and the circle, it is known that the best estimate of the minimum number of gods which forces positive entropy is two. But for a generic graph, the problem of determining this minimum remains open.
In this paper we restrict our attention to continuous maps defined on trees. Even in this case, there are still open questions related to the pantheon of the set of periods of a zero-entropy map. The most ambitious problem would be to determine, in terms of the topology of the tree, the maximum number of elements of the pantheon of the set of periods that a zero-entropy map can have. Clearly, this maximum has to be smaller than sΓ(s), wheresis the number of edges of the tree.Theorem 1.1gives also an upper bound, much better thansΓ(s) in general. One can easily construct simple examples showing that the exact computation of this maximum seems rather difficult and depends not only on the number of endpoints, vertices, and edges of the tree but also on the way they are organized.
and its topological entropy. One of them deals with the problem of determining the peri-ods forced by the existence of a given periodic orbit. For instance, in [9], it is proved that if the tree haseendpoints and f has an orbit of period larger thane, then f has also an orbit of periodmfor some 1< m≤e. See also [10] for similar results.
We are interested in zero-entropy maps defined on trees. Given anyn∈N, we de-termine the minimumisuch that there is a treeT withiendpoints and a zero-entropy map f :T→Thaving a periodic orbit of periodn. From now on, this minimum will be denoted byen. The main result of this paper is the following.
Theorem1.2. Letn∈Nwithn≥2. Letn=s1s2···sk be the decomposition ofninto a
product of primes such thatsi≤si+1for1≤i < k. Then,
en=s1s2···sk−
k
i=2
sisi+1···sk. (1.2)
Using the formula given byTheorem 1.2and a simple inductive argument, it is not difficult to check thaten≥2 for eachn≥2. Moreover,en=2 if and only if s1=s2=
··· =sk=2.
As a consequence ofTheorem 1.2, we obtain the following sufficient condition to de-cide whether a continuous self-map f defined on a treeThas positive entropy.
Corollary1.3. Let f :T→Tbe a tree map with a periodic orbit of periodn. Letebe the number of endpoints ofTand letn=s1s2···skbe the decomposition ofninto a product of
primes such thatsi≤si+1for1≤i < k. If
e < s1s2···sk−
k
i=2
sisi+1···sk, (1.3)
then the topological entropy of f is positive.
For instance, assume thatThas 9 endpoints andf is known to have an orbit of period n=15. Sincee15=3·5−5=10, it follows that the mapf has positive entropy. It is worth
observing that this fact cannot be directly obtained fromTheorem 1.1. Indeed,Thas at most 15 edges (it is well known that any tree witheendpoints has at most 2e−3 edges) and, thus, 15 is an admissible period for a zero-entropy map according toTheorem 1.1 since it is odd and is not larger than the number of edges ofTand all its prime divisors are not larger than the number of endpoints ofT.
We also remark thatTheorem 1.2, in the same way as the results of Blokh, Llibre, and Misiurewicz, states that the positive entropy of a map f is due to thegodsof its periods rather than to the periods themselves. To see it, taken∈Nand setn=n·2lfor some l≥0. Let n=s1s2···sk andn=r1r2···rm be the decompositions of nandn into a
product of primes such thatsi≤si+1andri≤ri+1. It is easy to check that
s1s2···sk−
k
i=2
sisi+1···sk=r1r2···rm−
m
i=2
Thus, in view ofTheorem 1.2,en=en. This implies that, for eachn∈N,en=egod(n). In other words,egod(n)endpoints are enough to assure that a zero-entropy map can be
de-fined exhibiting a periodic orbit of periodn.
2. Notation and basic definitions
Given any subsetXof a topological space, we will denote by Int(X) and Cl(X) the interior and the closure ofX, respectively. For a finite setA, we will denote its cardinality by|A|.
By aninterval, we mean any space homeomorphic to [0, 1]⊂R. Atreeis a uniquely arcwise connected space that is either a point or a union of finitely many intervals. Any continuous map from a tree into itself will be called atree map. The set of periods of all periodic orbits of a tree map f will be called theset of periods of f and will be denoted by Per(f).
IfT is a tree andx∈T, we define thevalence ofx to be the number of connected components ofT\ {x}. The valence ofxwill be denoted by valT(x) or simply by val(x). Each point of valence 1 will be called anendpointofTand the set of such points will be denoted by En(T). A point of valence different from 2 will be called avertexofT, and the set of vertices ofTwill be denoted byV(T). The closure of each connected component of T\V(T) will be called anedgeofT. The number of endpoints and the number of edges ofTwill be denoted, respectively, by en(T) and ed(T).
Any tree which is a union ofr intervals, withr >1, whose intersection is a unique pointxof valencer, will be called anr-star, andxwill be called itscentral point.
Given a treeTandP⊂T, we will define theconvex hullofP, denoted byPTor simply byP, as the smallest closed connected subset ofT containingP. WhenP= {x,y}, we will writex,y or [x,y] to denoteP. The notations (x,y), (x,y], and [x,y) will be understood in the natural way.
The notion oftopological entropy, introduced in [1], is defined for continuous maps on compact metric spaces and is a quantitative measure of the dynamical complexity of the map. It is an important topological invariant. The topological entropy of a mapf will be denoted byh(f).
3. Proof ofTheorem 1.2
This section is devoted to proveTheorem 1.2and, thus, to calculate the constantenfor anyn∈N. We recall that this constant has been defined inSection 1asen=mini∈Nthere is a tree map f :T−→Twith en(T)=i,h(f)=0,n∈Per(f).
In order to calculateen, we will use the characterization of the zero-entropy tree maps given in [4]. Next we introduce the necessary notions. Let f :T→T be a tree map and let P be a periodic orbit of f. The map fP:P → P defined by fP=r◦f, where r:T→ Pis the natural retraction, will be called thenatural restriction of f toP. Let y be a fixed point of fP. We will denote by ZP,y the connected component ofP \P
containingy, and byZ1P,y,Z
P,y
2 ,...,Z
P,y
l the connected components ofP \ZP,y.
LetPbe a periodic orbit of a tree map f :T→T. We say thatPhas a divisionif for a fixed pointyof fP, there exist{M1,M2,...,Mm}withm≥2, a partition ofP \ZP,y
for 1≤i < m, and f(Mm∩P)=M1∩P. The setsMiwill be called thebranches ofP.
Ob-serve that each branch contains|P|/mpoints ofP.
Remark 3.1. In the notation used in the definition of division, it is not difficult to check that
(i)Mi ∩Mj= ∅wheneveri=j. Hence,Mi∩P= Mi ∩P; (ii) En(Mi)∩En(P)= ∅;
(iii) there is at most one endpoint of Mi which is not an endpoint of P. Thus, m
i=1en(Mi)−m≤en(P)≤
m
i=1en(Mi);
(iv) if|P∩Mi| =1, thenMireduces to the only point ofP∩Mi. Thus, if each branch contains one point ofP, thenmi=1en(Mi)=en(P)= |P|.
The following result, which is [4, Corollary C], characterizes the zero-entropy tree maps in terms of their orbits.
Theorem3.2. Let f :T→Tbe a tree map. The following statements are equivalent. (1)h(f)=0.
(2)For everyn∈N, each periodic orbit of fnof a period larger than 1 has a division. (3) Per(f)⊂ {k·2l:k≤en(T)!odd, andl∈N∪ {0}}.
Theorems3.2and1.1 will be the main tools to calculate the numberenin general. However, it is easy to give simple expressions ofenin some particular cases.
Lemma3.3. Letn∈N. (1)Ifn=1, thenen=1.
(2)Ifn=2kfor somek≥1, thenen=2. (3)Ifnis prime, thenen=n.
Proof. Statement (1) is obvious. Statement (2) follows from the fact that, given anyk∈ N, one can define a zero-entropy interval map exhibiting a periodic orbit of period 2k. Indeed, by ˇSarkovs’ki˘ı’s theorem [13], there exists an interval map f such that Per(f)= {1, 2, 22,..., 2k}. ByTheorem 1.1, the topological entropy of f is zero.
Now let us prove (3). LetT be a tree with en(T)=en and let f :T→T be a zero-entropy map having a periodic orbitP of periodn. Without loss of generality, we can assume that En(T)⊂P. Indeed, otherwise the natural restriction of f onPwould be a zero-entropy map havingP as a periodic orbit and en(P)≤en(T). Therefore, from now on, we assume thatP =T. ByTheorem 3.2(2),Phas a division. Letm≥2 be the number of branches ofP and letkbe the number of points ofPin each branch. Since n=k·mandnis prime, it follows thatm=nandk=1. ByRemark 3.1(iv), each Mi reduces to the only point ofP∩Mianden=en(P)=mi=1en(Mi)=m=n.
Let us introduce a few more notions. LetΣbe the set of finite sequences (s1,s2,...,sk)
such thatsi∈Nandsi≥2 for 1≤i≤k. Let f :T→Tbe a zero-entropy tree map and let Pbe a periodic orbit of f. ByTheorem 3.2(2),Phas a division. Next we will associate an element ofΣto the orbitP. This element will be denoted bysP. Lett1≥2 be the number
of branches ofP and letM be a branch ofP. Observe thatP∩Mis a periodic orbit of ft1 of period|P|/t
1. If|P|/t1>1, thenP∩M has a division by Theorem 3.2(2). Thus,
branches ofP∩Mbyt2≥2. By iterating as far as possible, after a finite number of steps,
we get a sequence (tk,tk−1,...,t2,t1)∈Σ, which we denote bysP. We note that, since the
number of branches of a periodic orbit with a division is not uniquely determined, there is not a unique way of performing each step in the construction ofsP. Thus, sP is not uniquely determined. From now on, when we writesP=rfor a certainr∈Σ, we mean that there is a way of doing the above construction which yieldsr.
For eachs∈Σ, let us define inductively a tree which will be called ans-star. Fors= (s1), ans-staris simply ans1-star. Lets=(s1,s2,...,sk) withk >1. Consider ansk-starS
andsk-many copiesT1,T2,...,Tsk of an (s
1,s2,...,sk−1)-star. That is, for 1≤i≤k, there
exists a homeomorphismhi:Ti→T1. Letx
1,x2,...,xsk be the endpoints ofS. Choose an
endpointwofT1. Ans-staris the tree obtained when one glues togetherS,T1,T2,...,Tsk
by identifyingxiwithhi(w), for each 1≤i≤sk.
Observe that, since there are several choices for the endpointwin the construction above, for a givens∈Σ, there can be several nonhomeomorphics-stars.
Remark 3.4. Lets=(s1,s2,...,sk)∈Σ. Ifk=1, then the number of endpoints of ans
-star iss1. Whenk≥2, the number of endpoints of anys-star issk(p−1), wherepis the
number of endpoints of any (s1,s2,...,sk−1)-star. An inductive argument easily yields that
the number of endpoints of anys-star is
s1s2···sk−
k
i=2
sisi+1···sk. (3.1)
Lets∈Σ. A treeT will be calleds-admissibleif there exists a zero-entropy tree map f :T→Texhibiting a periodic orbitPsuch thatsP=s.
Lemma3.5. For eachs∈Σ, ans-star is ans-admissible tree.
Proof. We will prove the following claim: there exists a zero-entropys-star map exhibiting a periodic orbitP such thatsP=sand the endpoints of thes-star belong to P. Write s=(s1,s2,...,sk). Let us prove the claim by induction onk. Fork=1, the claim follows
by considering a rigid rotation of ans1-starTand the orbit composed by the endpoints
ofT.
Now letk≥2 and assume that the claim holds foriinstead ofk, for each 1≤i < k. LetRbe ans-star. By definition, there exists a subtreeS⊂Rsuch thatSis ansk-star and R\Sconsists ofsk-many disjoint connected components whose closuresT1,T2,...,Tsk are pairwise homeomorphic (s1,s2,...,sk−1)-stars. Letxibe the only endpoint ofSwhich
belongs toTi. For 1≤i < sk, letφi:Ti→Ti+1be a homeomorphism such thatφi(xi)=
xi+1and letφsk:Tsk→T1be a homeomorphism such thatφsk(xsk)=x1. In addition, we
choose the homeomorphisms in such a way thatφsk◦φsk−1◦ ··· ◦φ1is the identity map onT1. By the induction hypothesis, there exists a zero-entropy tree mapg:T1→T1with
a periodic orbitQsuch thatQ⊃En(T1) andsQ=(s
1,s2,...,sk−1). Now consider the map
f :R→Rdefined as follows:
(iii) denote by ythe central point ofS. Set f(y)=y. For each 1≤i≤sk, define f to map [y,xi] monotonically onto [y,f(xi)].
It is not difficult to check that f is well defined and continuous, fsk|T1=g, and for each n-periodic orbitX ofg,ski=0fi(X) is anskn-periodic orbit of f. Conversely, the period
of each periodic orbitX of f (excluding the fixed point y) has the formsknfor some n∈N, and the restriction ofX toT1is ann-periodic orbit ofg. Summarizing, there is
a bijection between the set ofn-periodic orbits ofgand the set ofskn-periodic orbits of f. It easily follows thath(f)=h(g)/skand thereforeh(f)=0. SetP=ski=0fi(Q). Then
En(R)⊂Pand it is not difficult to see thatsP=(s1,s2,...,sk), sincesQ=(s1,s2,...,sk−1).
This completes the induction step and proves the lemma.
Lemma3.6. For eachs∈Σ, ans-star has the minimum number of endpoints among the set ofs-admissible trees.
Proof. Sets=(s1,s2,...,sk) and letT be ans-admissible tree. That is, there exists a
pe-riodic orbitP of a zero-entropy map f :T→T such thatsP=s. It is enough to prove that en(T) is not smaller than the number of endpoints of ans-star. To prove this claim, we proceed by induction onk. Fork=1, we haves=(s1) and we must see thatT has
at leasts1endpoints. SincesP=(s1),Phass1branches and each branch contains exactly
one point ofP. ByRemark 3.1(iv), each branchMireduces to the only point ofP∩Mi and|P| =s1=en(P). Since en(P)≤en(T), the claim holds fork=1.
Now letk≥2 and assume that the claim holds foriinstead ofkfor each 1≤i < k. Since sP=(s1,s2,...,sk),P hassk branches. Choose a branchMofPwith the least number of
endpoints. ByRemark 3.1(iii), we have that
en P≥sken M−sk=sk en M−1. (3.2)
By definition of division,M∩Pis a periodic orbit of fsk of period|P|/sk. Moreover, sM∩P=(s1,s2,...,sk−1) sincesP=s. Let g be the natural restriction of fsk toM. By
standard results (see, e.g., [3, Lemma 4.3.1(a)]),h(g)≤h(fsk)=skh(f)=0. Summariz-ing, g:M → M is a zero-entropy tree map with a periodic orbitM∩P such that sM∩P=(s1,s2,...,sk−1). Letpbe the number of endpoints of any (s1,s2,...,sk−1)-star. By
the induction hypothesis, en(M) is not smaller than p. Since en(T)≥en(P), from (3.2) it follows that en(T)≥sk(p−1), which, byRemark 3.4, equals the number of
end-points of ans-star.
Lemma3.7. Lets=(s1,s2,...,sk)∈Σwithk≥2and assume thatsp> sp+1for somep∈
{1, 2,...,k−1}. Lets=(s1,s2,...,sp−1,sp+1,sp,sp+2,sp+3,...,sk). Then the number of
end-points of ans-star is smaller than the number of endpoints of ans-star.
Proof. Observe that, for eachN∈N, ((N−1)sp+1−1)sp<((N−1)sp−1)sp+1. The
state-ment follows easily from this fact,Remark 3.4, and a simple inductive argument. Lemma3.8. Lets=(s1,s2,...,sk)∈Σand assume thatsp=m·n, withn,m≥2, for some
p∈ {1, 2,...,k}. Let s=(s1,s2,...,sp−1,m,n,sp+1,sp+2,...,sk). Then the number of
Proof. Observe that, for eachN∈N, ((N−1)m−1)n <(N−1)sp. The statement follows easily from this fact,Remark 3.4, and a simple inductive argument.
Finally, we are ready to proveTheorem 1.2.
Proof ofTheorem 1.2. Let us call anyr=(r1,r2,...,rl)∈Σwithn=r1r2···rla
decompo-sition ofn. In particular,s=(s1,s2,...,sk) is a decomposition ofn. ByLemma 3.6, for each
decompositionrofn, anr-star has the minimum number of endpoints among the set of r-admissible trees. Therefore,
en=minen(T) :Tis anr-star for a decompositionrofn. (3.3)
Observe that there are only finitely many decompositions ofn. Using iteratively Lemmas 3.7and3.8finitely many times, one gets that the minimum in the right-hand side of (3.3) is reached whenr=s. Thus,enis the number of endpoints of ans-star and the theorem
follows from (3.1).
Acknowledgment
The authors have been partially supported by DGES Grants BFM2003-09504-C02-01 and BFM2002-01344.
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Esther Barrab´es: Departament d’Inform`atica i Matem`atica Aplicada, Universitat de Girona, Llu´ıs Santal ´o s/n, 17071 Girona, Spain
E-mail address:[email protected]
David Juher: Departament d’Inform`atica i Matem`atica Aplicada, Universitat de Girona, Llu´ıs San-tal ´o s/n, 17071 Girona, Spain