This paper introduces kneading sequences for a two-dimensional toy model of Hénon maps, demonstrating they serve as complete invariants for conjugacy classes and establishing a version of Singer's theorem.
Contribution
It extends the concept of kneading sequences to a 2D setting and proves their completeness as invariants for the toy model family of Hénon maps.
Findings
01
Kneading sequences are complete invariants for conjugacy classes.
02
A version of Singer's theorem is established for the toy model.
03
The study links combinatorial equivalence with topological conjugacy in 2D dynamics.
Abstract
The purpose of this article is to study the relation between combinatorial equivalence and topological conjugacy, specifically how a certain type of combinatorial equivalence implies topological conjugacy. We introduce the concept of kneading sequences for a setting that is more general than one-dimensional dynamics: for the two-dimensional toy model family of H\'enon maps introduced by Benedicks and Carleson, we define kneading sequences for their critical lines, and prove that these sequences are a complete invariant for a natural conjugacy class among the toy model family. We also establish a version of Singer's Theorem for the toy model family.
t_{F}(x,y)=\left\{\begin{array}[]{ll}L&\textrm{, if }\pi_{1}(x,y)<0\\
0^{-}&\textrm{, if }\pi_{1}(x,y)=0^{-}\\
0^{+}&\textrm{, if }\pi_{1}(x,y)=0^{+}\\
R&\textrm{, if }\pi_{1}(x,y)>0.\end{array}\right.
t_{F}(x,y)=\left\{\begin{array}[]{ll}L&\textrm{, if }\pi_{1}(x,y)<0\\
0^{-}&\textrm{, if }\pi_{1}(x,y)=0^{-}\\
0^{+}&\textrm{, if }\pi_{1}(x,y)=0^{+}\\
R&\textrm{, if }\pi_{1}(x,y)>0.\end{array}\right.
q(x)=-x^{2},\ \ f(x)=\left\{\begin{array}[]{rl}x&\textrm{, if }x\leq 0\\
-x&\textrm{, if }x\geq 0\end{array}\right.,\ \ g(x)=\left\{\begin{array}[]{rl}-\sqrt{-x}&\textrm{, if }x\leq 0\\
-\sqrt{x}&\textrm{, if }x\geq 0.\end{array}\right.
q(x)=-x^{2},\ \ f(x)=\left\{\begin{array}[]{rl}x&\textrm{, if }x\leq 0\\
-x&\textrm{, if }x\geq 0\end{array}\right.,\ \ g(x)=\left\{\begin{array}[]{rl}-\sqrt{-x}&\textrm{, if }x\leq 0\\
-\sqrt{x}&\textrm{, if }x\geq 0.\end{array}\right.
KQ(Lc)=KF(Lc)=KG(Lc).
KQ(Lc)=KF(Lc)=KG(Lc).
Peer Reviews
No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.
Videos
No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.
Full text
\psset
algebraic
Kneading sequences for toy
models of Hénon maps
Ermerson Araujo
Ermerson Araujo, Departamento de Matemática, Centro de Ciências, Campus do Pici,
Universidade Federal do Ceará (UFC), Fortaleza – CE, CEP 60440-900, Brasil
The purpose of this article is to study the relation between combinatorial equivalence and
topological conjugacy, specifically how a certain type of combinatorial equivalence implies
topological conjugacy. We introduce the concept of kneading
sequences for a setting that is more general than one-dimensional dynamics:
for the two-dimensional toy model family of Hénon maps introduced by Benedicks and Carleson,
we define kneading sequences for their critical lines, and
prove that these sequences are a complete invariant
for a natural conjugacy class among the toy model family.
We also establish a version of Singer’s Theorem for the toy model family.
One of the main questions in dynamical systems is to identify when two
systems are ‘the same’, where the term ‘the same’ means
some type of equivalence between the systems.
One of the simplest notions of equivalence is topological conjugacy.
Let X and Y be topological spaces, and let f:X→X and g:Y→Y be
continuous maps. We say that f,g are topologically
conjugate (or simply conjugate) if there is a
homeomorphism h:X→Y satisfying the conjugacy equation h∘f=g∘h.
Topological conjugacy heavily depends on class of maps. For orientation preserving
circle homeomorphisms, Poincaré introduced in the late XIX century the notion of rotation numbers,
which essentially characterises all conjugacy classes: if f:S1→S1
is an orientation preserving
homeomorphism with irrational rotation number ρ, then f is topologically semiconjugate to
the rigid rotation Rρ(x)=x+ρ.
Furthermore, if f contains a dense orbit then f is indeed topologically conjugate to Rρ.
For interval homeomorphisms, topological conjugacy is
trivial, due to the following fact: every orbit is either periodic
or converges to a periodic orbit. Now, if we consider interval
endomorphisms, then the situation is much richer.
For this class, Milnor and Thurston developed a theory that substitutes the theory of rotation
numbers of Poincaré [MT]. Nowadays known as kneading theory, it
plays an important role in one-dimensional dynamics. It allows, for example,
to prove that topological entropy varies continuously in the the quadratic family, see [dMvS] for details.
The setting considered by Milnor and Thurston is the following: let I be an interval,
and let f:I→I be a map with finitely many turning points (i.e. a point x∈I where
f changes monotonicity). For each turning point, they defined a sequence, called kneading sequence,
that encodes the trajectory of this point. Then they proved that if two interval maps do not have wandering intervals,
periodic attractors neither intervals of periodic points, then they are topologically conjugate
if and only if their kneading sequences are the same.
When we pass to two-dimensional maps, the situation becomes much more complicated.
One of the main maps considered is Ha,b:R2→R2 given by
Ha,b(x,y)=(a−x2−by,x), a,b∈R, known as Hénon map.
Over the last years, there has been various attempts to construct a similar theory in this context,
but none of them is yet completely satisfactory. The difficulty comes from two facts: the plane does not possess
a natural order as the interval, and there is no sufficiently nice dynamical notion of critical point.
In this work we introduce kneading sequences for a particular two-dimensional family.
This family was considered by Benedicks and Carleson as a toy model for the study of Hénon maps [BC].
Each of these maps acts on a two-dimensional rectangle via the expression F(x,y)=(f(x,y),K(x,y)),
where f is a family of unimodal maps and K is an inverse branch of a Cantor map. We call
each map of this form a toy model.
Matheus, Moreira and Pujals proved that, among the toy models, Smale’s Axiom A property is
C1-dense and, on the C2-topology, there exists an open subset satisfying a
Newhouse-like phenomenon [MMP]. These properties indicate that the toy models exhibit
rich dynamical properties. It is for them that we characterize topological conjugacy in terms
of kneading sequences.
Theorem A**.**
Let F,G be two toy models, and assume that
neither of them has wandering intervals, interval of periodic points nor
weakly attracting periodic points. Then F,G are fiber topologically conjugate
if and only if they have the same kneading sequences.
The hardest part in Theorem A is the reverse implication, that
same kneading sequences lead to fiber topological conjugacy. To construct the conjugacy,
we employ ideas similar to those used in
one-dimensional dynamics: we define it in preimages of turning points and
then extend to the closure. This approach also leads to the same result for more general toy models,
see Theorem 2.8.
Theorem A has an underlying motivation that we now explain.
For dissipative surface diffeomorphisms, Crovisier and Pujals introduced
a new class of diffeomorphisms, called strongly dissipative, whose
dynamical properties behave as an intermediate level between one-dimensional dynamics
and two-dimensional dynamics [PC]. They proved that this class is open and non-empty (it contains
Hénon maps with Jacobian in (−1/4,1/4)), and that every strongly dissipative diffeomorphism of the disc
(in particular Hénon maps) is equivalent to a one-dimensional structure (it is semiconjugate to an
endomorphism defined on a tree). If two strongly dissipative diffeomorphisms on the disc are topologically
conjugate, then their one-dimensional structures are also conjugate. It is unknown
if this one-dimensional structure carries all relevant dynamical information, i.e., if the
one-dimensional dynamics are conjugate, are the strongly dissipative diffeomorphisms also conjugate?
Theorem A shows that, on the
presence of a good definition of turning points, we can recover information about the conjugacy class
of the toy model by means of an appropriate combinatorial structure. We hope that the methods
introduced in the proof of Theorem A will be used to study the conjugacy
classes of strongly dissipative diffeomorphisms of the disc.
It is worth mentioning that kneading sequences are not the only method to characterize
topological conjugacy. Based on numerical experiments [cvi], Cvitanović introduced
in [civ2] the concept of pruning fronts and conjectured that every map
Ha,b in the Hénon family can be understood as a pruned horseshoe:
if F:R2→R2 is Smale’s horseshoe map,
then after pruning (destroying) some orbits of F we construct
a new map F that is equivalent, in some sense, to Ha,b.
This is known as the Pruning Front Conjecture.
Mendonza proved that the Pruning Front Conjecture holds in an open set of
the parameter space [men1], and Ishii proved it for the Lozi family [ishii1].
For more detail on the theory of pruning fronts and its relationship with kneading theory,
see [carvalho2, carvalho3].
On a different direction of kneading sequences, we also establish for the toy models
a version of the classical Singer theorem. Let us explain this. Singer proved that,
for C3 interval maps with negative Schwarzian derivative, the basin of any
attracting periodic point contains either a critical point or a boundary point of the interval.
Using that the toy models do preserve horizontal lines,
we prove the following theorem, which is a version of Singer’s theorem for the toy models.
Theorem B**.**
Let F(x,y)=(f(x,y),K(x,y)) be a toy model.
If each interval map f(⋅,y) has negative Schwarzian
derivative, then the closure of the immediate
basin of any strongly attracting periodic orbit contains either a
point of the critical line or a boundary point of the rectangle.
This paper is organized as follows. Section 2 introduces in full details the
toy models and consider their kneading sequences. It also states the main results, establishes
some basic properties of the toy models, and discusses examples.
Section 3 contains the proofs of the main results stated in Section 2.
Acknowledgements. This article is part of my Ph.D thesis at IMPA
under the supervision of Enrique Pujals. I deeply thank him for his constant
support, patience, encouragement and availability. I also would like to thank
Yuri Lima for the carefully reading and several helpful comments. This
work was supported by CNPq and CAPES.
2. Toy models and kneading sequences
Results in one-dimensional dynamics heavily rely on the total order of the line.
For example, given two multimodal interval maps, the preimages
of the turning points divide the interval from left to right, and if these preimages
are combinatorially equivalent then there is a monotone bijection
between them. The family of two-dimensional maps we consider also inherits this property:
as we will see below, the one-dimensional order structure
is preserved along “unstable leaves”.
2.1. The Toy Model
Given a continuous interval map f:[a,b]→[a,b], we call it
a unimodal map if f(a)=f(b)=a and if there exists a
unique point c=a,b, called turning point, for which f is
increasing on [a,c] and decreasing on [c,b]. We consider two-dimensional maps
with fiber dynamics given by unimodal maps. We follow [MMP] for the description.
One-Parameter Family of Unimodal Maps: Consider a one-parameter family
[TABLE]
such that:
∘
y↦f(y) is continuous,
2. ∘
each f(y) is a unimodal map with f(y)(−1)=f(y)(1)=−1 and [math] as a turning point.
Fix a<b in (0,1).
Cantor Map: A Cantor map is a differentiable map k:[0,a]∪[b,1]→[0,1] such that
k(0)=k(1)=0, k(a)=k(b)=1 and ∣k′∣>γ>1.
Each Cantor map defines a Cantor setKF given by
[TABLE]
Let K+=(k↾[0,a])−1 and K−=(k↾[b,1])−1 be the inverse branches of k,
and define K(x,y) by
[TABLE]
Toy Model: A toy model is a map F of the form
[TABLE]
The map F is not defined on the vertical line {0}×[−1,1], and this
may introduce serious difficulties to analyse F and apply compactness arguments.
To bypass this, we duplicate each (0,y) to two points (0−,y) and (0+,y)
and define F on them by F(0±,y)=(f(y)(0),K±(y)).
Critical Line of F: It is the set
[TABLE]
equal to the disjoint union of two vertical segments {0−}×[0,1] and {0+}×[0,1].
Each element of Lc(F) is called a turning point of F.
Therefore F is now defined on (([−1,1]\{0})×[0,1])∪Lc(F),
call this set Dom(F). We define a natural topology on Dom(F) as follows:
∘
If (x,y)∈Dom(F)\Lc(F), then
the neighborhoods of (x,y) are the standard Euclidean neighborhoods.
2. ∘
If x=0− then the neighborhoods
of (x,y) are U∩([−1,0−]×[0,1]), where U is a standard
Euclidean neighborhood of (0,y).
3. ∘
If x=0+ then the neighborhoods
of (x,y) are U∩([0+,1]×[0,1]), where U is a standard
Euclidean neighborhood of (0,y).
We remark that, although F is now defined on a compact set, it is not continuous.
2.2. Notations and preliminaries
For y∈[0,1], write f−(y) for the restriction f(y)↾[−1,0−] and
f+(y) for the restriction f(y)↾[0+,1], see Figure 2.
Also, let F− and F+ be the restrictions of F defined by
[TABLE]
and
[TABLE]
Every orbit of F is therefore obtained by compositions of F− and F+:
for each (x,y)∈Dom(F) there is a sequence
j(x,y)=(j1j2⋯jm⋯)∈{−,+}N
such that
[TABLE]
where
[TABLE]
Let I(y):=([−1,0−]∪[0+,1])×{y}, and
call J⊂I(y) an interval if there are
x1,x2∈[−1,0−] or x1,x2∈[0+,1] such that
[TABLE]
We will sometimes denote J as above by [(x1,x2),y]. If
[TABLE]
then we write J=[[x1,x2],y].
Given a periodic point (p,q) of F, set
[TABLE]
We call (p,q) a weakly attracting periodic point if
B(p,q) contains an interval. If B(p,q)
contains an open subset, then call (p,q) a
strongly attracting periodic point.
When OF(p,q)∩Lc(F)=∅, then
the immediate basin B0(p,q) of O(p,q) is the union of the
connected components of B(p,q) that contain of the set
from {(p,q),F(p,q),…,Fn−1(p,q)},
where n is the period of (p,q). Hence B0(p,q)∩Lc(F)=∅,
and so there is a sequence j1⋯jn∈{−,+}n
such that for each (x,y)∈B(p,q) we have
[TABLE]
where B(p,q)⊂B0(p,q) is the
component containing (p,q).
Clearly, if (p,q) is a strongly attracting
periodic point, then it is also a weakly attracting periodic point.
Given y∈[0,1], an interval J⊂I(y) is called an
interval of periodic points if every (x,y)∈J is a
periodic point of F.
Wandering Interval: An interval J⊂Dom(F) is called
wandering if
∘
Fn(J)∩Fm(J)=∅, ∀n=m;
2. ∘
Fn(J)∩Lc(F)=∅, ∀n≥0.
A point (x,y) is called non-wandering if
every neighborhood V⊂Dom(F) containing (x,y)
has an iterate ℓ≥1 such that Fℓ(V)∩V=∅.
The non-wandering setΩ(F) is the set of all non-wandering points.
It is well known that if f:I→I is an unimodal map without
intervals of periodic points nor attracting periodic points,
then the set of pre-images of the turning point is dense in Ω(F).
We establish the same for toy models. Define
[TABLE]
Proposition 2.1**.**
Let F be a toy model, and assume that F has no weakly attracting periodic point nor intervals of
periodic points. Then Ω(F)⊂C(F).
Proof.
The proof is by contradiction.
Suppose that (x,y)∈Ω(F)\C(F).
Let V be a neighborhood of (x,y) such that
Fn(V)∩Lc(F)=∅ for all n≥0.
Without loss of generality, we can assume that V=I×J,
where x∈I, y∈J, and I,J are open intervals.
Fix m≥1 with Fm(V)∩V=∅, and put
[TABLE]
Since, for each r=0,…,m−1, the union
⋃j≥0Fjm+r(V) is connected, the set
L has finitely many connected components.
Set L=L1⊔⋯⊔Ls, where L1,…,Ls are connected.
Using that F(L)⊂L, for each i
there exists j such that F(Li)⊂Lj.
Let L be the connected component containing (x,y).
We have Fm(L)∩L=∅, hence Fm(L)⊂L.
Take ℓ=min{d≥1;Fd(L)⊂L}.
Hence, we have that L is a domain such that Fℓ(L)⊂L,
Fi(L)∩Lc(F)=∅ for i=0,1,…,ℓ−1, and
Fi(L)∩Fj(L)=∅
for i,j=0,1,…ℓ−1 with i=j.
Therefore there exists a sequence j0j1⋯jℓ−1∈{−,+}ℓ
such that for any (a,b)∈L and n≥0
it holds Fnℓ(a,b)=(anℓ,b(jℓ−1⋯j1j0)n), where
b(jℓ−1⋯j1j0)n=(Kjℓ−1∘⋯∘Kj0)n(b)
and
[TABLE]
Let π2 be the projection on the second coordinate, π2(x,y)=y.
Since K:=Kjℓ−1∘⋯∘Kj0
is a contraction, there exists w∈π2(L) such that K(w)=w.
We claim that w=y. To see that,
for each n≥1 take Vn:=In×JnI×J
neighborhood of (x,y) such that
In+1×Jn+1In×Jn with
∣In∣→0 and ∣Jn∣→0 as n→∞.
As before, for each n≥1 there are
(xn,yn)∈Vn and kn≥1 such that
Fkn(xn,yn)∈Vn and
Fkn(xn,yn)→(x,y) as n→∞.
We may suppose that kn→∞ as n→∞.
Furthermore, by the construction of L,
we have kn=rnℓ for some rn≥1.
We already know that K is a contraction,
hence there is λ<1 such that
[TABLE]
Note that
[TABLE]
therefore
[TABLE]
Using that Fkn(xn,yn)→(x,y) as n→∞, we conclude that w=y, as claimed.
Now, for every (v,y)∈L∩I(y) and n≥1 we get
[TABLE]
where fjℓ−1(yjℓ−2⋯j0)∘⋯∘fj0(y) is a strictly monotone map.
Without loss of generality, we assume that
fjℓ−1(yjℓ−2⋯j0)∘⋯∘fj0(y) is strictly increasing.
Let {Ii}i be the connected components of L∩I(y).
Claim: There exists i0 such that Fℓ(Ii0)⊂Ii0.
Proof of the claim. Assume, by contradiction, that it does not hold.
Since the composition fjℓ−1(yjℓ−2⋯j0)∘⋯∘fj0(y)
is strictly increasing, for every i there exists σ(i) such that
Fℓ(Ii)⊂Iσ(i) and supIi<infIσ(i).
Fixed any k, this implies that the sequence {Fℓn(Ik)}n≥0
is formed by disjoint intervals satisfying supFℓn(Ik)<infFℓ(n+1)(Ik) and ∣Fℓn(Ik)∣→0 as n goes to infinity.
This implies that there is (v0,y) such that Fℓ(v0,y)=(v0,y) and
Fℓn(Ik)→(v0,y), contradicting the non-existence of
weakly attracting periodic point. The claim is proved.
Now we complete the proof of the proposition. Let i0 be given by the claim. The
map Fℓ:Ii0→Ii0 is strictly increasing, hence either Ii0 contains
an interval of periodic points for Fℓ , or some open subinterval
of Ii0 converges to a single periodic point. Both possibilities contradict the
underlying assumptions on F.
∎
2.3. Kneading sequences and main results
Let π1 be the projection on first coordinate, π1(x,y)=x. Let F be a toy model.
Consider the alphabet A={L,0−,0+,R}. The letter L means “left”
and R means “right”.
Address of a point: The address of a point (x,y)∈Dom(F) is
the letter tF(x,y)∈A defined by:
[TABLE]
Itinerary of a point: The itinerary of a point (x,y)∈Dom(F)
is the sequence TF(x,y)∈A{0,1,2,…} defined by
[TABLE]
Kneading Sequences: The kneading sequences of F are all the itineraries
of elements of Lc(F). We denote this set of itineraries by KF(Lc).
In the following proposition, we construct conjugacies between the Cantor sets appearing in
the second coordinate of two toy models. The proof is standard, but we include it for completeness.
Proposition 2.2**.**
Let F(x,y)=(f(y)(x),Ksign(x)F(y)) and
G(x,y)=(g(y)(x),Ksign(x)G(y))
be two toy models. There exists a strictly increasing continuous
map ψ:[0,1]→[0,1] such that
[TABLE]
Moreover, ψ↾KF is independent of the choice of ψ.
Proof.
Let kF:[0,aF]∪[bF,1]→[0,1] and kG:[0,aG]∪[bG,1]→[0,1]
the Cantor maps of F and G.
If KF and KG are the Cantor sets generated by kF and kG respectively, then
there is a unique strictly increasing bijection
ψ:KF→KG such that ψ∘kF=kG∘ψ. To see this,
observe that if we order {0,1}N lexicographically, then KF and KG
are isomorphic to {0,1}N through an order preserving bijection.
Hence, for each x∈KF we take ψ(x) to be
the unique point of KG with the same symbolic representation
as x in KF.
Now we extend ψ continuously to gaps of KF. Take any
strictly increasing homeomorphism h:(aF,bF)→(aG,bG).
Denote by k+F,G the restriction kF,G↾[0,aF,G] and
by k−F,G the restriction kF,G↾[bF,G,1].
Given a gap J⊂[0,1] of KF, there exists a unique n≥1 and
a sequence j1⋯jn∈{−,+}n such that
(kjnF∘⋯∘kj1F)(J)=(aF,bF).
Thus, for each x∈J we define ψ(x) by
[TABLE]
Clearly ψ:[0,1]→[0,1] is a strictly increasing continuous
map with ψ∘K±F=K±G∘ψ.
By construction, ψ↾KF does not depend on the choice of ψ.
∎
Definition 2.3**.**
Consider two toy models F(x,y)=(f(y)(x),Ksign(x)F(y)) and
G(x,y)=(g(y)(x),Ksign(x)G(y)), and let ψ:[0,1]→[0,1] be the
map given by Proposition 2.2.
We say that F* and G have the same kneading sequences*, and
write KF(Lc)=KG(Lc),
if
[TABLE]
Observe that the above definition does not actually depend on the choice of ψ.
Indeed, the invariant set defined by F is
([−1,0−]∪[0+,1])×KF, as we know that in KF
the map ψ is uniquely defined. We are now ready to state
the main tool to be used in the proof of Theorem A.
Theorem 2.4**.**
Let F,G be two toy models, and assume that KF(Lc)=KG(Lc).
Then there exists a bijective map
H:C(F)→C(G) such that:
(i)
H∘F=G∘H* on C(F)\Lc(F).*
2. (ii)
H(C(F)∩I(y))⊂C(G)∩I(ψ(y))* for all y∈[0,1].*
3. (iii)
H↾C(F)∩I(y)* is strictly increasing for all y∈[0,1].*
The proof of Theorem 2.4 will be given in the next section.
When the map H given by Theorem 2.4 exists, we say that
F and G are combinatorially equivalent.
Definition 2.5**.**
Consider two toy models F(x,y)=(f(y)(x),Ksign(x)F(y)) and
G(x,y)=(g(y)(x),Ksign(x)G(y)), and let ψ:[0,1]→[0,1] be the
map given by Proposition 2.2.
We say that F and G are fiber topologically semiconjugate
if there is a continuous surjective map H:Dom(F)→Dom(G)
which is increasing on intervals, such that
H∘F=G∘H and H(I(y))⊂I(ψ(y)) for all y∈[0,1].
If H is a homeomorphism, then we say that F and G are fiber topologically conjugate.
If H conjugates F and G as above, then H(0±,y)=(0±,ψ(y)) for all y∈[0,1].
In particular, H(Lc(F))=Lc(G).
The following proposition states that kneading sequences are a topological invariant
for the notion introduced in Definition 2.5.
Proposition 2.6**.**
Let F,G be two toy models. If F,G are fiber
topologically conjugate then KF(Lc)=KG(Lc).
Proof.
Let H:Dom(F)→Dom(G) be conjugacy between F and G.
We will show that TF(0±,y)=TG(0±,ψ(y)) for all y∈[0,1].
Fix y∈[0,1].
Clearly tF(0±,y)=tG(0±,ψ(y)).
Since F,G are fiber topologically conjugate,
[TABLE]
Since H is strictly increasing on intervals,
[TABLE]
Therefore TF(0±,y)=TG(0±,ψ(y)).
∎
Notice that the above proposition implies the first part of Theorem A.
Moreover, within the notion of topological conjugacy given by Definition 2.5,
Theorem A establishes that kneading sequences are a complete topological invariant.
It would be natural to find, inside the class of toy models,
a family of representatives for each combinatorial equivalence class.
To better understand this question, we remind a classical result
in one-dimensional dynamics. Consider the quadratic family{fμ}0≤μ≤4
of maps fμ:[0,1]→[0,1] defined by fμ(x)=μx(1−x).
This family is universal for unimodal maps in the following sense: if g:[0,1]→[0,1]
is a unimodal map then g is combinatorially equivalent to fμ for some μ=μ(g).
Using this analogy, it is natural to find a universal family for toy models.
At the moment, it is not clear how to construct such a family, left alone its existence.
Let F be a toy model.
Wandering Domain: A domain V⊂Dom(F) is called wandering if
∘
Fn(V)∩Fm(V)=∅, ∀n=m;
2. ∘
Fn(V)∩Lc(F)=∅, ∀n≥0.
For y∈[0,1] and n≥1, let
[TABLE]
Consider the map ϕnF:[0,1]→2Dom(F) defined by
ϕnF(y):=CnF(y). Under some regularity of this function,
we can state a weaker version of Theorem A.
Theorem 2.7**.**
Let F,G be two toy models. Assume that
F,G have no wandering domain and that KF(Lc)=KG(Lc).
If the family {ϕnG}n≥1 is equicontinuous
then F,G are fiber topologically semiconjugate.
Above, equicontinuity means the following: given ε>0 there is
δ>0 such that for all y1,y2∈[0,1] with ∣y1−y2∣<δ it holds
dH(CnG(y1),CnG(y2))<ε for alln≥1,
where dH is the Hausdorff distance.
At the moment, we observe that the assumptions that G has no wandering domain and that
{ϕnG} is equicontinuous imply that G has no wandering intervals, therefore
Theorem 2.7 follows from Theorem A. Nevertheless,
we will give an independent proof Theorem 2.7 in the next section.
2.4. Examples
We now give some examples of toy models, and discuss the
hypotheses required in Theorem A and Theorem 2.7.
Example 1**.**
Let f:[−1,1]→[−1,1] be the tent map, i.e.
f(x)=1+2x for x∈[−1,0] and f(x)=1−2x for x∈[0,1]. Let k:[0,1]→[0,1] be the Cantor
map defined by k(y)=3y if y∈[0,1/3] and k(y)=3−3y if y∈[2/3,1].
The map F(x,y):=(f(x),K(x,y)) satisfies the hypotheses of Theorems A
and 2.7.
Example 2**.**
Let f:[−1,1]→[−1,1] be a unimodal map with [math] as turning point and containing
a wandering interval I0. Let k be the Cantor map of example 1.
The toy model F(x,y):=(f(x),K(x,y)) has wandering intervals I0×{y}.
Example 3**.**
Let q,f,g:[−1,1]→[−1,1] be three unimodal maps defined by
[TABLE]
Let k be the Cantor map defined in example 1, and define the toy models
Q(x,y)=(q(x),K(x,y)), F(x,y)=(f(x),K(x,y)), and G(x,y)=(g(x),K(x,y)). Then
[TABLE]
By Theorem 2.4, there exists HQ,F:C(Q)→C(F),
HF,G:C(F)→C(G) and HG,Q:C(G)→C(Q).
On the other hand, they are not conjugate
since the turning point of q,f,g is respectively an attracting,
neutral and repelling fixed point.
2.5. Generalized toy models
In this section we introduce a larger class of toy models, changing unimodal to multimodal maps.
We say that an interval map f:[0,1]→[0,1] is an ℓ-modal map if it has
exactly ℓ turning points c1<⋯<cℓ. This means that
the intervals [0,c1],[c1,c2],…,[cℓ,1] are the largest intervals in which f is monotone.
Fixed c1<⋯<cℓ, consider a one-parameter family
[TABLE]
of ℓ-modal maps with turning points c1<⋯<cℓ such that y∈[0,1]↦f(y) is
continuous. We write c0=0,cℓ+1=1, and
I1=[c0,c1),I2=(c1,c2),…,Iℓ+1=(cℓ,cℓ+1] for the intervals of monotonicity.
For each y∈[0,1], let fi(y) be f(y)↾Ii, see Figure 3.
Let 0=a1<b1<a2<b2<⋯<aℓ+1<bℓ+1=1, and let
k:[a1,b1]∪⋯∪[aℓ+1,bℓ+1]→[0,1]
be a differentiable map such that
k({ai,bi})⊂{0,1} and ∣k′∣>γ>1.
We define K(x,y) by
[TABLE]
where Ki=(k↾[ai,bi])−1, i=1,…,ℓ+1, are the inverse branches of k.
Generalized Toy Model: It is the map defined by
[TABLE]
We can similarly introduce extra points to make the domain of F become compact.
For each i=1,…,ℓ, introduce points cii and cii+1 and extend F via the formula
F(cii,y)=(fi(y)(ci),Ki(y)) and F(cii+1,y)=(fi+1(y)(ci),Ki+1(y)), see Figure 4.
For i=1,…,ℓ+1, let Ii be an interval on the construction of the family y↦f(y).
We make the following convention.
If f(y) is strictly increasing on Ii, we put
[TABLE]
If f(y) is strictly decreasing on Ii, we put
[TABLE]
For each (x,y)∈Dom(F), there exists
j(x,y)=(j1j2⋯jm⋯)∈{1,…,ℓ+1}N
such that
[TABLE]
where ym=(Kjm∘⋯∘Kj1)(y)=:yjm⋯j1 and
xm=fjm(yjm−1⋯j1)∘⋯∘fj2(yj1)∘fj1(y)(x).
Consider the alphabet A={I1,c11,c12,I2,…,cℓℓ,cℓℓ+1,Iℓ+1}.
Address of a point: The address of a point (x,y)∈Dom(F) is
the letter tF(x,y)∈A defined by
[TABLE]
Itinerary of a point: The itinerary of a point (x,y)∈Dom(F)
is the sequence TF(x,y)∈A{0,1,2,…} defined by
[TABLE]
Critical Line: The critical line of F is the set
[TABLE]
Kneading Sequences: The kneading sequences of F are all the itineraries
of elements of Lc(F). We denote this set of itineraries by KF(Lc).
With these more general definitions, a result similar to Theorem A holds.
Theorem 2.8**.**
Let F,G be two generalized toy models, and assume that
neither of them has wandering intervals, interval of periodic points nor
weakly attracting periodic points. Then F,G are fiber topologically conjugate
if and only if they have the same kneading sequences.
The proof of Theorem 2.8 follows exactly the same ideas and arguments
as in the proof of Theorem A, but it demands a much heavier notation. Therefore we decided
not to include its proof and leave the details to the interested reader.
3. Proofs of the main results
We now present the proofs of our main results. Given a set X, let ∂X denote its boundary.
A similar proof of Theorem 2.4 can be found in [Rand, Thm1].
Remind that
[TABLE]
and that for y∈[0,1] and n≥1 we have
[TABLE]
Setting
[TABLE]
we have
[TABLE]
We can similarly define the sets C(G),CnG(y),CG(y).
Suppose that, for all y∈[0,1] and n≥1, there exists a strictly
increasing bijection Hn(y):CnF(y)→CnG(ψ(y)) such that:
∘
Hn(y)∘F=G∘Hn(y) on CnF(y)\{(0±,y)},
2. ∘
Hn(y)↾Cn−1F(y)=Hn−1(y).
Since ψ is a bijection, we have
[TABLE]
Therefore, we can define H:C(F)⟶C(G) by
H(x,y)=Hn(y)(x,y) where n is some (any) integer for which (x,y)∈CnF(y).
It follows directly that H∘F=G∘H on C(F)\Lc(F).
Hence the proof of Theorem 2.4 is reduced to
constructing the maps Hn(y). This is what we will now do.
Let PnF(y) be the partition of I(y) induced by CnF(y).
More specifically,
[TABLE]
where knF(y):=#PnF(y) and the increasing order of the index of
IiF(y) is the same as the intervals are placed in I(y). Note that the partition PnF(y)
includes two intervals, one with left endpoint 0+ and one with
right endpoint 0−.
Given a sequence
j=j1j2⋯jm∈{−,+}m, we will use the following notation
[TABLE]
where yjℓ⋯j1=(Kjℓ∘⋯∘Kj1)(y)
for 1≤ℓ≤m.
Given (x,y)∈CnF(y)\{(0±,y)}, we will identify the pair (x,y)
with x(y)j(x,y)F, where j(x,y)=j1j2⋯jk∈{−,+}k is the minimal sequence such that
[TABLE]
for some 1≤k≤n−1.
We will also abuse notation
by considering x(y)j(x,y)F either as an ordered pair in order to apply F or as a real number
in order to apply unimodal maps.
For each IiF(y)∈PnF(y) there exists a unique minimal sequence
j(IiF(y))=j1ij2i⋯jni∈{−,+}n
such that the function
[TABLE]
is strictly monotone. Hence
[TABLE]
for some 1≤k≤n−1. For each n≥1, set
[TABLE]
We similarly define PnG(ψ(y)) and AnG(ψ(y)).
Now, we will use induction on n to complete the proof of the theorem.
Our underlying assumptions imply that we can define
H1(y):C1F(y)→C1G(ψ(y)) with
(0±,y)↦(0±,ψ(y)).
By induction, suppose that there exists a strictly increasing map
Hn(y):CnF(y)→CnG(ψ(y)) such that
Hn(y)∘F=G∘Hn(y) on CnF(y)\{(0±,y)} for every y∈[0,1].
In particular, #PnF(y)=#PnG(ψ(y)) and
AnF(y)=AnG(ψ(y)) for all y∈[0,1].
Let IiF(y)∈PnF(y). There is a unique sequence
j=j1j2⋯jn∈{−,+}n with
[TABLE]
for some 1≤k≤n−1. We analyse the possibilities for the interval IiF(y).
The first possibility is
[TABLE]
for 1≤k=ℓ≤n−1. By the induction hypothesis, we have
[TABLE]
Moreover, Hn(y)(x(y)j1⋯jkF)=x(ψ(y))j1⋯jkG
and Hn(y)(x(y)j1⋯jℓF)=x(ψ(y))j1⋯jℓG.
Hence Fn[IiF(y)] is equal to the union
[TABLE]
where Λ(IiF(y))={Fn−k(0jk,yjk⋯j1),Fn−ℓ(0jℓ,yjℓ⋯j1)}, and Gn[IiG(ψ(y))] is equal to the union
[TABLE]
where Λ(IiG(ψ(y)))={Gn−k(0jk,ψ(y)jk⋯j1),Gn−ℓ(0jℓ,ψ(y)jℓ⋯j1)}.
Since KF(Lc)=KG(Lc), we have
[TABLE]
If Fn[IiF(y)]∩Lc(F)=∅,
then IiF(y)∈Pn+1F(y).
On the other hand, if
Λ(IiF(y))∩Lc(F)=∅
we get IiF(y)∈Pn+1F(y) since
F is injective. In any case, we also have
IiG(ψ(y))∈Pn+1G(ψ(y)).
Now, if (Fn[IiF(y)]\Λ(IiF(y)))∩Lc(F)=∅ then there is a unique point
[TABLE]
and so (Gn[IiG(ψ(y))]\Λ(IiG(ψ(y))))∩Lc(G)=∅, which in turn implies the existence of a unique point
[TABLE]
Therefore we can define
[TABLE]
The other possibilities IiF=[[−1,x(y)j1j2⋯jkF],y],
IiF=[[x(y)j1j2⋯jkF,1],y],
and IiF=[[x(y)j1j2⋯jkF,0±],y], 1≤k≤n−1, are treated similarly.
Finally, defining Hn+1(x,y)=Hn(x,y) for (x,y)∈CnF(y), we have just constructed
a strictly increasing map
[TABLE]
such that:
∘
Hn+1(y)∘F=G∘Hn+1(y) on Cn+1F(y)\{(0±,y)}, and
2. ∘
Let y↦f(y) be the one-parameter family of unimodal maps, as in the definition of toy models.
For every y∈[0,1] and ε>0
there is δ>0 satisfying the following: if y′∈[0,1], x,x′∈[−1,1]
are such that d(y,y′)<δ, d(x,x′)<δ and there exist
x(y)=fj−1(y)(x) and x′(y′)=fj−1(y′)(x′), where j=−or+,
then d(x(y),x′(y′))<ε.
Proof.
The proof is by contradiction.
Suppose there is ε>0 such that for all n≥1 there are
yn∈[0,1] and xny,xnyn∈[−1,1] with d(y,yn)<n1
and d(xny,xnyn)<n1 such that xn(y)=fj−1(y)(xny) and
xn(yn)=fj−1(yn)(xnyn) exist and d(xn(y),xn(yn))≥ε.
By compactness, we may suppose that limxny=limxnyn=x0,
limxn(y)=x1 and limxn(yn)=x2.
Note that x1=x2 and that either x1,x2∈[−1,0] or x1,x2∈[0,1].
In particular f(y)(x1)=f(y)(x2). But, using that f(yn)
converges uniformly to f(y), we have
f(y)(x1)=limf(y)(xn(y))=limxny=limxnyn=limf(yn)(xn(yn))=f(y)(x2),
a contradiction.
∎
Let ξ−:([−1,0−]∪(0+,1])×[0,1]→[−1,0−] and
ξ+:([−1,0−)∪[0+,1])×[0,1]→[0+,1]
be two maps defined by
Let (x,y)∈Dom(ξj) and ε>0. Assume that
x∈[−1,fj(y)(0j)). Since the family y↦f(y)
depends continuously on y, there exists δ>0 such that
for every (x′,y′)∈Dom(ξj) satisfying d(x,x′)<δ and d(y,y′)<δ,
we have x′∈[−1,fj(y′)(0j)).
From Lemma 3.1, if δ>0 is small enough then
[TABLE]
Suppose now that x>fj(y)(0j). Again by continuity, there exists
δ>0 such that for all (x′,y′)∈Dom(ξj) satisfying d(x,x′)<δ
and d(y,y′)<δ we have x′>fj(y′)(0j). Therefore,
ξj(x,y)=0j=ξj(x′,y′) and so
[TABLE]
The case x=fj(y)(0j) is treated analogously.
∎
The following lemma is fundamental to prove the
continuity of the conjugacy in Theorem A.
Lemma 3.3**.**
For every (x,y)∈C(F) there exists a continuous curve
γF:[0,1]→C(F) of the form γF(w)=(γF(w),w)
such that γF(y)=(x,y).
Proof.
Given (x,y)∈C(F) there exists n≥1 and a
sequence (j1⋯jn)∈{−,+}n
such that (x,y)=(fj1−1(y)∘⋯∘fjn−1(yjn−1⋯j1)(0jn),y).
Consider a sequence of curves defined inductively as follows:
∘
γj1F:[0,1]→Dom(F) is defined by
γj1F(w)=(ξj1(0j1,w),w).
2. ∘
γj1j2F:[0,1]→Dom(F) is defined by
γj1j2F(w)=(ξj1(π1(γj2F(wj1)),w),w).
3. ∘
⋯
4. ∘
γj1⋯jnF:[0,1]→Dom(F) is defined by
γj1⋯jnF(w)=(γj1⋯jnF(w),w),
where γj1⋯jnF(w):=ξj1(π1(γj2⋯jnF(wj1)),w).
From Corollary 3.2, each γj1⋯jnF:[0,1]→Dom(F) is a continuous curve.
By construction, γj1⋯jnF(y)=(x,y).
Now it is quite easy to see that γj1⋯jnF([0,1])⊂C(F).
For each w∈[0,1], put
[TABLE]
Hence, if ℓj1⋯jn(w) exists then
[TABLE]
On the other hand, if ℓj1⋯jn(w) does not exist, then
[TABLE]
In any case we have γj1⋯jnF([0,1])⊂C(F).
∎
Here is a direct consequence of Lemma 3.3:
if (x,y),(z,w)∈C(F) are such that ∃n≥1 and
(j1⋯jn)∈{−,+}n
with (x,y)=(fj1−1(y)∘⋯∘fjn−1(yjn−1⋯j1)(0jn),y) and
(z,w)=(fj1−1(w)∘⋯∘fjn−1(wjn−1⋯j1)(0jn),w),
then there is a continuous curve
γF:[0,1]→C(F) such that γF(y)=(x,y) and γF(w)=(z,w).
Because of Proposition 2.6, we only need to prove the reverse implication.
We will employ similar notation to the one used in the proof of Theorem 2.4.
For y∈[0,1], let H(y):CF(y)→CG(ψ(y)) be the map defined by
H(y)(x,y):=Hn(y)(x,y),
where n is some (any) integer such that (x,y)∈CnF(y). We will extend
H(y) to I(y). Take (z,y)∈CF(y)\CF(y)
and suppose that there exist (zn,y)j∈CF(y), j=1,2, such that (zn,y)1↑(z,y)
and (zn,y)2↓(z,y). The cases where we have only
(zn,y)↑(z,y) or (zn,y)↓(z,y) are similar. Since G has no wandering
intervals, no intervals of periodic points and no weakly attracting
periodic point, CG(ψ(y)) is dense on I(ψ(y)).
Thus the limits limH(y)((zn,y)1)=:H1(y)(z,y) and
limH(y)((zn,y)2)=:H2(y)(z,y) exist and H1(y)(z,y)=H2(y)(z,y).
Therefore, for (z,y)∈CF(y)\CF(y) we can define
[TABLE]
where (zn,y)∈CF(y) is some (any) sequence such that (zn,y)→(z,y).
Since CF(y) is dense on I(y), we obtain the map H:Dom(F)→Dom(G)
defined by H(x,y):=H(y)(x,y). Note that H is strictly increasing on each fiber I(y).
Claim: The map H is continuous.
Proof of the claim..
Let (x,y)∈Dom(F)\{(±1,y),(0±,y):y∈[0,1]} (the case
(x,y)∈{(±1,y),(0±,y):y∈[0,1]} can be treated similarly).
Let Bε(H(x,y)) be a neighborhood of H(x,y) for some ε>0.
Since CG(ψ(y))=I(ψ(y)),
there are z1<x<z2 such that (z1,y),(z2,y)∈C(F) and
H(zi,y)∈Bε(H(x,y))∩I(ψ(y)), i=1,2. Note that
π1[H(z1,y)]<π1[H(x,y)]<π1[H(z2,y)].
By Lemma 3.3, there are
continuous curves γj1⋯jℓG and γe1⋯ekG
such that γj1⋯jℓG(ψ(y))=H(z1,y) and
γe1⋯ekG(ψ(y))=H(z2,y). By continuity, there are
y1<y<y2 so that the domain D(H(x,y)) whose boundary is formed by the two curves
γj1⋯jℓG([ψ(y1),ψ(y2)]),
γe1⋯ekG([ψ(y1),ψ(y2)]) and the two intervals
[(γj1⋯jℓG(ψ(y1)),γe1⋯ekG(ψ(y1))),ψ(y1)],
[(γj1⋯jℓG(ψ(y2)),γe1⋯ekG(ψ(y2))),ψ(y2)]
contains H(x,y) and satisfies D(H(x,y))⊂Bε(H(x,y)), see Figure 5.
Again by Lemma 3.3, for every sequence
t1⋯tm∈{−,+}m and w∈[0,1] we get
H(γt1⋯tmF(w))=γt1⋯tmG(ψ(w)).
Let D(x,y) be the domain whose boundary is formed by the curves
γj1⋯jℓF([y1,y2]),
γe1⋯ekF([y1,y2]) and the intervals
[(γj1⋯jℓF(y1),γe1⋯ekF(y1)),y1],
[(γj1⋯jℓF(y2),γe1⋯ekF(y2)),y2].
Such domain contains a neighborhood of (x,y).
Since H is monotone on fibers, we have H(D(x,y))=D(H(x,y)).
This proves the claim.
∎
At the moment, we know that H is injective (by construction) and surjective (by the claim).
It remains to show that H∘F=G∘H. From Theorem 2.4,
we already have H∘F=G∘H on C(F)\Lc(F).
Now, take (z,y)∈Dom(F)\C(F)
and let (zn,y)∈CF(y)\{0±,y} such that lim(zn,y)=(z,y)
(the case (z,y)∈Lc(F) is treated analogously).
Using that F is continuous on Dom(F)\C(F) and that
H(z,y)∈CG(ψ(y))\CG(ψ(y)), we conclude that
[TABLE]
This finishes the proof of the theorem.
∎
We finish this section remarking that the same proof above can be used
to prove a variant of Theorem A, with assumptions only on G:
Let F,G be two toy models, and assume that G has no wandering intervals,
no interval of periodic points nor weakly attracting periodic points.
If F,G have the same kneading sequences, then they are fiber topologically semiconjugate.
We begin with a lemma whose proof is similar to that of Proposition 2.1.
Lemma 3.4**.**
Let F be a toy model. If F has no wandering domain, then
[TABLE]
Proof.
By contradiction, assume that there exists an open set V⊂Dom(F) such that
Fm(V)∩Lc(F)=∅ for all m≥0.
Construct, as in the proof of Proposition 2.1,
a connected set L containing V such that Fℓ(L)⊂L for some 0<ℓ≤m,
Fi(L)∩Lc(F)=∅ and Fi(L)∩Fj(L)=∅
for all i,j=0,…,ℓ−1 with i=j.
Therefore, there exists a sequence j0j1⋯jℓ−1∈{−,+}ℓ
such that, for any (a,b)∈L and n≥0
it holds Fnℓ(a,b)=(anℓ,b(jℓ−1⋯j1j0)n), where
b(jℓ−1⋯j1j0)n:=(Kjℓ−1∘⋯∘Kj0)n(b)
and anℓ=fjℓ−1(bjℓ−2⋯j0(jℓ−1⋯j0)n−1)∘⋯∘fj0(b(jℓ−1⋯j0)n−1)∘⋯∘fjℓ−1(bjℓ−2⋯j0)∘⋯∘fj1(bj0)∘fj0(b)(a). Since Kjℓ−1∘⋯∘Kj0
is a contraction, there is w∈π2(L) such that
(Kjℓ−1∘⋯∘Kj0)(w)=w. For each (x,y)∈V, we can
prove that y=w. Therefore V⊂I(w),
which contradicts the fact that V is an open set.
∎
For n≥1, let
[TABLE]
For each y∈[0,1], let Hn(y):CnF(y)→CnG(ψ(y))
be the map constructed in the proof of Theorem 2.4.
Let Hn(y):I(y)→I(ψ(y)) be the homeomorphism such that:
∘
Hn(y)(±1,y)=(±1,ψ(y)),
2. ∘
Hn(y)↾CnF(y)=Hn(y), and
3. ∘
Hn(y) is linear in each connected component of I(y)\CnF(y).
Now define Hn:Dom(F)→Dom(G) by
Hn(x,y):=Hn(y)(x,y).
Lemma 3.5**.**
For each n≥1, Hn is continuous.
Proof.
We start invoking the curves constructed in the proof of Lemma 3.3:
for each j1j2⋯jℓ∈{−,+}ℓ with 0≤ℓ≤n−1, let
γj1j2⋯jℓF:[0,1]→CnF
be the continuous curve appearing in the construction of CnF.
Note that CnF is the union of finitely many such curves:
[TABLE]
Take (x,y)∈Dom(F)\CnF, and take
δ>0 such that Bδ(x,y)∩CnF=∅.
Let x1<x<x2 such that (x1,y),(x2,y)∈CnF and
CnF(y)∩[(x1,x2),y]=∅ (if (x1,y) or (x2,y) does not
exist, we take x1=−1 or x2=1 accordingly).
Take curves γj1j2⋯jℓF and
γe1e2⋯ekF such that
γj1j2⋯jℓF(y)=(x1,y) and
γe1e2⋯ekF(y)=(x2,y), for some 1≤ℓ,k≤n−1.
For any curve γt1t2⋯tiF with 1≤i≤n−1, we have
γt1t2⋯tiF(y)∩[(x1,x2),y]=∅.
See Figure 6.
Choose y1<y<y2 and two continuous curves
α1,α2:[y1,y2]→CnF such that:
∘
αi(y)=(xi,y), i=1,2.
2. ∘
If D(x,y) is the domain whose boundary is formed by the two curves α1([y1,y2]),
α2([y1,y2]) and the two intervals
[(α1(y1),α2(y1)),y1], [(α1(y2),α2(y2)),y2],
then D(x,y)∩CnF=α1([y1,y2])∪α2([y1,y2]).
Then Hn(D(x,y))=:D(Hn(x,y)) is a domain such that
By the definition of Hn, for (z,w)∈D(x,y) we have
Hn(z,w)=(ξx,y(z,w),ψ(w)), where
[TABLE]
In particular, Hn is continuous at (x,y).
Indeed, Hn is continuous on the union D(x,y)∪∂D(x,y).
Suppose now that (x,y)∈CnF. Let s∈{−,+}ℓ
such that (x,y)=(γsF(y),y) for some 0≤ℓ≤n−1,
and consider (xj,yj)→(x,y). For each j≥1, there are curves
γajF,γbjF such that
(xj,yj) belongs to the interval [[γajF(yj),γbjF(yj)],yj]
and [(γajF(yj),γbjF(yj)),yj]∩CnF(y)=∅.
To prove that
Hn(xj,yj)→Hn(x,y),
it is enough to prove that
any subsequence (xjk,yjk)
contains a sub-subsequence (xjkℓ,yjkℓ) such that
Hn(xjkℓ,yjkℓ)→Hn(x,y).
Let (xjk,yjk) be a subsequence of (xj,yj).
There is a sub-subsequence with
(xjkℓ,yjkℓ)∈[[γaF(yjkℓ),γbF(yjkℓ)],yjkℓ] for fixed a,b. Since
γaF(yjkℓ)≤xjkℓ≤γbF(yjkℓ) for every ℓ,
when passing to the limit ℓ→∞ we obtain γaF(y)≤γsF(y)≤γbF(y).
If γaF(y)=γbF(y)
then Hn(xjkℓ,yjkℓ)→Hn(x,y), since
[TABLE]
and
[TABLE]
Assume now that γaF(y)=γbF(y). Then either
γsF(y)=γaF(y) or γsF(y)=γbF(y).
In either case we can build a domain D(x,y) such that (xjkℓ,yjkℓ)∈D(x,y)
for all ℓ≥1. Hence Hn(xjkℓ,yjkℓ)→Hn(x,y).
It follows that Hn is continuous.
∎
Our next goal is to prove that {Hn}n≥1 is a Cauchy sequence.
Lemma 3.6**.**
The sequence {Hn}n≥1 is Cauchy.
Proof.
Let ε>0. By the equicontinuity of the family {ϕnG}, there is
δ>0 such that if ∣y1−y2∣<δ then
dH(CnG(y1),CnG(y2))<ε for all n≥1.
By Lemma 3.4,
C(G) is dense in Dom(G), so there is N>0 large enough
such that {Bδ(x,y);(x,y)∈CNG} covers Dom(G).
Let (x,y)∈Dom(F) and m,n≥N.
Since Hn is surjective, we can take x1<x<x2 such that
∣Hn(x1,y)−Hn(x,y)∣=∣Hn(x2,y)−Hn(x,y)∣=2ε. Take
(x1′,y′),(x2′,y′′)∈CNF such that
Hn(x1,y)∈Bδ(HN(x1′,y′))
and Hn(x2,y)∈Bδ(HN(x2′,y′′)).
Again by equicontinuity, there are
(x1,y),(x2,y)∈CNF such that
Hn(x,y) belongs to the interval [[HN(x1,y),HN(x2,y)],ψ(y)]
and ∣HN(x1,y)−HN(x2,y)∣<6ε. Since
Hℓ↾CNF=HN↾CNF∀ℓ≥N,
we obtain that Hn(x,y),Hm(x,y) both belong to the interval
[[HN(x1,y),HN(x2,y)],ψ(y)]. Hence
[TABLE]
and so
[TABLE]
This proves that {Hn}n≥1 is a Cauchy sequence.
∎
Now we finish the proof of Theorem 2.7. By the previous lemma,
H:Dom(F)→Dom(G) defined by H:=limHn is
continuous and surjective, so it remains to prove that H semiconjugates F and G.
Since H agrees with Hn on
CnF, we have H∘F=G∘H on C(F)\Lc(F).
Take (x,y)∈Dom(F)\C(F) and (xn,yn)∈C(F)
such that lim(xn,yn)=(x,y).
Since F is continuous on Dom(F)\C(F) and
H(x,y)∈Dom(G)\C(G),
[TABLE]
Take now (x,y)∈Lc(F), say (x,y)=(0−,y) (the case (x,y)=(0+,y) is similar).
If F(0−,y)∈/Lc(F), then for any sequence
(xn,yn)∈C(F)\Lc(F) with lim(xn,yn)=(0−,y) we have limF(xn,yn)=F(0−,y). Hence
(H∘F)(0−,y)=(G∘H)(0−,y). Now assume that F(0−,y)∈Lc(F), then
[TABLE]
where in the third passage we used that the kneading sequences
KF(Lc) and KG(Lc) are the same.
Therefore H∘F=G∘H on Dom(F),
which concludes the proof.
Let f:I→I be a C3 interval map and let x∈I
such that Df(x)=0.
Schwarzian Derivative: The Schwarzian derivativeSf(x) of f at x is
[TABLE]
We say that f has negative Schwarzian derivative if Sf(x)<0
for all x∈I except, possibly, the turning points.
At these points we define Sf(x)=−∞.
It is easily seen from the definition that if f,g:I→I are C3 and
x∈I satisfies Df(x)=0 and Dg(f(x))=0 then
[TABLE]
This implies the following: if f,g have negative Schwarzian derivative,
then g∘f also has negative Schwarzian derivative.
Lemma 3.7** (Minimum Principle).**
Let I=[a,b] and f:I→I a C3 map with negative Schwarzian derivative.
If Df(x)=0 for all x∈I, then
[TABLE]
Proof.
See [dMvS, Lemma 6.1].
∎
We call c∈I a critical point of f if Df(c)=0. The following theorem, due to Singer,
is a consequence of the Minimum Principle.
Theorem** (Singer).**
If f:I→I is a C3 map with negative Schwarzian
derivative then the immediate basin of any
attracting periodic orbit contains either a critical
point of f or a boundary point of I.
Proof.
See [dMvS, Thm 6.1].
∎
Let us restate Theorem B,
which is a version of Singer’s theorem for the toy models.
Theorem B**.**
Let F(x,y)=(f(y)(x),Ksign(x)F(y)) be a toy model.
If each interval map f(y) has negative Schwarzian
derivative, then the closure of the immediate
basin of any strongly attracting periodic orbit contains either a
point of the critical line or a point of
Λ:={−1,1}×[0,1].
Before proving this theorem, let us introduce some notation. Write the derivative
of F(x,y)=(f(y)(x),Ksign(x)(y)) by
[TABLE]
For (x,y)∈Dom(F), let j(x,y)=(j1⋯jm⋯)∈{−,+}N
such that
Let (p,q)∈Dom(F) be a strongly attracting periodic point of period m,
and let B be the closure of its immediate basin.
By contradiction, we assume that B does not contain points of
Λ neither points of the critical line.
Let B0 be the interior of the connected component of B that contains (p,q). Then
Fm(B0)⊂B0. Let [(a,b),q] be the connected
component of B0∩I(q) containing (p,q), see Figure 8.
Since Fm(p,q)=(p,q) and F preserves fibers,
Fm([(a,b),q])⊂([(a,b),q]).
Let j=(j1⋯jm)∈{−,+}m be the sequence that gives the iterations of F up to order m inside
[(a,b),q], i.e. Fm(x,q)=(g(x),q) for x∈(a,b), where
g=fjm(qjm−1⋯j1)∘⋯∘fj2(qj1)∘fj1(q).
By assumption, we also have Am(x,q)=0.
Claim: g({a,b})⊂{a,b}.
Proof of the claim..
Suppose that g(a)∈(a,b) (the case g(b)∈(a,b) is treated analogously).
Since Fm is continuous at (a,q), there is a neighborhood V∋(a,q)
such that Fm(V)⊂B0. This contradicts the assumption that B0 is the
connected component of the immediate basin containing (p,q). The claim is proved.
∎
We can assume that Am(x,q)>0 for all x∈(a,b)
and that g(α)=α for α∈{a,b}
(if this is not the case, take A2m(x,q) instead of Am(x,q)).
Since (a,q),(b,q) are not attracting periodic points, we have
Am(α,q)≥1 for α∈{a,b}. Since g is the composition of
maps with negative Schwarzian derivative, it also has negative Schwarzian
derivative. By the Minimum Principle, Am(x,q)>1
for all (x,q)∈[(a,b),q], which contradicts the equality Fm([(a,b),q])=[(a,b),q].
∎
We finish this paper with a question. One of the assumptions for Theorem A
is the non-existence of wandering intervals. For unimodal maps,
negative Schwarzian derivative guarantees the absence of wandering intervals.
This is a famous theorem of Guckenheimer.
Theorem** (Guckenheimer).**
Let f:I→I be a C3 unimodal map with
negative Schwarzian derivative. If f′′(c)=0 at the unique critical point c of f,
then f has no wandering intervals.
Proof.
See [dMvS, Thm 6.3].
∎
We do not know if such implication also holds for toy models, so we ask the following.
Question: Let F(x,y)=(f(y)(x),Ksign(x)F(y)) be a toy model.
If each unimodal map f(y) has negative Schwarzian
derivative, can F have wandering intervals?