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Research Article

Renormalization of translated cone exchange transformations

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Received 09 Feb 2024, Accepted 03 Jul 2024, Published online: 16 Jul 2024

Abstract

In this paper, we investigate a class of non-invertible piecewise isometries on the upper half-plane known as Translated Cone Exchanges. These maps include a simple interval exchange on a boundary we call the baseline. We provide a geometric construction for the first return map to a neighbourhood of the vertex of the middle cone for a large class of parameters, then we show a recurrence in the first return map tied to Diophantine properties of the parameters, and subsequently prove the infinite renormalizability of the first return map for these parameters.

2020 Mathematics Subject Classifications:

1. Introduction

Piecewise isometries (PWIs) are a class of maps that can be generally described as a ‘cutting-and-shuffling’ action of a metric space, specifically a partitioning of the phase space into at most countably many convex pieces called atoms, which are each moved according to an isometry. The phase space of these maps can be partitioned into two (or three) subsets based on the dynamics – a polygon or disc packing of periodic islands known as the regular set, and its complement, the set of points whose orbit either lands on, or accumulates on, the discontinuity set. Some authors choose to further distinguish those points in the pre-images of the discontinuity and those points which accumulate on it. The most well-known and well-understood examples of such maps are the interval exchange transformations (IETs), which arise as return maps to cross-sections of some measured foliations [Citation20] and also as generalizations of circle rotations [Citation7, Citation19, Citation28] and their encoding spaces generalise Sturmian shifts [Citation12]. Furthermore, interval exchanges which aren't irrational rotations are known to be almost always weakly mixing [Citation5] but never strongly mixing [Citation16]. Piecewise isometries in general, however, are not as well-known and as a subset of this class, interval exchanges are in many ways exceptional, due in part to being one-dimensional, as well as the invariance of Lebesgue measure.

In the more general setting, although the inherent lack of hyperbolicity restricts the variety of possible behaviours, for example it is known that all piecewise isometries have zero topological entropy [Citation8], piecewise isometries are still capable of quite complex behaviour; many examples show the presence of unbounded periodicity and an underlying renormalizability which structures the dynamics near the discontinuities [Citation1–3, Citation13, Citation14, Citation17, Citation24]; numerical evidence suggests the existence of invariant curves in the exceptional set which seem fractal-like and form barriers to ergodicity [Citation2–4, Citation18]; there are conjectured conditions for piecewise isometries to have sensitive dependence on initial conditions [Citation15].

Renormalization in theoretical physics and nonlinear dynamical systems has a longstanding history, see for example [Citation9–11, Citation21, Citation25–27], driven by the problem of understanding phenomena that occur simultaneously at many spatial and temporal scales, particularly near phase transitions, periodic points, or in the case of piecewise isometries, the set of discontinuities.

In this paper, we investigate the renormalizability of a class of piecewise isometries called Translated Cone Exchanges on the closure of the upper half-plane H¯. This family of maps was introduced in [Citation4] and has since been investigated in [Citation22, Citation23]. In particular, we use a geometric construction to describe the action of a first return map to a subset containing the origin, and show that this map displays renormalizable behaviour locally to the origin in accordance with Diophantine approximation of one of its parameters. These results go beyond [Citation22, Citation23] in that they are much less constrained in the continued fraction expansion associated with the baseline translation.

This paper is organized as follows. In Section 2, we introduce the family of maps we will investigate, namely, Translated Cone Exchange transformations. In Section 3 we will develop some tools that will be useful in the next section. Section 4 presents the preliminary results that lead to the main result of this paper, Theorem 4.7, which gives an explicit form of renormalization for the first return maps of maps in our class to a neighbourhood of 0. Finally, in Section 5 we present an example for fixed values of the parameters.

2. Translated cone exchange transformations

Let HC denote the upper half plane, and let H¯ be its closure in C, that is H¯={zC:Im(z)0}.A Translated Cone Exchange transformation ( TCE ) [Citation4] is a PWI (C,Fκ) defined on the closed upper half plane H¯. For any integer d>0, let Bd+2 be the set Bd+2={α=(α0,,αd+1)(0,π)d+2:j=0d+1αj=π},Next, for some α=(α0,,αd+1)Bd+2, partition the interval [0,π] by subintervals Wj={[0,α0),ifj=0,[α0,α0+α1],ifj=1,(k=0j1αk,k=0jαk],ifj{2,,d+1}.We then define the partition C as C={Cj:j{0,,d+1}},where C1={0}{zH:ArgzW1}and Cj={zH¯:ArgzWj}forj1.

The mapping Fκ is defined as a composition (1) Fκ(z)=GE(z),(1) where E is a permutation of the cones C1,,Cd, G is a piecewise horizontal translation, and κ is a tuple of the parameters. Formally, let τ be a permutation of {1,,d}, that is a bijection τ:{1,,d}{1,,d}, and let θj(α,τ)=τ(k)<τ(j)αkk<jαk.When α and τ are unambiguous, we may refer to θj(α,τ) simply as θj. The map E:H¯H¯ is then defined as (Figures  and ) E(z)={zifzC0Cd+1zeiθjifzCj,j{1,,d}.

Figure 1. An example of a partition of the closed upper half plane H¯ into 6 cones.

Figure 1. An example of a partition of the closed upper half plane H¯ into 6 cones.

Figure 2. The same partition from Figure  after applying the cone exchange map E.

Figure 2. The same partition from Figure 1 after applying the cone exchange map E.

Figure 3. The example partition from Figure after the TCE F=GE is performed. Note the overlapping of the cones in a region containing 0.

Figure 3. The example partition from Figure 1 after the TCE F=G∘E is performed. Note the overlapping of the cones in a region containing 0.

Note that E is invertible Lebesgue-almost everywhere in H¯. We define the middle cone Cc of Fκ as Cc=j=1dCj=H(C0Cd+1).The map G:H¯H¯ is defined as (2) G(z)={z+λifzCd+1,zηifzCc,zρifzC0,(2) where ρ,λ(0,) are rationally independent, i.e. λ/ρRQ, and λ<η<ρ. Finally, we collect the parameters into the tuple κ=(α,τ,λ,η,ρ). See Figures  and  for plots of the orbit structure for Fκ for example choices of parameters κ.

Figure 4. A plot of the first 3000 elements of the forward orbits of 1000 points (omitting the first 1500 to remove transients) chosen uniformly in the box [ρ,λ]×[0,1] under a TCE with parameters α=(0.5,π/7,π/4,17π/280.5), τ:12,21, λ=2/2, η=1λ and ρ=1. Each orbit is given a (non-unique) colour to illustrate the trajectories.

Figure 4. A plot of the first 3000 elements of the forward orbits of 1000 points (omitting the first 1500 to remove transients) chosen uniformly in the box [−ρ,λ]×[0,1] under a TCE with parameters α=(0.5,π/7,π/4,17π/28−0.5), τ:1↦2,2↦1, λ=2/2, η=1−λ and ρ=1. Each orbit is given a (non-unique) colour to illustrate the trajectories.

Figure 5. A similar plot to Figure , this time of the first 3000 elements (excluding the first 1500 for transients) of the forward orbits of 1500 points chosen uniformly in the box [ρ,λ]×[0,1] under a TCE with parameters α=(π/2+0.1,π/8,0.2,π/50.1,7π/400.2), τ:13,22,31, λ=π/4, and η=1λ. Note that the phase space is very sparse compared to Figures and .

Figure 5. A similar plot to Figure 4, this time of the first 3000 elements (excluding the first 1500 for transients) of the forward orbits of 1500 points chosen uniformly in the box [−ρ,λ]×[0,1] under a TCE with parameters α=(π/2+0.1,π/8,0.2,π/5−0.1,7π/40−0.2), τ:1↦3,2↦2,3↦1, λ=π/4, and η=1−λ. Note that the phase space is very sparse compared to Figures 4 and 9.

For a>0, let Fκ denote the TCE with parameters κ=(α,τ,λ/a,η/a,ρ/a). Define sa:H¯H¯ as uniform scaling by a about the origin (3) sa(z)=az.(3)

Proposition 2.1

We have the following conjugacy: (4) Fκ(z)=sa1Fκsa(z).(4)

Proof.

Firstly, observe that for all j{0,,d+1}, aCj=Cj,from which we can deduce that (5) azCjif and only ifzCj.(5) Let zH¯. From (Equation3), we get sa1Fκsa(z)=1aFκ(az),and by expanding Fκ as in (Equation1), we have sa1Fκsa(z)={1a(az+λ)ifazCd+11a(eiθj(az)η)ifazCj,j{1,,d}1a(azρ)ifazC0.Recalling (Equation5), distributing the multiplication by 1/a, we finally see that sa1Fκsa(z)={z+λ/aifzCd+1eiθjzη/aifzCj,j{1,,d}zρ/aifzC0=Fκ(z).

Clearly from this proposition, we can normalize ρ=1 without any loss of generality. Indeed, normalizing in this way proves very helpful for establishing recurrence results, as we shall see.

Simulations of the orbits of points under some TCEs appears to reveal complex behaviour even at small scales close to the real line, such as in Figures and . One way to investigate this behaviour is by applying the tools of renormalization. In particular, let h:CcN{0} denote the first return time of zCc to Cc under Fκ, that is (6) h(z)=inf{n>0:Fκn(z)Cc}.(6) The first return map Rκ:CcCc of Fκ to Cc is then defined as (7) Rκ(z)=Fκh(z)(z).(7) Observe that for all zCc, Rκ(z)=Fκh(z)(z)=Gh(z)E(z), since E is the identity outside of Cc.

We will now state the main theorem of our paper in a simplified form, which we shall restate in more detail later as Theorem 4.7, after establishing some terminology and preliminary results.

Theorem 2.2

Let αBd+2, τ:{1,,d}{1,,d} be a bijection, λ[0,1)Q, λ<η=p<1 for some p,qN{0} and set κ=(α,τ,λ,η,1). Then there exist λ[0,1)Q, ηR of the form λ<η=pqλ<1 for some p,qN{0}, and a convex, positive area set VCc containing 0 such that Rκ(z)=11λ1λRκ((1λ1λ)z),for all zV, where κ=(α,τ,λ,η,1).

3. Tools

In this section we prove some preliminary results that will serve as tools for more detailed investigation of the renormalization of TCEs. Firstly, we note the following smaller observations.

Proposition 3.1

Let Fκ be as in (Equation1) and G as in (Equation2). Then for all xR, Fκ(x)=G(x).

Proof.

This is clear from the observation that E is the identity on R.

Proposition 3.2

The dynamics of Fκ on R and on H are separate in the sense that Fκ(R)R and Fκ(H)H.

Proof.

The statement Fκ(R)R is clear from Proposition 3.1 and the fact that G is a horizontal translation.

We now prove that Fκ(H)H. Suppose that zH such that Fκ(z)R. Then E(z)R, as G is a horizontal translation. If zHCc then RE(z)=zH, which is a contradiction. On the other hand, if zCc, then E(z)Cc But RCc={0}, meaning that E(z)R implies E(z)=0. But this is only the case if z=0R, which is a contradiction.

Let ι:H¯{1,0,1}N denote the itinerary for Fκ, defined by ι(z)=ι0ι1ι2,where ιn=ιn(z)={1,ifFκn(z)Cd+1,0,ifFκn(z)Cc,1,ifFκn(z)C0.This map is similar to, but distinct from the true notion of the encoding map, since here we do not distinguish between the cones C1, C2,…, Cd within the middle cone Cc. The next Lemma provides a crucial tool in the proof of Theorem 4.5, since the dynamics on the interval [ρ,λ) is that of a rotation of the circle (except at the point 0, in which case Fκ(0)=η), which is more easily understood than that of an arbitrary point in Cc.

Lemma 3.3

Let αBd+2, τ:{1,,d}{1,,d} be a bijection, λ,ρR such that λ/ρQ, and λ<η<1. If zCc, then for all 1jh(z), Fκj(z)=E(z)+Fκj(0).

Proof.

Suppose not, for a contradiction. Then there is some n with 0nh(z) such that Fκn(z)E(z)+Fκn(0), and without loss of generality assume that n is the smallest such integer. Clearly n>1, so for all 0jn1, Fκj(z)E(z)=Fκj(0), and therefore ιj(z)=ιj(0) for all 0jn2, but ιn1(z)ιn1(0). Since nh(z), we cannot have Fκj(z)Cc for all 1jn1. Hence for addresses of the (n1)th iterates of z and 0 to disagree, one of two cases must occur:

  1. Fκn1(z)Cd+1 and Fκn1(0)CcC0; or

  2. Fκn1(z)C0 and Fκn1(0)CcCd+1.

Since the orbit of 0 is restricted to R and since E(0)=0, the second parts of each case become Gn1(0)0 and Gn1(0)0, respectively.

Suppose E(z)=z, and let ε1=Im(z)cot(αd+1) and ε2=Im(z)cot(α0). Then zCc if and only if ε1Re(z)ε2. Note that since G is a horizontal translation, Fκn1(z)=Gn1(E(z))=Gn1(z)Cd+1 if and only if Re(Gn1(z))<ε1. Similarly Gn1(z)C0 if and only if Re(Gn1(z))>ε2. The two above cases above can thus be reformulated as:

  1. Re(Gn1(z))<ε1 and Gn1(0)0; or

  2. Re(Gn1(z))>ε2 and Gn1(0)0.

In case 1, we get 0Gn1(0)=Gn1(z)z=Re(Gn1(z))Re(z). Hence Re(Gn1(z))Re(z) and thus ε1Re(z)Re(Gn1(z))<ε1,which is a contradiction. Case 2 leads to a similar contradiction. Therefore there is no such n.

3.1. Continued fractions

Recall from the theory of continued fractions that the nth convergent to a positive, irrational real number λ=[λ0;λ1,λ2,] is a fraction pn/qn=[λ0;λ1,,λn], where pn,qn are coprime integers and qn>0. The numbers pn,qn can be generated by the recursive relations: (8) p0=λ0,q0=1,p1=λ1λ0+1,q1=λ1,pn=λnpn1+pn2,qn=λnqn1+qn2.(8) Furthermore, the convergents to λ satisfy the property that for all positive integers s<qn+1 and all rZ, |qnλpn||r|,with equality only when (r,s)=(pn,qn). Also observe that we can use the recurrence relation in (Equation8) to set (9) p1=1,q1=0.(9) Let g:[0,1][0,1] denote the Gauss map, given by g(x)=1x1x.In particular, if λ=[0;λ1,λ2,λ3,][0,1], then g(λ)=1λλ1=[0;λ2,λ3,].Let λ=[0;λ1,λ2,][0,1)Q. To start, let Nλ={(m,n)N2:0nλm+1},and define the function wλ:NλN by wλ(m,n)={nifm=0,λ1++λm+nifm>0.Note that wλ is surjective and in fact (10) wλ(m+1,0)=wλ(m,λm+1).(10) Furthermore, if we define the subset Nλ<Nλ to be Nλ<={(m,n)N2:0n<λm+1},then wλ|Nλ< is a bijection.

From now on, we denote the jth coefficient of the continued fraction expansion of gm(λ) by gm(λ)j. The next proposition gives us a nice relationship between the set Nλ and the Gauss map g.

Proposition 3.4

Let j,m,nN. Then (m+j,n)Nλ if and only if (j,n)Ngm(λ). Moreover, wλ(m+j,n)=λ1++λm+wgm(λ)(j,n).

Proof.

We have that (m+j,n)Nλ is equivalent to 0nλm+j+1. We also have that λm+j+1=gm(λ)j+1, so 0ngm(λ)j+1. This is equivalent to (j,n)Ngm(λ).

If m = j = 0, then the second part of our lemma is clearly true.

Assume j = 0 and m>0. Then wλ(m+j,n)=wλ(m,n)=λ1++λm+n=λ1++λm+wgm(λ)(0,n),where the final equality is true since (m,n)Nλ is equivalent to (0,n)Ngm(λ).

Finally, suppose m, j>0. Then wλ(m+j,n)=λ1++λm+λm+1++λm+j+n=λ1++λm+gm(λ)1++gm(λ)j+n.Note that since (m+j,n)Nλ is equivalent to (j,n)Ngm(λ), we have wλ(m+j,n)=λ1++λm+wgm(λ)(j,n).

The bijection wλ mainly serves as a way to show that there is a ‘natural’ well-ordering for the set Nλ<, which allows us to meaningfully index sequences by Nλ< and, as we shall see later, define the notion of a maximal element of a finite subset of Nλ.

We define the one-sided convergents (or semiconvergents) to λ as the fractions Pm,n(λ)Qm,n(λ)={[0;λ1,,λm]ifn=0,[0;λ1,,λm,n]ifn>0,Indeed, this formula is compatible with the indexing wλ in that Pm,0(λ)Qm,0(λ)=Pm1,λm(λ)Qm1,λm(λ).A standard result in the theory of continued fractions is that for (m,n)Nλ<, we have (11) Pm,n(λ)Qm,n(λ)={npm(λ)+pm1(λ)nqm(λ)+qm1(λ),ifn0pm(λ)qm(λ)ifn=0.(11) One way to interpret these fractions is as being the best rational approximates of λ from one ‘direction’. In particular, borrowing notation from the beginning of Section 2.2 in [Citation23], we have (Pm,nQm,n)(m,n)Nλ<,meven=(pkqk)kN,and(Pm,nQm,n)(m,n)Nλ<,modd=(pkqk)kN,where (pk/qk)k are the best rational approximates from above in the sense that pk/qk>λ and for all rational numbers r/spk/qk such that r/s>λ and 1s<qk+1, we have (12) 0<|qkλpk|<|r|,(12) and in a similar fashion (pk/qk)k are the best rational approximates from below in the same sense except that pk/qk<λ and (13) 0<|qkλpk|<|r|,(13) holds for r/spk/qk such that r/s<λ and 1s<qk+1.

We define the sequence (Δm,n(λ))(m,n)Nλ by (14) Δm,n(λ)=Qm,n(λ)λPm,n(λ).(14) We see immediately from the above discussion that Pm,n(λ)/Qm,n(λ)<λ if and only if m is odd and n>0 or m is even and n = 0, that is (15) Δm,n(λ)>0if and only if{mis even andn=0,ormis odd andn>0.(15) By expanding the definitions of Pm,n and Qm,n, we see (16) Δm,n(λ)=n(qmλpm)+qm1λpm1=nΔm,0(λ)+Δm1,0(λ),(16) for (m,n)Nλ with m1. Moreover, by expanding the recurrence relation for pm and qm and rearranging terms, we have the additional property Δm,0(λ)=qmλpm=λm(qm1λpm1)+qm2λpm2=λmΔm1,0(λ)+Δm2,0(λ),for mN,m2.

Note that in agreement with the function wλ, we have Δm1,λm(λ)=Δm,0(λ). A result by Bates et al. [Citation6] presents an interesting connection between iterates of the Gauss map and consecutive errors in the approximation of λ by its convergents.

Lemma 3.5

Theorem 10 of [Citation6]

Let λ=[0;λ1,λ2,][0,1)Q. For all mN, (17) gm(λ)=qmλpmpm1qm1λ.(17)

Equation (Equation17) can be equivalently formulated as (18) gm(λ)=Δm,0(λ)Δm1,0(λ).(18)

Lemma 3.6

Let λ=[0;λ1,λ2,][0,1)Q. For all (m,n)Nλ with m1, (19) Δ0,n(gm(λ))=Δm,n(λ)Δm1,0(λ).(19)

Proof.

For n = 0, Δ0,0(gm(λ))=gm(λ), so (Equation19) holds.

Next, observe that Δj,0(gm(λ))=gm(λ)jΔj1,0(gm(λ))+Δj2,0(gm(λ)).Since gm(λ)j=λm+j, this becomes Δj,0(gm(λ))=λm+jΔj1,0(gm(λ))+Δj2,0(gm(λ)).We thus see that Δ0,n(gm(λ))=n(q0gm(λ)p0)+q1gm(λ)p1,and by recalling p1, q1, p0, and q0 from (Equation8), we get Δ0,n(gm(λ))=ngm(λ)1.Using (Equation18), we can substitute gm(λ) and rearrange terms to get Δ0,n(gm(λ))=n(Δm,0(λ)Δm1,0(λ)+1)=nΔm,0(λ)+Δm1,0(λ)Δm1,0(λ).Finally, from this and (Equation16) we get (Equation19).

Corollary 3.7

For all (m+j,n)Nλ, where m,jN, m1, we have (20) Δ0,n(gm+j(λ))=Δm,n(gj(λ))Δm1,0(gj(λ))(20)

Proof.

This follows from Lemma 3.6 with gj(λ) instead of λ.

Our next Lemma is an important tool for determining scaling properties of these errors.

Lemma 3.8

Let λ=[0;λ1,λ2,][0,1)Q. For all (m+j,n)Nλ such that m,jN and m1, (21) Δj,n(gm(λ))=Δm+j,n(λ)Δm1,0(λ).(21)

Proof.

Let us prove first that (Equation21) holds for n = 0. By multiplying and dividing by Δm+k1,0 for all 0kj, we get Δm+j,0(λ)Δm1,0(λ)=k=0jΔm+k,0(λ)Δm+k1,0(λ)Then, using (Equation18), we get Δm+j,0(λ)Δm1,0(λ)=k=0jΔ0,0(gm+k(λ)).Rearranging this last expression and using (Equation20), we get Δm+j,0(λ)Δm1,0(λ)=(1)j+1Δ0,0(gm(λ))k=1jΔk,0(gm(λ))Δk1,0(gm(λ))We then simplify the product by cancelling terms in the numerator and denominator to get Δm+j,0(λ)Δm1,0(λ)=(1)j+1Δ0,0(gm(λ))((1)jΔj,0(gm(λ))Δ0,0(gm(λ)))=Δj,0(gm(λ)).Finally, for general (m+j,n)Nλ, m,jN, m1, we have Δm+j,n(λ)Δm1,0(λ)=Δm+j,n(λ)Δm+j1,0(λ)Δm+j1,0(λ)Δm1,0(λ)Using (Equation19) and (Equation21) for n = 0, we get Δm+j,n(λ)Δm1,0(λ)=Δ0,n(gm+j(λ))Δj1,0(gm(λ)),and then using (Equation20) gives us Δm+j,n(λ)Δm1,0(λ)=Δj,n(gm(λ))Δj1,0(gm(λ))Δj1,0(gm(λ))=Δj,n(gm(λ)),as required.

With these properties in mind, we will now define the sets which will partition a neighbourhood of the middle cone Cc. Recall that Nλ< denotes the subset of Nλ defined by Nλ<={(m,n)N2:0n<λm+1}.For (m,n)Nλ<, let Sm,n(λ) be the set defined by (22) Sm,n(λ)={(C0Δm,0(λ))Cc(CcΔm,n+1(λ))(Cd+1(nΔm,0(λ)+Δm1,0(λ))),ifmis even,(C0(nΔm,0(λ)+Δm1,0(λ)))Cc(CcΔm,n+1(λ))(Cd+1Δm,0(λ)),ifmis odd.(22) For brevity, we will drop the argument λ from Sm,n(λ) if it is unambiguous. Additionally for the purposes of the case that m = 0, and recalling (Equation9), we have Δ1,0(λ)=q1λp1=1.Recall from (Equation15) that for (m,n)Nλ<, Δm,n>0 if and only if m is odd and n>0 or m is even and n = 0. Thus we can clearly see that Sm,n for all (m,n)Nλ<. Additionally, since every point in Sm,n has positive imaginary part, the boundary of Sm,n consists of segments of the non-horizontal boundary lines of C0, Cc, and Cd+1, and all of these lines either have angle α0 or παd+1. Thus, Sm,n is a quadrilateral, and its opposing sides must be parallel, so it is a parallelogram.

Indeed, since opposite edges of Sm,n are parallel, the side lengths of Sm,n are uniquely determined by the horizontal distances between opposing edges. In the case that m is even, these are precisely the distances between the vertices of the pairs of cones C0Δm,0 and Cc, and CcΔm,n+1 and Cd+1Δm,n. In the case that m is odd, the horizontal distances are determined by the distance between the vertices of pairs of cones C0Δm,n and CcΔm,n+1, and Cc and Cd+1Δm,0. Since Δm,n+1Δm,n=Δm,0, we know that these distances are equal. Therefore, the side lengths of opposing edges of Sm,n are equal and can be calculated as (23) Δm,nsinα0sin(α0+αd+1)andΔm,nsinαd+1sin(α0+αd+1)(23) These sidelengths are equal only when α0=αd+1, in which case Sm,n is a rhombus for all (m,n)Nλ<. See Figure  for an example of the geometry of the construction of sets Sm,n.

Figure 6. An illustration of the construction of the parallelogram S0,1(λ) for the parameters in Figure . Here, the angles shown indicate the cones used to construct S1,0(λ). In this case, the vertices of these cones can be verified via (Equation22) to be Δ0,0(λ)=λ, Δ1,1(λ)=12λ, 0 and (Δ0,0(λ)+Δ1,0)=1λ.

Figure 6. An illustration of the construction of the parallelogram S0,1(λ) for the parameters in Figure 4. Here, the angles shown indicate the cones used to construct S1,0(λ). In this case, the vertices of these cones can be verified via (Equation22(22) Sm,n(λ)={(C0−Δm,0(λ))∩Cc∩(Cc−Δm,n+1(λ))∩(Cd+1−(nΔm,0(λ)+Δm−1,0(λ))),ifmis even,(C0−(nΔm,0(λ)+Δm−1,0(λ)))∩Cc∩(Cc−Δm,n+1(λ))∩(Cd+1−Δm,0(λ)),ifmis odd.(22) ) to be −Δ0,0(λ)=−λ, −Δ1,1(λ)=1−2λ, 0 and −(Δ0,0(λ)+Δ−1,0)=1−λ.

An interesting property of these sets can be found by an application of Lemma 3.8.

Theorem 3.9

Let λ[0,1)Q. For all (m+j,n)Nλ< such that m,jN and m2 is even, 1|Δm1,0(λ)|Sm+j,n(λ)=Sj,n(gm(λ)).

Proof.

Firstly, recall from (Equation15) that since m is even, Δm1,0(λ)<0, and thus 1|Δm1,0(λ)|=1Δm1,0(λ)>0.Hence, 1|Δm1,0(λ)|(Ckx)=Ck+xΔm1,0(λ),for all k{1,,d} and all xR. Thus, we have 1|Δm1,0(λ)|Sm+j,n(λ)={(C0+Δm+j,0(λ)Δm1,0(λ))(Cc+Δm+j,n+1(λ)Δm1,0(λ))Cc(Cd+1+nΔm+j,0(λ)+Δm+j1,0(λ)Δm1,0(λ)),ifm+jis even,(C0+nΔm+j,0(λ)+Δm+j1,0(λ)Δm1,0(λ))Cc(Cc+Δm+j,n+1(λ)Δm1,0(λ))(Cd+1+Δm+j,0(λ)Δm1,0(λ)),ifm+jis odd.Using (Equation21) many times give us 1|Δm1,0(λ)|Sm+j,n(λ)={(C0Δj,0(gm(λ)))(CcΔj,n+1(gm(λ)))Cc(Cd+1(nΔj,0(gm(λ))+Δj1,0(gm(λ)))),ifjis even,(C0(nΔj,0(gm(λ))+Δj1,0(gm(λ))))Cc(CcΔj,n+1(gm(λ)))(Cd+1Δj,0(gm(λ))),ifjis odd.Finally, comparing this with (Equation22) gets 1|Δm1,0(λ)|Sm+j,n(λ)=Sj,n(gm(λ)).

This theorem seems to suggest the possibility of infinite renormalizability of the first return maps to Cc for a whole class of TCEs. At the very least, if indeed the first return map of a TCE to Cc is an isometry on Sm,n(λ) for (m,n)Nλ< with mm0 and some m0N, then at the very least the partition matches with its potential renormalization (Figures and ).

Figure 7. An illustration of the E-image of the partition for the first return map Rκ of the TCE Fκ in Figure . Observe the alternating stacks of parallelograms decreasing in size and cascading towards the origin.

Figure 7. An illustration of the E-image of the partition for the first return map Rκ of the TCE Fκ in Figure 4. Observe the alternating stacks of parallelograms decreasing in size and cascading towards the origin.

Figure 8. A similar illustration to Figure  of the E-image of the partition for Rκ, but with the parameters from Figure .

Figure 8. A similar illustration to Figure 7 of the E-image of the partition for Rκ, but with the parameters from Figure 5.

4. Renormalization around zero

In this section we investigate renormalizability of TCEs around the origin for a broad range of values of λ and η.

Let λ[0,1)Q and ηR such that λ<η=p<1 for some p,qN. Note that Nλ< has a well-ordering < induced by the indexing function wλ so that (m,n)<(r,s)if and only ifwλ(m,n)<wλ(r,s).Thus, the notion of a ‘maximal element’ of a finite subset of Nλ< is well-defined.

Let (m0,n0) be the largest element of Nλ< such that (24) Pm0,n0<porQm0,n0<q.(24) The pair (m0,n0) is well-defined since p,q1 but P0,0=0. Thus, wλ(m0,n0)wλ(0,0).

Note that for all (m,n)Nλ<, Δm,n=Qm,nλPm,n=η+(Qm,nq)λ(Pm,np).Define the sequence (hm,n)(m,n)Nλ< of positive integers by (25) hm,n=(Qm,nq)+(Pm,np)+1.(25) We can establish some recurrence relations for the sequence hm,n using those of Pm,n and Qm,n.

Proposition 4.1

Let (m,n)Nλ<. Then hm,n+1=(n+1)hm,0+hm1,0+(n+1)(p+q1).Moreover, if wλ(m,n)>wλ(m0,n0), then hm,n>0.

Proof.

Recall from the definition of Pm,n and Qm,n in (Equation11) that Qm,n+1=(n+1)Qm,0+Qm1,0,and Pm,n+1=(n+1)Pm,0+Pm1,0.By applying this to the formula (Equation25), we get hm,n+1=(Qm,n+1q)+(Pm,n+1p)+1=((n+1)Qm,0+Qm1,0q)+((n+1)Pm,0+Pm1,0p)+1=(n+1)(Qm,0+Pm,0)+((Qm1,0q)+(Pm1,0p)+1).Recalling the formula (Equation25) for hm,0, we get hm,n+1=(n+1)((Qm,0q)+(Pm,0p)+1+(p+q1))+hm1,0=(n+1)hm,0+hm1,0+(n+1)(p+q1).Recall that (m0,n0) is the largest pair in Nλ< such that either Qm0,n0<q or Pm0,n0<p. Thus, if wλ(m,n)>wλ(m0,n0), then necessarily Qm,nq and Pm,np, which further implies hm,n>0 by (Equation25).

We won't attempt to find recursive relations for all iterates of Fκ at 0, but we will at least calculate the orbit of 0 at iterates given by the sequence (hm,n)(m,n)Nλ<.

Lemma 4.2

Let (m,n)Nλ< such that wλ(m,n)>wλ(m0,n0). Then Fκhm,n(0)=Δm,n.

Proof.

Suppose, for a contradiction, that Fκhm,n(0)Δm,n. Let (at)tN and (bt)nN denote the sequences defined by (26) Fκt(0)=η+btλat.(26) Note that since η=p, we have Fκt(0)=p+btλat=(bt+q)λ(at+p),which is never equal to 0 since λ is irrational and bt+q and at+p are both positive. Observe that (at)t and (bt)t are both non-decreasing and obey the following rule: (at+1,bt+1)={(at,bt+1),if Fκt(0)<0,(at+1,bt),if Fκt(0)>0.Given that a0=a1=b0=b1=0, we can deduce that the sequences (at)t and (bt)t achieve every non-negative integer value, that is, for any NN, there is some tN such that at=N, and similarly there is some tN such that bt=N. Moreover, a simple inductive argument shows that for all integers t1, at+bt+1=t.Next, observe that Fκhm,n(0)Δm,n is equivalent to the statement that ahm,nPm,n and bhm,nQm,n. However, notice that (Qm,nq)+(Pm,np)+1=hm,n=ahm,n+bhm,n+1.This implies one of two cases.

  1. ahm,n<Pm,np and bhm,n>Qm,nq; or

  2. ahm,n>Pm,np and bhm,n<Qm,nq.

Suppose case (1) holds. Since (bt)t is non-decreasing and takes every non-negative integer value, we know that there is some non-negative integer t<hm,n such that bt=Qm,nq. Thus, Fκt(0)=η+(Qm,nq)λat.Note that at<at<Pm,np. Now, suppose Δm,n>0. Then for any 0jPm,npat, we have Fκt(0)j=η+(Qm,nq)λ(at+j)>(Qm,nq)λ(Pm,np)=Δm,n>0.Thus, Fκt+j(0)=Fκt(0)j for 0jPm,npat. In particular, Fκt+Pm,npat(0)=η+(Qm,nq)λ(Pm,np).But then t+Pm,npat=(Qm,nq)+(Pm,np)+1=hm,n.And this implies that Fκhm,n(0)=Fκt+Pm,npat(0)=η+(Qm,nq)λ(Pm,np)=Δm,n.But this contradicts our assumption that Fκhm,n(0)Δm,n.

Now suppose that Δm,n<0. Let 0j<Pm,npat. Then either Fκt(0)j>0 or Δm,n<Fκt(0)j<0. But Δm,n<Fκt(0)j<0 implies |Qm,nλ(at+j+p)|<|Δm,n|=|Qm,nλPm,n|,and at+j+p<Pm,n. This contradicts the ‘best approximate’ property of the semiconvergent Pm,n/Qm,n. Thus Fκt(0)j>0 for all 0j<Pm,npat. Thus, by a similar argument to before, we reach the contradiction that Fκhm,n(0)=Δm,n.

In case (2), ahm,n>Pm,np and bhm,n<Qm,nq. We can reach a similar contradiction as above, by using a similar argument where the roles of (at)t and (bt)t are interchanged.

This exhausts all cases, so our assumption that Fκhm,n(0)Δm,n must be false.

Lemma 4.3

Let mN such that wλ(m,0)>wλ(m0,n0). Then, for all 1t<hm+1,0, |Fκt(0)||Δm,0|.

Proof.

recall that Fκt(0)=η+btλat, for some at,btN. Since (at)t and (bt)t are non-decreasing and t<hm+1,0, we have that btqm+1q and atpm+1p, not both equal. Hence either bt+q<qm+1 or at+p<pm+1 and bt+qqm+1. In either case, by the best approximate property of convergents |Fκt(0)|=|(bt+q)λ(at+p)||qmλpm|=|Δm,0|.

Lemma 4.4

Let (m,n)Nλ<. Then for all 1t<hm,n+1, Fκt(0)Δm,0orFκt(0)nΔm,0+Δm1,0ifmis even,Fκt(0)Δm,0orFκt(0)nΔm,0+Δm1,0ifmis odd.

Proof.

From (Equation12), recall that if m is even and n>0, then Pm,n/Qm,n>λ is a best approximate from above, which implies (27) aQm,nλPm,n<0,(27) for all a,bZ with 0<b<Qm,n+1 such that Pm,n/Qm,na/b>λ.

Let (at)tN and (bt)tN be the sequences described by (Equation26). Suppose that 1thm,n with Fκt(0)<0. Then btQm,n+1qandatPm,n+1p,since (at)t and (bt)t are non-decreasing. Thus, (28) bt+qQm,n+1andat+pPm,n+1,(28) not both equal. Therefore, from (Equation26) we know that Fκt(0)=(bt+q)λ(at+p),and by (Equation27) with (Equation28), we have Fκt(0){Qm,nλPm,n,ifn0,qm1λpm1,ifn=0.Recalling the definition of Δm,n as in (Equation14), we get Fκt(0){Δm,n,ifn0,Δm1,0,ifn=0.Finally, by using (Equation16), we have Fκt(0)nΔm,0+Δm1,0.If Fκt(0)>0, then by Lemma 4.3, we have Fκt(0)Δm,0.From (Equation13), recall that if m is odd and n>0, then Pm,n/Qm,n<λ is a best approximate from below, that is aQm,nλPm,n>0,for all a,bZ with 0<b<Qm,n+1 such that Pm,n/Qm,na/b<λ. Thus, in the case that m is odd and Fκt(0)>0, then bt+qQm,n+1 and at+pPm,n+1,not both equal. Thus, similarly to the above case where m is even, we have Fκt(0)=(bt+q)λ(at+p){Qm,nλPm,nifn0qm1λpm1ifn=0=nΔm,0+Δm1,0.On the other hand, if Fκt(0)<0, then by Lemma 4.3, Fκt(0)Δm,0.

In order to prove the next theorem, we will distinguish between the following two cases and we will prove them separately. We will first prove that ifzE1(Sm,n),thenh(z)=hm,n+1,and then we will prove that ifh(z)=hm,n+1,thenFκh(z)(z)=E(z)+Δm,n+1.

Theorem 4.5

Let αBd+2, τ:{1,,d}{1,,d} be a bijection, λ[0,1)Q, λ<η=p<1 for some p,qN{0} and set κ=(α,τ,λ,η,1). For all (m,n)Nλ< with wλ(m,n)>wλ(m0,n0), h(E1(Sm,n)) exists and is equal to hm,n+1. Moreover, let zE1(Sm,n). Then (29) Rκ(z)=E(z)+Δm,n+1(λ).(29)

Proof.

Let zCc. Assume zE1(Sm,n). Observe that E(z)+Fκhm,n+1(0)Cc, so h(z)hm,n+1. By Lemma 3.3, we know that Fκh(z)(z)=E(z)+Fκh(z)(0).Suppose, for a contradiction, that h(z)<hm,n+1. We will prove the contradiction for even and odd m separately, starting with the case that m is even. By Lemma 4.4, we know that either Fκh(z)(0)Δm,0 or Fκh(z)(0)nΔm,0+Δm1,0. Since m is even, recall that Sm,n=(C0Δm,0)(CcΔm,n+1)Cc(Cd+1(nΔm,0+Δm1,0)).Observe that Fκh(z)(z)Sm,n+Fκh(z)(0)(C0+(Fκh(z)(0)Δm,0))(Cd+1+(Fκh(z)(0)(nΔm,0+Δm1,0))).Therefore, if Fκh(z)(0)Δm,0, then Fκh(z)(z)C0+(Fκh(z)(0)Δm,0)C0.But by the definition of h(z) as in (Equation6), we have Fκh(z)(z)=Rκ(z)Cc,which reveals a contradiction. Similarly, if Fκh(z)(0)nΔm,0+Δm1,0, then Fκh(z)(z)Cd+1+(Fκh(z)(0)(nΔm,0+Δm1,0))Cd+1,which also contradicts Fκh(z)(z)Cc.

Now suppose m is odd. Then Sm,n=(C0(nΔm,0+Δm1,0))Cc(CcΔm,n+1))(Cd+1Δm,0).By Lemma 4.4, we know that either Fκh(z)(0)nΔm,0+Δm1,0 or Fκh(z)(0)Δm,n.Clearly, either of these cases give similar contradictions as before. Therefore our assumption that h(z)<hm,n+1 must be false, so in fact h(z)=hm,n+1.

Now, suppose h(z)=hm,n+1. By Lemma 3.3, Fκh(z)(z)=E(z)+Fκh(z)(0),and by Lemma 4.2 we know that Fκhm,n+1(0)=Δm,n+1.Combining these two with our assumption gives us that if h(z)=hm,n+1, then Fκh(z)(z)=E(z)+Δm,n+1.

With this Theorem, as well as Theorem 3.9, we can prove the existence of a renormalization scheme around the point 0. First, we will find the definition of the first return map Rκ on the rest of Cc.

Lemma 4.6

Let Sm,n be as in (Equation22) and (m0,n0)Nλ< as in (Equation24). Define the set U(κ) U(κ)={0}(m,n)Nλ<wλ(m,n)wλ(m0,n0)Sm,n.Then U(κ)={(C0Δm0,0)Cc(Cd+1(n0Δm0,0+Δm01,0)),ifm0is even,(C0(n0Δm0,0+Δm01,0))Cc(Cd+1Δm0,0),ifm0is odd,and U(κ) is convex.

Proof.

For brevity we will drop the parameters κ when they are unambiguous. We will first prove the equality U(κ)={(C0Δm0,0)Cc(Cd+1(n0Δm0,0+Δm01,0)),ifm0is even,(C0(n0Δm0,0+Δm01,0))Cc(Cd+1Δm0,0),ifm0is odd..Observe that for all (m,n)Nλ<, Sm,nCc.Additionally, if (m,n)Nλ< such that wλ(m,n)=j+k01, then Sm,n{m=m0(C0Δm,0),ifm0is even,wλ(m,n)wλ(m0,n0)(C0(nΔm,0+Δm1,0)),ifm0is odd.Note that C0(nΔm+1,0+Δm,0)C0Δm,0 for all (m+1,n)Nλ<. Thus, we have Sm,n{C0Δm0,0,ifm0is even,C0(n0Δm0,0+Δm01,0),ifm0is odd..We also know that Sm,n{wλ(m,n)wλ(m0,n0)(Cd+1(nΔm,0+Δm1,0)),ifm0is even,m=m0+1(Cd+1Δm,0),ifm0is odd.Thus, with a similar argument as above, we can show that Sm,n{Cd+1(n0Δm0,0+Δm01,0),ifm0is even,Cd+1Δm0,0,ifm0is odd.,Altogether, we deduce that U(κ){(C0Δm0,0)Cc(Cd+1(n0Δm0,0+Δm01,0)),ifm0is even,(C0(n0Δm0,0+Δm01,0))Cc(Cd+1Δm0,0),ifm0is odd..Suppose m0 is even, and let z(C0Δm0,0)Cc(Cd+1(n0Δm0,0+Δm01,0)),with z0. Then there is some (m,n)Nλ< with m is odd and wλ(m,n)wλ(m0,n0) such that zC0(nΔm,0+Δm1,0),and there is some (m,n)Nλ< with m even and wλ(m,n)wλ(m0,n0) such that zCd+1(nΔm,0+Δm1,0).Suppose that (m,n) and (m,n) are the largest such pairs, which is well-defined since z0. Since m is odd and m is even, we either have m<m or m<m. Suppose m<m. Then z(C0(nΔm,0+Δm1,0))Cc(Cd+1Δm,0).Since (m,n) is the largest pair such that zC0(nΔm,0+Δm1,0), and noting that (n+1)Δm,0+Δm1,0=Δm,n+1 since n + 1>0, we have that z(CcΔm,n+1)orz(Cd+1Δm,n+1).However, Cd+1Δm,n+1Cd+1 since m is odd and so Δm,n+1>0. Importantly, Cc(Cd+1Δm,n+1)=,and thus zCcΔm,n+1. Therefore, z(C0(nΔm,0+Δm1,0))(CcΔm,n+1)Cc(Cd+1Δm,0)=Sm,n,so clearly zU(κ). In the case that m<m we can use a similar argument to prove that z(C0Δm,0)Cc(CcΔm,n+1)(Cd+1(nΔm,0+Δm1,0))=Sm,n,so that zU(κ). If m0 is odd and z(C0(n0Δm0,0+Δm01,0))Cc(Cd+1Δm0,0), with z0, then we can use a similar argument to prove that zU(κ). Hence, we have U(κ)={(C0Δm0,0)Cc(Cd+1(n0Δm0,0+Δm01,0))ifm0is even(C0(n0Δm0,0+Δm01,0))Cc(Cd+1Δm0,0)ifm0is odd.To show that U is convex, one must note that the cones C0, Cc, and Cd+1 and all their translates are convex sets, and that the intersection of convex sets is also convex.

Define (30) Uk,l(κ)={0}(m,n)Nλ<wλ(m,n)wλ(k,l)Sm,n,(30) for wλ(k,l)wλ(m0,n0) (omitting the arguments where unambiguous), and let κ=(α,τ,λ,η,ρ),where λ<η=pqλ<ρ is such that pp and qq. If (m0,n0)Nλ< is the maximal element of Nλ< such that Pm0,n0<p or Qm0,n0<q, then wλ(m0,n0)wλ(m0,n0). Furthermore, (31) Rκ|Um0,n0(κ)(z)=Rκ|Um0,n0(κ)(z).(31) Given αB, τ:{1,,d}{1,,d} be a bijection, λ[0,1)Q, ρ=1, λ<η=p<1 for some p,qN, let (m0,n0)Nλ< be as in (Equation24). Let p,qN{0} be defined by p=Pm0,0(g2(λ)),andq=Qm0,0(g2(λ)),and let (32) η=pqg2(λ).(32) Clearly by the definition of the one-sided convergents in (Equation11), we have g2(λ)<η<1.With this in mind, we have the following Theorem.

Theorem 4.7

Let αBd+2, τ:{1,,d}{1,,d} be a bijection, λ[0,1)Q and λ<η=p<1 for some p,qN. Set the tuples κ=(α,τ,λ,η,1) and κ=(α,τ,λ,η,1), where λ=g2(λ) and η is as in (Equation32). Then for all zUm0,0(κ), Rκ(z)=11λ1λRκ((1λ1λ)z).

Proof.

We begin by noting that the quantity (33) (m0,n0)=max{(m,n)Ng2(λ)<:Pm,n(g2(λ))<porQm,n(g2(λ))<q},(33) satisfies (m0,n0)=(m0,0) and, importantly for the reasons of (Equation31), we have (34) wλ(m0+2,n0)wλ(m0+2,0).(34) Here we recall the equality in Proposition 3.4 since (m0,n0) is being considered an element of Ng2(λ)< and (m0,0) an element of Nλ<.

Recall that by Theorem 3.9, we have 11λ1λSj+2,n(λ)=Sj,n(g2(λ)),for all (j+2,n)Nλ< with wλ(j+2,n)wλ(m0+2,0), i.e. jm0. Note that by Proposition 3.4, wλ(m0+2,n0)wλ(m0+2,0),is equivalent to the statement that wg2(λ)(m0,n0)wg2(λ)(m0,0).Hence, 11λ1λUm0+2,0(κ)=Um0,0(κ).Let zCc so that (1λ1λ)zSm+2,n(λ) with wλ(m+2,n)wλ(m0+2,0). Also note that since aCj=Cj for all a>0 and E consists of rotations about 0, we have (35) E(az)=aE(z),(35) for a>0. Therefore zSm,n(g2(λ)) and by expanding Rκ as in (Equation29) we get 11λ1λRκ((1λ1λ)z)=11λ1λ(E((1λ1λ)z)+Δm+2,n+1(λ)).Using Equation35, we get 11λ1λRκ((1λ1λ)z)=11λ1λ((1λ1λ)E(z)+Δm+2,n+1(λ))=E(z)+(Δm+2,n+1(λ)λ1λ1).Now, recall that p1=1 and q1=λ1, so that Δ1,0(λ)=λ1λ1. Using (Equation21) and comparing with the formula for Rκ as in (Equation29), we see 11λ1λRκ((1λ1λ)z)=E(z)+(Δm+2,n+1(λ)Δ1,0(λ))=E(z)+Δm,n+1(g2(λ))=Rκ(z).

The immediate consequence of Theorem 4.7 is that when λ[0,1)Q and λ<η=p<1, one can renormalize infinitely ‘towards’ 0 in the sense that the domains of each renormalization are shrinking neighbourhoods of 0 in Cc. Additionally, the renormalizations are determined by the orbit of λ under the square of the Gauss map g2(x). We shall now apply our results in this and the previous section to an example (Figure ).

Figure 9. A plot of 1500 iterates of 500 uniformly chosen points within the box [1,λ]×[0,1.1] under the TCE with parameters α=(π/20.6,0.5,0.7,π0.6), τ:12,21, λ=Φ, η=Φ2, and ρ=1. The first 400 points of each orbit are omitted to remove transients.

Figure 9. A plot of 1500 iterates of 500 uniformly chosen points within the box [−1,λ]×[0,1.1] under the TCE with parameters α=(π/2−0.6,0.5,0.7,π−0.6), τ:1↦2,2↦1, λ=Φ, η=Φ2, and ρ=1. The first 400 points of each orbit are omitted to remove transients.

5. An example

As an example inspired by Peres and Rodrigues [Citation23], we will set αBd+2, λ=Φ, ρ=1, and η=Φ2, where (36) Φ=512.(36) In this case the the space Nλ<=N×{0}N since λ=[0;1¯] and so λk=1 for all kN. Thus, wΦ(m,n)=wΦ(m)=m. Also due to λk=1 for each kN, the semiconvergents Pm/Qm simply coincide with the convergents pm/qm, and the convergents are in this case defined by p0=0,q0=1,p1=1,q1=1,pm=pn1+pn2,qm=qm1+qm2.It is thus clear that pm=Fibm and qm=Fibm+1, where (Fibm)mN is the Fibonacci sequence with Fib0=0 and Fib1=1.

The first return times (hm)mN in the case of λ=Φ are given by hm=(qm1)+(pm1)+1=Fibm+1+Fibm1=Fibm+21for all mN. Observe that hm=hm1+hm2+1 for all m2, and h0=2, h1=4. Now note that η=Φ2=1Φ=1λ, and thus we have m0=max{mN:pm<1orqm<1}=0.The errors of the convergents Δm(Φ) are given by (37) Δm(Φ)=qmλpm=Fibm+1ΦFibm.(37)

Proposition 5.1

We have Δm(Φ)=(Φ)m+1

Proof.

Observe that Δm(Φ)=Fibm+1ΦFibm=FibmΦ+Fibm1ΦFibm=Fibm1Φ(1Φ)Fibm=Φ(FibmΦFibm1)=ΦΔm1(Φ).Recall that p0=0 and q0=1. Then a simple inductive argument shows us that Δm(Φ)=(Φ)mΔ0(Φ)=(Φ)m(q0Φp0)=(Φ)m+1

Noting that the recurrence relations for pm and qm give us p1=1 and q1=0 and so we can set Δ1=1=(Φ)0, which remains consistent with (Equation37). With this proposition in mind, for mN we can determine the sets Sm as Sm={(C0Δm)Cc(CcΔm+1)(Cd+1Δm1),ifmis even,(C0Δm1)(CcΔm+1)Cc(Cd+1Δm),ifmis odd.={(C0+(Φ)m+1)Cc(Cc+(Φ)m+2)(Cd+1+(Φ)m),ifmis even,(C0+(Φ)m)(Cc+(Φ)m+2)Cc(Cd+1+(Φ)m+1),ifmis odd.These are rhombi, as can be seen in Figure , and as can be deduced from the discussion around (Equation23) since α0=αd+1. It is also clear to see that for all mN. (38) Sm+2=Φ2Sm.(38) In this case, it is simple to find a partition for the entirity of the middle cone Cc for the map Rκ. In particular, define the sets X and Y to be (39) X=Cc(Cc(λη))(Cd+1+η)=Cc(CcΦ3)(Cd+1+Φ2),(39) and (40) Y=Cc(Cc+η)=Cc(Cc+Φ2).(40) As we will see soon, the collection {X,Y,Sn:n2} forms a partition of Cc. We are interested in the pre-image of these sets under E. In particular, define the partition C={E1(S)Cj:j{1,,d},S{Y,X,S2,S3,,}}.

Figure 10. A partition of Cc in the case where α=(1,0.5,π2.5,1), τ:12,21, λ=Φ, η=Φ2, and ρ=1. A cascading pattern towards the origin can be seen, but its geometric structure becomes clearer after we apply the cone exchange E.

Figure 10. A partition of Cc in the case where α=(1,0.5,π−2.5,1), τ:1↦2,2↦1, λ=Φ, η=Φ2, and ρ=1. A cascading pattern towards the origin can be seen, but its geometric structure becomes clearer after we apply the cone exchange E.

Figure 11. The same partition as in Figure , but after an application of E, which reveals an alternating pattern of rhombi.

Figure 11. The same partition as in Figure 10, but after an application of E, which reveals an alternating pattern of rhombi.

This partition can be seen in Figure . As a consequence of the next theorem, (C,Rκ) is a PWI with a countably infinite number of atoms. We also define a separate family of sets Q={Qn,j=E1(Sn)Cj:nN,j{1,,d}},which includes only the rhombi, and thus forms a partition of only a subset of the middle cone. Note that since λm=1 for all mN and m0=0. We see that the set U(κ) from Lemma 4.6 is given by U(κ)={0}m=0Sm=(C0Δ0)Cc(Cd+1Δ1)=(C0Φ)Cc(Cd+1+1).Also observe that by removing S0 and S1 we get U2,0(κ)={0}m=2Sm=(C0Φ3)Cc(Cd+1+Φ2).From this, we notice that U2(κ)X=Cc(Cd+1+Φ2)((CcΦ3)(C0Φ3)),but since Cd+1Φ3Cd+1, we know that Cc(Cd+1Φ3)=, so U2(κ)X=Cc(Cd+1+Φ2)((Cd+1Φ3)(CcΦ3)(C0Φ3))=Cc(Cd+1+Φ2)H¯=Cc(Cd+1+Φ2).Therefore, we can see that U2(κ)XY=Cc((Cc+Φ2)(Cd+1+Φ2)),and a similar argument tells us that Cc(C0+Φ2)= and so finally we get, U2(κ)XY=Cc((C0+Φ2)(Cc+Φ2)(Cd+1+Φ2))=CcH¯=Cc.Therefore, C is a partition of Cc up to a set of Lebesgue measure 0.

Note that for all (m,n)NΦ<, wΦ(m,1)=wΦ(m+1,0), and so by recalling that wΦ(m,0)=wΦ(m)=m, the condition that wΦ(m,1)>wΦ(m0,0) is equivalent to the condition that wΦ(m+1,0)>wΦ(m0,0),which is itself equivalent to m>m01=1.With this in mind, Theorem 4.5 tells us that for all mN, if zE1(Sm), then (Figure ) h(z)=hm,1=hm+1,0=hm+1=Fibm+31,and Rκ(z)=E(z)+Δm+1=E(z)(Φ)m+2,

Figure 12. The same partition of Cc as in Figure , after an application of Rκ, which has shifted the rhombi alternately. Note that there is an overlap between the cone and ribbon (both of which are more clearly seen in Figure ) on the top of the figure, causing an unavoidable overlap of the colours.

Figure 12. The same partition of Cc as in Figure 10, after an application of Rκ, which has shifted the rhombi alternately. Note that there is an overlap between the cone and ribbon (both of which are more clearly seen in Figure 11) on the top of the figure, causing an unavoidable overlap of the colours.

Observe that λ=Φ is a special case of irrational number within [0,1] in the sense that it is a fixed point of the Gauss map g. Thus, g2(Φ)=Φ and thus we can choose η=η=1Φ=Φ2 and Theorem 4.7 tells us that the first return map Rκ exhibits exact self-similarity within U(κ). In particular, for all zU0,0(κ)=U(κ), we have the following conjugacy Rκ(z)=1Δ1Rκ((Δ1)z)=1Φ2Rκ(Φ2z).One consequence of this is that if there exists a periodic point zSm of period k, then z is a periodic point of Rκ with period k/hm+1. The self-similarity shows that for all nZ such that 2nm, Φ2nz is a periodic point of Rκ, thus also a periodic point of Fκ whose period is an integer multiple of hm+2n+1. In particular, there is a sequence (zn)nN given by (41) zn=Φ2nmz,(41) so that for all nN, znS2n+m~ and the period of zn is an integer multiple of h2n+m~+1, where {0,1}m~m(mod2).

Given a map f:XX, let Of+(x) denote the forward orbit of xX under f, that is Of+(x)={fn(x):nN}.

Proposition 5.2

Suppose there exists a periodic point zSm for some mN, and let (zn)nN be the sequence of periodic points given by (Equation41). Then the sequence (OFκ+(zn))nN of periodic orbits accumulates on the interval [1,Φ].

Proof.

Let nN. Note that (42) {Fκj(zn):1jh(z)}OFκ+(zn).(42) Lemma 3.3 tells us that for all 1jh(zn), Fκj(zn)=E(zn)+Fκj(0).Therefore, (43) |Fκj(zn)Fκj(0)|=|E(zn)|=|zn|.(43) Let HN. Then there exists an NN such that h(zn)H for all nN, and thus (Equation43) holds for all 1jH. Now let ε>0 be small. Then there exists an NN such that for all integers nN such that (44) |zn|<ε.(44) Set N=max{N,N}. Then for all integers nN, both (Equation43) holds for all 1jH and (Equation44) holds. Hence, for all nN we have |Fκj(z)Fκj(0)|<ε,for all 1jH. Since H and ε are independent and arbitrary, we conclude that the sequence (OFκ+(zn))nN accumulates on the set OFκ+(0).

By Proposition 3.1, Fκ is a 2-IET everywhere on the interval [1,λ] except on the preimages of 0, since Fκ(0)=η=1λ, contrary to Fκ(x)=x+λ for x[1,0) and Fκ(x)=x1 for x(0,λ).

Therefore Fκ is conjugate to an irrational rotation almost everywhere (with respect to one-dimensional Lebesgue measure), since λ=Φ is irrational. In particular, since λ is irrational and η=1λ, we know, by for example Lemma 4.3, that Fκj(0)=Fκj1(η) is bounded away from 0 for all integers j>0, so Fκj(0)0 for any j>0.

Hence, the orbit of Fκ(0)=η under Fκ is also the orbit under an irrational rotation, and thus the orbit of 0 is dense in the interval [1,λ], i.e. OFκ+(0)¯=[1,λ].Therefore, the sequence (OFκ+(zn))nN accumulates on the interval [1,λ].

Remark 5.3

Although extending Proposition 5.2 to periodic continued fractions λ should follow from a similar strategy to the proof used here, an extension to aperiodic continued fractions seems to require nothing short of assuming/proving that every TCE has at least one periodic point in its ‘renormalizable domain’ U(κ).

6. Discussion

Translated cone exchanges, introduced first in [Citation4] and investigated in [Citation22, Citation23], are an interesting and largely unexplored family of parametrized PWIs. They contain an embedding of a simple IET on the baseline and as such they are an interesting tool to understand more general PWIs by gaining leverage from known results for IETs. In this paper we go beyond results in [Citation4, Citation22, Citation23] to show that for a dense subset of an open set in the parameter space of TCEs there is a mapping (κκ in Theorem 4.7) that determines a renormalization scheme for the first return map Rκ of Fκ to the vertex 0 of the middle cone Cc. This helps us describe the small-scale, long-term behaviour of Fκ near the baseline [1,λ] via the large-scale, short-term behaviour of Fκ with κ=(α,τ,λ,η,1), λ=g2k(λ), a large enough integer k>0 and some suitably chosen η described by (Equation32). Proposition 5.2 is an example of this, where a periodic disk of small period for Fκ and the periodicity of the continued fraction coefficients of λ=Φ give rise to an countable collection of periodic disks of arbitrarily high period clustering on [1,λ], through the renormalizability established by Theorem 4.7.

Although these results give a glimpse into the dynamics for orbits close to the baseline, there remains a lot more to do to understand the dynamics of these TCEs near general points in exceptional set, but this seems to be a far more complex task to undertake, especially as the dynamics near the baseline primarily consists of horizontal translations, whereas in general the effect of the rotations will be inextricably linked to translations.

Supplemental material

Acknowledgments

We thank Pedro Peres and Arek Goetz for discussions about this research. NC and PA thank the Mittag-Leffler Institute for their hospitality and support to visit during the ‘Two Dimensional Maps’ programme of early 2023.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Data availability statement

For the purpose of open access, the authors have applied a Creative Commons Attribution (CC BY) licence to any Author Accepted Manuscript version arising from this submission.

No new data were generated or analysed during this study. The figures in this study were produced by python programming code written by NC. This code, namely the ‘pyTCE’ program, is publicly available at the following link: https://github.com/NoahCockram/pyTCE/tree/main.

Additional information

Funding

This work was supported by the Engineering and Physical Sciences Research Council.

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