Multiple tilings associated to d-Bonacci beta-expansions

# Multiple tilings associated to d-Bonacci beta-expansions

Tomáš Hejda LIAFA, CNRS UMR 7089, Université Paris Diderot – Paris 7, Case 7014, 75205 Paris, France
Dept. Math. FCE, University of Chemistry and Technology Prague, Studentská 2031/6, 16000 Prague, Czechia
Dept. Algebra. FMF, Charles University, Sokolovská 49/83, 18600 Prague, Czechia
###### Abstract.

Let be a Pisot unit and consider the symmetric -expansions. We give a necessary and sufficient condition for the associated Rauzy fractals to form a tiling of the contractive hyperplane. For a -Bonacci number, i.e., Pisot root of we show that the Rauzy fractals form a multiple tiling with covering degree .

###### Key words and phrases:
beta-expansions, Rauzy fractals, tiling, multiple tiling
###### 2010 Mathematics Subject Classification:
11A63 52C23 (11R06 37B10)
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m s o m \IfBooleanTF#2(#4)_\ExInd#1 \IfNoValueTF#3(#4)_\ExInd#1 #3(#4#3)_\ExInd#1

## 1. Introduction

Tilings arising from -expansions were first studied in the 1980s by A. Rauzy [rauzy_1982] and W. Thurston [thurston_1989]. They consider the greedy -expansions that are associated to the transformation . S. Akiyama [akiyama_2002] showed that the collection of -tiles forms a tiling if and only if satisfies the so-called weak finiteness property (W). M. Barge [barge_2016b, barge_2016a] proved that all Pisot numbers satisfy property (W); he actually proves that the -substitution associated to the greedy transformation has pure discrete spectrum.

If we drop the “greedy” hypothesis, things are getting more interesting. C. Kalle and W. Steiner [KS] showed that the symmetric -expansions for two particular cubic Pisot numbers  induce a double tiling — i.e., a multiple tiling such that almost every point of the tiled space lies in exactly two tiles. More generally, they proved that every “well-behaving” -transformation with a Pisot unit induces a multiple tiling. The method of Barge cannot be straightforwardly extended to the symmetric -expansions, because for these, there is no direct link to Pisot substitutions. The provided examples of multiple tilings therefore do not disprove the general Pisot substitution conjecture that all Pisot irreducible substitutions have pure discrete spectrum. Actually, multiple tilings are also considered for combinatorial substitutions as done e.g. by S. Ito and H. Rao [ito_rao_2006]. We refer to a mémoire by A. Siegel and J. Thuswaldner [siegel_thuswaldner_2009] for a thorough survey on substitution tilings and the Pisot conjecture.

In this paper we concentrate on the symmetric -expansions associated to the transformation . This transformation was studied before e.g. by S. Akiyama and K. Scheicher in the context of shift radix systems [akiyama_scheicher_2007]. We consider and we define on two intervals . We show the following theorem about the multiple tiling:

###### Theorem 1.

Let , , and let be the -Bonacci number, i.e., the Pisot number satisfying . Then the symmetric -expansions induce a multiple tiling of with covering degree equal to .

It was shown before by H. Rao, Z.-Y. Wen and Y.-M. Yang [rao_wen_yang_2014, Theorem 1.6] that the covering degree is a multiple of . We also note that for any particular and any particular transformation, the degree of the multiple tiling can be computed from the intersection (or boundary) graph, eventually multi-graph, as defined for instance by A. Siegel and J. Thuswaldner [siegel_thuswaldner_2009]; however, such an algorithmic approach is not usable for an infinite number of cases.

We also characterize the tiles that form the distinct layers of the multiple tiling:

###### Theorem 2.

Let , , and let be the -Bonacci number. Let . Then the collection of tiles , where

 (1.1) Lh\coloneqq(⟦h⟧∩[1−β2,12))∪(⟦h−1⟧∩[−12,β2−1)),

forms a tiling of , that is, it is a layer of the multiple tiling guaranteed by Theorem 1. Here we denote .

The two results rely substantially on the knowledge of the purely periodic integer points of :

###### Theorem 3.

Let , , and let be the -Bonacci number. Let denote the set of non-zero such that for some . Then

 P∪{0}={±\scalebox0.8$∙$0p2p3⋯pd:pi∈{0,1}}={±d∑i=2piβ−i:pi∈{0,1}}.

(We exclude from as it does not lie in the support of the invariant measure of .)

Last but not least, for general Pisot units we give a necessary and sufficient condition on the tiling property for the symmetric -expansions:

###### Theorem 4.

Let be a Pisot unit of degree . Then the symmetric -expansions induce a tiling of if and only if the following two conditions are satisfied:

1. is an algebraic unit (i.e., );

2. the balanced -expansions induce a tiling of (the balanced expansions are defined in § 2.2).

The paper is organized as follows. In the following section we define all the necessary notions. The theorems are proved in Section 3. We conclude by a pair of related open questions in Section 4.

## 2. Preliminaries

### 2.1. Pisot numbers

An algebraic integer is a Pisot number iff all its Galois conjugates, i.e., the other roots of its minimal polynomial, lie inside of the unit complex circle. As usual, denotes the ring of integer combinations of powers of , and denotes the field generated by the rational numbers and by .

Suppose that is of degree and has complex Galois conjugates (where denotes the complex conjugate of ) and real ones . Denote the corresponding Galois isomorphisms. Then we put

 Φ:Q(β)→d−e−1∏j=1Q(β(j)),x↦(σ(1)(x),…,σ(d−e−1)(x)).

Since , we consider that . We have the closure properties ; this follows from the Strong Approximation Theorem [neukirch_1999, Ch. 3, § 1, Exercise 1] and from the fact that has finite index in the ring of integers of .

In this paper, we focus on -Bonacci numbers. For a -Bonacci number is the Pisot root of the polynomial . A. Brauer [brauer_1951] showed that this polynomial is irreducible and has a Pisot root. This root satisfies because and have the opposite signs.

We say that two numbers are congruent modulo iff . By , for , we denote the congruence class modulo that contains , i.e., . If is a -Bonacci number, then the norm of is . Therefore there are exactly distinct classes modulo and we can take numbers as their representatives, i.e.,

 Z[β]=d−1⋃h=1⟦h⟧=d−1⋃h=1h+(β−1)Z[β].

### 2.2. β-expansions

We fix . Let be a union of intervals and be a piecewise constant function (digit function) such that for all . Then the map is a -transformation. The -expansion of is then the (right-infinite) sequence , where . We say that is -admissible iff it is the expansion of some .

We define two particular -transformations:

1. Let and (we denote for convenience). This defines the symmetric -expansions. We denote the transformation and the expansion of .

2. Let and iff and otherwise. This defines the balanced -expansions. We denote and accordingly.

Both and are plotted in Figure 1 for the Tribonacci number.

Besides expansions, we consider arbitrary representations. Any bounded sequence of integers is a representation of .

factor of a sequence is any finite word with . A tail of a sequence is any of the infinite words for . A sequence is periodic iff . It is purely periodic iff .

### 2.3. Rauzy fractals

We consider the symmetric -transformations for Pisot units . The symmetric -transformation possesses a unique absolutely continuous invariant measure (w.r.t. the Lebesgue measure). This follows from the work of T.-Y. Li and J. Yorke [li_yorke_1978, Theorem 1], because while has more than one discontinuity point when , we have that for all these points, therefore each in the theorem statement must contain in its iterior; this means that there is only . For any , we define the -tile (or Rauzy fractal) as the Hausdorff limit

 R(x)\coloneqqlimn→∞Φ(βnT−n\ExIndS(x))⊂Rd−1.

Note that for all and all (this holds as the boundary points of the intervals are not in , nor are their images under ), therefore .

The Rauzy fractals induce a multiple tiling, as will follow from the work of Kalle and Steiner [KS, Theorem 4.10]. We recall that the family of tiles is a multiple tiling iff the following is satisfied:

1. The tiles take only finitely many shapes (i.e., are only finitely many modulo translations in ).

2. The family is locally finite, i.e., for every bounded set , only finitely many tiles from intersect .

3. The family covers , i.e., for every there exists such that .

4. Every tile is a closure of its interior.

5. There exists an integer such that almost every point in lies in exactly tiles from ; this is called the covering degree of .

If , we say that is a tiling. Every multiple tiling with covering degree is a union of tilings; we call these tilings layers of the multiple tiling.

## 3. Proofs

First, we establish a strong relation between the symmetric and the balanced expansions in Lemma 1; this works for all .

Then, we suppose that is an integer and is the -Bonacci number. In Lemma 2 we show that the support of the invariant measure of is the whole ; from this, we conclude that is a multiple tiling [KS, Theorem 4.10]. Then we investigate arithmetic properties of the balanced expansions in Lemmas 3, 4 and 5. We use these properties to determine the degree of the multiple tiling, which is done in Lemmas 6, 7 and 8. The proof of Theorem 3 is given after Lemma 5, the proofs of Theorems 1 and 2 are after Lemma 8.

We close this section by the proof of Theorem 4.

###### Lemma 1.

Let . Define a bijection

 ψ:X\ExIndS→X\ExIndB,x↦⎧⎪⎨⎪⎩1β−1xif x∈[1−β2,12),1β−1(x+1)if x∈[−12,β2−1).

Suppose that . Then . Moreover, is purely periodic if and only if is, and the length of the periods is the same.

###### Proof.

The transformations and are conjugated via , i.e., the following diagram commutes:

(see Figure 1). We partition into , , and . We also denote . Then we have for .

We similarly partition , as depicted in Figure 1 right. Then

 ψI¯1=J10,ψI0−=J11,ψI0+=J00,ψI1=J01,

therefore we see that if then . Finally, we see that for we have that and hence . This means that if then and also hence .

The periodicity is preserved because . ∎

###### Lemma 2.

Let be a -Bonacci number. The support of the invariant measure of is the whole domain .

###### Proof.

Denote . Put and for . Similarly, put and for , see Figure 1.

Define a measure by

 dμ(x)=f(x)dx\coloneqq(1β+1β2+⋯+1βk)dxfor x∈Y±k, 1≤k≤d.

Then we verify that for any , we have

 μ([x,x+dx))=f(x)dx=1βdx∑y∈X\ExIndST\ExIndSy=xf(y)=μ(T−1\ExIndS[x,x+dx)),

because

 (3.1) T\ExIndSY±k={Y∓1∪Y∓2∪⋯∪Y∓dif k=d,Y±(k+1)otherwise.

Therefore is the invariant measure of . ∎

###### Lemma 3.

Let be a -Bonacci number. A sequence is -admissible if and only if it contains neither nor as a factor and it does not have as a tail.

###### Proof.

We will rely on the generalized Parry condition [KS, Theorem 2.5]. We have that is -admissible if and only if for all we have

 \ExpBl (Ai)⪯xixi+1xi+2⋯(Bi)≺\Exp~Bl+12if xi=0, \ExpBl+12 (Ci)⪯xixi+1xi+2⋯(Di)≺\Exp~Bl+1if xi=1,

where and is the expansion of w.r.t. transformation defined on with digit function if and otherwise. Here we denote the lexicographic ordering on . We have that

 \ExpBl=(0d1)ω,\Exp~Bl+12=(01d)ω,\ExpBl+12=(10d)ω,\Exp~Bl+1=(1d0)ω.

Note that and .

Direction (). Suppose does not contain either of the two forbidden factors nor the forbidden tail. We need to show that conditions and are satisfied. Fix . From the absence of we know that either or it has a prefix with . Either way, is satisfied. Similarly, from the absence of and we derive that is satisfied. Therefore the sequence is -admissible.

Direction (). We know that and are satisfied by the -expansion of any . Now, forbids as a factor since any sequence starting with is lexicographically smaller than . Similarly, forbids . The forbidenness of the tail follows from the strict inequality in . ∎

###### Lemma 4.

Let be a -Bonacci number. Suppose that the balanced expansion of has the form

 \ExpBx=x1x2x3⋯xn(xn+1⋯xn+d)ω.

Then for any such that , the balanced expansion of has the form

 \ExpBx+z=y1y2y3⋯ym(ym+1⋯ym+d)ω,

where, moreover, .

###### Proof.

Clearly it is enough to consider the simplest case for some , since any is a finite sum of powers of . Then , where for , and . Let be the balanced expansion of .

Denote

where we put . Then and , and we have that ; we will denote this relation by a labelled arrow .

Consider . Then the only possible values of and possible arrows are:

 (3.2)

Consider . On one hand, we know that because , which makes unreachable. On the other hand, we have some additional arrows labelled , namely

For , we have . We have to check where the possible values lead us; we get:

 0−−→±\scalebox0.8$∙$0q−11r (1≤q,r≤d−1 and q+r≤d), ∓1−−→∓\scalebox0.8$∙$1q−10r1d−q−r+1 (1≤q,r≤d−1 and q+r≤d), ∓1−−→±\scalebox0.8$∙$1q−10r1t (1≤q,r,t≤d−1 and q+r+t≤d).

(We easily verify that no other arrows are reachable by showing that for any other pair of , where is already included in the lists above, we get that does not lie in the required intervals.)

Let us now investigate the properties of the graph of all possible arrows. A schematic view of the arrows is given in Figure 2. Note that the inner arrows that live inside each “cloud” do not form cycles, therefore sooner or later, any walk through the graph exits a cloud. Based on the solid arrows in the graph, we conclude that either eventually, or infinitely many times. If eventually, we get that , which finishes the proof. Otherwise, fix such that . Then is purely periodic, i.e., for some .

There are two cases. First, suppose . Then

 yi+1yi+2yi+3⋯=p1p2…pq1rpq+r+1⋯pd(p1⋯pd)ω.

Second, suppose . Then we can find unique with and such that

 ptpt+1⋯pq=01q−tandpq+upq+u+1⋯pq+r=10r−u

(if we had , it would be a contradiction with ). Then the new pre-period and period are

 (3.3) yi+1yi+2⋯yi+d=p1⋯pt−110q−tpq+1⋯pq+u−101r−upq+r+1⋯pd,yi+d+1yi+d+2⋯=(p1⋯pt−11pt+1\cleaders⋅pq+u−10pq+u+1\cleaders⋅pd)ω,

because this value of the sequence is -admissible and satisfies that

In either case, the sum of the elements of the period is preserved. ∎

###### Example 1.

We apply the lemma to an example , and . Then and . The computation is as follows:

(this computation follows the arrows in Figure 2 middle). For , we have that and is purely periodic. Therefore we have and and . We have ; we get and . From (3.3) we confirm that .

###### Lemma 5.

Let be a -Bonacci number. Let . Then the set contains exactly such that the balanced expansion of has the form

 (3.4) \ExpB[]|x|β−1=x1x2⋯xn(xn+1xn+2⋯xn+d)ω withxn+1+xn+2+⋯+xn+d=⎧⎨⎩hif x>0,d−1if x<0 and h=d−1,d−1−hif x<0 and 1≤h≤d−2.
###### Proof.

We start by proving that whatever we take, it satisfies (3.4). As for all , there exists such that

Since , the result follows from Lemma 4.

We finish by proving other direction. Suppose satisfies (3.4). Without the loss of generality, suppose that the length of the pre-period is a multiple of , and put . Then

Therefore . The result for follows from the fact that . ∎

###### Proof of Theorem 3.

Let . By Lemmas 1 and 5, the symmetric expansion is periodic with period . Suppose it is purely periodic. Then by Lemma 1, is also purely periodic; we denote it . Therefore, since , we have that . The fact that follows from .

On the other hand, any satisfies that and is purely periodic, therefore . ∎

###### Lemma 6.

Let be a -Bonacci number. There exists a number such that lies exactly in tiles.

Before we prove this lemma, let us recall a helpful result by C. Kalle and W. Steiner:

###### Lemma 7.

[KS, Proposition 4.15] Suppose . Let be an integer such that for all , the expansions and have a common prefix at least as long as the period of .

Then lies in a tile for if and only if

 x=Tk\ExIndS(y+β−kz)for some y∈P.
###### Proof of Lemma 6.

We put . Let us fix . Then we can write as , where

 pi={yiif y>0,1−yiif y<0

(we put ). Note that . Let

 t\coloneqqψ(y+β−d2z)=1β−1×\scalebox0.8$∙$p1p2⋯pd(0d−11)(0d−11)⋯(0d−11)d−1 times.

Defining , we get that

where