Finite beta-expansions with negative bases Zuzana Krčmáriková1,∗, Wolfgang Steiner2,†, Tomáš Vávra1,∗ 1 Department of Mathematics FNSPE 7 1 Czech Technical University in Prague 0 2 Trojanova 13, 120 00 Praha 2, Czech Republic n zuzka.krcmarikova@gmail.com, vavrato@gmail.com a J 2 IRIF, CNRS UMR 8243, Université Paris Diderot – Paris 7 7 1 Case 7014, 75205 Paris Cedex 13, France ] steiner@irif.fr T N . h t a Abstract m [ Thefiniteness property is an importantarithmetical property of beta-expansions. 1 We exhibit classes of Pisot numbers β having the negative finiteness property, that is v the set of finite (−β)-expansions is equal to Z[β 1]. For a class of numbers including − 9 theTribonaccinumber,wecomputethemaximallengthofthefractionalpartsarising 0 6 in the addition and subtraction of (−β)-integers. We also give conditions excluding 4 the negative finiteness property. 0 . 1 0 1 Introduction 7 1 : Digital expansions in real bases β > 1 were introduced by Rényi [23]. Of particular v i interest are bases β satisfying the finiteness property, or Property (F), which means that X each element of Z[β 1]∩[0,∞) has a finite (greedy) β-expansion. We know from Frougny r − a and Solomyak [13] that each base with Property (F) is a Pisot number, but the converse is not true. Partial characterizations are due to [13, 16, 1]. In [2], Akiyama et al. exhibited an intimate connection to shift radix systems (SRS), following ideas of Hollander [16]. For results on shift radix systems (with the finiteness property), we refer to the survey [18]. ∗Supported by the Czech Science Foundation, grant No. 13-03538S, and by the Grant Agency of the Czech Technical University in Prague,grant No. SGS14/205/OHK4/3T/14 †Supported by the ANR-FWF project “Fractals and Numeration” (ANR-12-IS01-0002, FWF I1136) and the ANR project “Dyna3S” (ANR-13-BS02-0003) 1 Numeration systems with negative base −β < −1, or (−β)-expansions, received con- siderable attention since the paper [17] of Ito and Sadahiro in 2009. They are given by the (−β)-transformation T : [ℓ ,ℓ +1) → [ℓ ,ℓ +1), x 7→ −βx−⌊−βx−ℓ ⌋, with ℓ = β ; β β β β β β β β−+1 − see Section 2 for details. Certain arithmetic aspects seem to be analogous to those for positive base systems [12,20], others aredifferent, e.g., both negative and positive numbers have (−β)-expansions; for β < 1+√5, the only number with finite (−β)-expansion is 0. We 2 say that β > 1 has the negative finiteness property, or Property (−F), if each element of Z[β 1] has a finite (−β)-expansion. By Dammak and Hbaib [10], we know that β must − be a Pisot number, as in the positive case. It was shown in [20] that the Pisot roots of x2−mx+n, with positive integers m,n, m ≥ n+2, satisfy the Property (−F). This gives a complete characterization for quadratic numbers, as β does not possess Property (−F) if β has a negative Galois conjugate, by [20]. First, we give other simple criteria when β doesnot satisfy Property (−F).Surprisingly, this happens when ℓ has a finite (−β)-expansion, which is somewhat opposite to the β positive case, where Property (F) implies that β is a simple Parry number. Theorem 1. If Tk (ℓ ) = 0 for some k ≥ 1, or if β is the root of a polynomial p(x) ∈ Z[x] β β with |p(−1)| = 1,−then β does not possess Property (−F). The main tool we use is a generalization of shift radix systems. We show that the (−β)-transformation is conjugated to a certain α-SRS. Then we study properties of this dynamical system. We obtain a complete characterization for cubic Pisot units. Theorem 2. Let β > 1 be a cubic Pisot unit with minimal polynomial x3 −ax2 +bx−c. Then β has Property (−F) if and only if c = 1 and −1 ≤ b < a, |a|+|b| ≥ 2. Considering Pisot numbers of arbitrary degree, we have the following results. Theorem 3. Let β > 1 be a root of xd − mxd 1 − ··· − mx − m for some positive − integers d,m. Then β has Property (−F) if and only if d ∈ {1,3,5}. Theorem 4. Let β > 1 be a root of xd − a xd 1 + a xd 2 + ··· + (−1)da ∈ Z[x] with 1 − 2 − d a ≥ 0 for i = 1,...,d, and a ≥ 2+ d a . Then β has Property (−F). i 1 i=2 i These theorems are proved in SecPtion 3. In Section 4, we give a precise bound on the number of fractional digits arising from addition and subtraction of (−β)-integers in case β > 1 is a root of x3 −mx2 −mx−m for m ≥ 1. This is based on an extension of shift radix systems. The corresponding numbers for β-integers have not been calculated yet, although they can be determined in a similar way. 2 2 (−β)-expansions For β > 1, any x ∈ [ℓ ,ℓ +1) has an expansion of the form β β ∞ x x = i with x = ⌊−βTi 1(x)−ℓ ⌋ for all i ≥ 1. (−β)i i −β β − i=1 X N This gives the infinite word d (x) = x x x ··· ∈ A with A = {0,1,...,⌊β⌋}. Since the β 1 2 3 base is negative, we can repre−sent any x ∈ R without the need of a minus sign. Indeed, let k ∈ N be minimal such that x ∈ (ℓ ,ℓ +1) and d ( x ) = x x x ···. Then the ( β)k β β β ( β)k 1 2 3 − − − (−β)-expansion of x is defined as x ···x x •x x ··· if k ≥ 1, 1 k 1 k k+1 k+2 hxi β = − − (0•x1x2x3··· if k = 0. Similarly to positive base numeration systems, the set of (−β)-integers can be defined using the notion of hxi , by β − Z = {x ∈ R : hxi = x ···x x •0ω} = (−β)kT k(0), β β 1 k 1 k −β − − − − k 0 [≥ where0ω istheinfiniterepetitionofzeros. Theset ofnumbers withfinite(−β)-expansionis Z Fin(−β) = {x ∈ R : hxi−β = x1···xk−1xk •xk+1···xk+n0ω} = (−−ββ)n . n 0 [≥ If hxi = x ···x x •x ···x 0ω with x 6= 0, then fr(x) = n denotes the length β 1 k 1 k k+1 k+n k+n of the−fractional pa−rt of x; if x ∈ Z , then fr(x) = 0. β − 3 Finiteness In this section, we discuss the Property (−F) for several classes of Pisot numbers β. Note that Fin(−β) is a subset of Z[β 1] since β is an algebraic integer, hence Property (−F) − means that Fin(−β) = Z[β 1], i.e., Fin(−β) is a ring. We start by showing that bases β − satisfying d (ℓ ) = d d ...d 0ω, which can be considered as analogs to simple Parry β β 1 2 k − numbers, do not possess Property (−F). This was conjectured in [19] and supported by the fact that d (ℓ ) = d d ...d 0ω with d ≥ d + 2 for all 2 ≤ j ≤ k implies that β β 1 2 k 1 j d (β−1−d ) =−(d +1)(d +1)···(d +1)1ω. However, the assumption d ≥ d +2 is not β 1 2 3 k 1 j − necessary for showing that Property (−F) does not hold. We also prove that a base with Property (−F) cannot be the root of a polynomial of the form a xd +a xd 1 +···+a with | d (−1)ia | = 1. 0 1 − d i=0 i P 3 Proof of Theorem 1. If Tk (ℓ ) = 0, i.e., d (ℓ ) = d d ...d 0ω, then we have β β β β 1 2 k − − −β d d d 1 2 k = + +···+ β +1 −β (−β)2 (−β)k and thus 1 ∈ Z[β 1]. However, we have 1 ∈/ Fin(−β) since T ( 1 ) = 1 , i.e., β−+1 − β−+1 β β−+1 β−+1 − d ( 1 ) = 1ω. Hence β does not possess Property (−F). β β−+1 − If p(β) = 0 with |p(−1)| = 1, then write p(x−1) = xf(x)+p(−1), with f(x) ∈ Z[x]. Then we have β+11 = |f(β + 1)| ∈ Z[β] and thus −(β−+β1)−j ∈ Z[β−1] for some j ≥ 0. Now, d β(−β(β+)1−j) = 0j1ω implies that β does not have the Property (−F). − The main tool we will be using in the rest of the paper are α-shift radix systems. An α-SRS is a dynamical system acting on Zd in the following way. For α ∈ R, r = (r0,r1,...,rd 1) ∈ Rd, and z = (z0,z1,...,zd 1) ∈ Zd, let τr,α be defined as − − τr,α(z0,z1,...,zd 1) = (z1,...,zd 1,zd), − − where z is the unique integer satisfying d 0 ≤ r z +r z +···+r z +z +α < 1. (1) 0 0 1 1 d 1 d 1 d − − Alternatively, we can say that τr,α(z0,z1,...,zd 1) = (z1,...,zd 1,−⌊rz+α⌋), − − where rz stands for the scalar product. The usefulness of α-SRS with α = 0 for the study of finiteness of β-expansions was first shown by Hollander in his thesis [16]. His approach was later formalized in [2] where the case α = 0 was extensively studied. The symmetric case with α = 1 was then studied 2 in [4]. Finally, general α-SRS were considered by Surer [24]. We say that τr,α has the finiteness property if for each z ∈ Zd there exists k ∈ N such that τrk,α(z) = 0. The finiteness property of τr,α is closely related to the Property (−F), thus it is desirable to study the set D0 = {r ∈ Rd : ∀z ∈ Zd,∃k,τk (z) = 0}. d,α r,α The following proposition shows the link between (−β)-expansions and α-SRS. Proposition 5. Let β > 1 be an algebraic integer with minimal polynomial xd+a xd 1+ 1 − ···+a x+a . Set α = β and let (r ,r ,...,r ) ∈ Rd 1 be such that d 1 d β+1 0 1 d 2 − − − xd +(−1)a xd 1 +···+(−1)da = (x+β)(xd 1 +r xd 2 +···+r x+r ), 1 − d − d 2 − 1 0 − a a i.e., ri = (−1)d−i d−i +···+ d for i = 0,1,...,d−2. β βi+1 (cid:18) (cid:19) Then β has Property (−F) if and only if (r ,r ,...,r ) ∈ D0 . 0 1 d 2 d 1,α − − 4 Proof. Let r = (r ,r ,...,r ). First we show that for φ : z 7→ rz−⌊rz+α⌋ the following 0 1 d 2 commutation diagram holds,−i.e., the systems (τr,α,Zd−1) and (T β,Z[β]∩[ℓβ,ℓβ +1)) are − conjugated. Zd 1 −−τr−,α→ Zd 1 − − φ φ Z[β]∩[ℓ,ℓ +1) −−T−−β→ Z[β]∩[ℓ,ℓ +1) yβ β yβ β Since r = (−1)d i 1(βd i 1 + a βd i 2 + ··· + a ) for 0 ≤ i ≤ d − 2, the set {r : i −− −− 1 −− d i 1 i 0 ≤ i < d} with r = 1 forms a basis of Z[β],−h−ence φ is a bijection. Moreover, we d 1 have −βr = r +−c with c ∈ Z and r = 0. For z = (z ,z ,...,z ), we have i i 1 i i 1 0 1 d 2 φ(z) = di=−01rizi−with zd 1 = −⌊rz+α⌋, thu−s − − P d 1 − T β(φ(z)) = −βφ(z)+n = ri 1zi +n′ = φ(z1,...,zd 2,zd 1) = φ(τr,α(z)), − − − − i=1 X wherenandn areintegers; forthethirdequality, wehaveusedthatT (φ(z)) ∈ [ℓ ,ℓ +1). ′ β β β Therefore, we have r ∈ D0 if and only if for each x ∈ Z[β]∩[ℓ−,ℓ +1) there exists d 1,α β β k ≥ 0 such that Tk (x) = 0. S−ince for each x ∈ Z[β 1]∩[ℓ ,ℓ +1) we have Tn (x) ∈ Z[β] β − β β β for some n ∈ N, P−roperty (−F) is equivalent to r ∈ D0 . − d 1,α − Thus the problem of finiteness of (−β)-expansions can be interpreted as the problem of finiteness of the corresponding α-SRS. This problem is often decidable by checking the finiteness of α-SRS expansions of a certain subset of Zd. A set of witnesses of r ∈ Rd is a set V ⊂ Zd that satisfies 1. ±e ∈ V where e denotes the standard basis of Rd, i i 2. if z ∈ V, then τr,0(z),−τr,0(−z) ∈ V. The following proposition is due to Surer [24] and Brunotte [7]. Proposition 6. Let α ∈ [0,1) and r ∈ Rd. Then r ∈ D0 if and only if there exists a set d,α of witnesses that does not contain nonzero periodic elements of τr,α. Setsofwitnessesforseveralclassesofr ∈ Rd werederivedin[3]. Exploitingtheirexplicit form, several regions of finiteness can be determined; see in particular [3, Theorems 3.3– 3.5]. An α-SRS analogy of some of those regions was given by Brunotte [7]. Brunotte’s result, however, is unsuitable for our purposes. The next proposition gives several regions of finiteness of α-SRS. Proposition 7. Let r = (r ,r ,...,r ) ∈ Rd and α ∈ [0,1). 0 1 d 1 − 1. If di=−01|ri| ≤ α and ri<0ri > α−1, then r ∈ Dd0,α. 2. If P0 ≤ r ≤ r ≤ ··· ≤Pr ≤ α, then r ∈ D0 . 0 1 d 1 d,α − 5 3. If di=−01|ri| ≤ α and ri < 0 for exactly one index i = d−k, then r ∈ Dd0,α if and only if P r > α−1. (2) d jk − 1 j d/k ≤X≤ Proof. 1. The set V = {−1,0,1}d is closed under τr,0(z) and −τr,0(−z), hence it is a set of witnesses. For any z ∈ V we have |rz| ≤ α, thus ⌊rz + α⌋ ∈ {0,1}. Hence any periodic point of τr,α is in {0,−1}d. For z ∈ {0,−1}d we have rz+α ≤ − r +α < 1. Therefore ⌊rz+α⌋ = 0, so the only period is the trivial one. rj<0 j 2. InPthis case we take as a set of witnesses the elements of {−1,0,1}d with alternating signs, i.e., z z ≤ 0 for any pair of indices i < j such that z = 0 for each i < k < j. i j k For any z ∈ V we have again |rz| ≤ α, thus ⌊rz + α⌋ ∈ {0,1} and τr,α(z) ∈ V. Therefore, we have τn (z) = (−1,0,...,0) for some n ≥ 0, hence τn+1(z) = 0. r,α r,α 3. In this case we have V = {−1,0,1}d. As above, all periodic points of τr,α are in {0,−1}d. If z = (z ,z ,...,z ) is a periodic point with z = −⌊rz + α⌋ = −1, 0 1 d 1 d − then we must have z = −1, and consequently z = −1 for all 1 ≤ j ≤ d/k. d k d jk − − Then z = −1 also implies that − r +α ≥ 1, i.e., (2) does not hold. On d 1 j d/k d jk ≤ ≤ − the other hand, if (2) holds, then the vector (z ,z ,...,z ) with z = −1 for 0 1 d 1 d jk P − − 1 ≤ j ≤ d/k, zi = 0 otherwise, is a periodic point of τr,α. Next we prove Property (−F) when β is a root of a polynomial with alternating coefficients, where the second highest coefficient is dominant. Proof of Theorem 4. Let β > 1 be a root of p(x) = xd−a xd 1+a xd 2+···+(−1)da ∈ 1 − 2 − d Z[x] with a ≥ 0 for i = 1,...,d, and a ≥ 2+ d a . As d (p(x)x d) ≥ a1 − a1 2 > 0 i 1 i=2 i dx − x2 x−3 for x > 1, the polynomial p(x) has a unique root β > 1, and we have β > a − 1 since 1 P p(a −1) ≤ −(a −1)d 1+(a −2)(a −1)d 2 < 0. By Proposition 5, Property (−F) holds 1 1 − 1 1 − if and only if (r ,r ,...,r ) ∈ D0 , with r = a β 1−a β 2+a β 3−···+ 0 1 d 2 d 1,α i d i − d i+1 − d i+2 − (−1)d ia β d+i 1. We hav−e − − − − − d − − a −2 a −2 a −2 a −2 1 1 1 1 1 − r ≤ + +···+ ≤ < i β2 β4 β2 d/2 2 β2 −1 β +1 ⌈ ⌉− rXi<0 6 and d 1 β +1 − β +1 a +···+a a +···+a a 2 d 3 d d |r | ≤ + +···+ i β β β β2 βd 1 i=0 (cid:18) − (cid:19) X a +···+a a +2a +···+2a a +2a +···+2a a +2a a 2 d 2 3 d 3 4 d d 1 d d = + + +···+ − + β β2 β3 βd 1 βd − a −2 2(a −2)−a 2(a −a −2)−a 2(a −a −···−a −2)−a 1 1 2 1 2 3 1 2 d 1 d ≤ + + +···+ − β β2 β3 βd 1 a −2 a −a −2 a −a −···−a −2 1 1 2 1 2 d 1 ≤ 1−2 − − −···− − β β2 β3 βd (cid:18) (cid:19) 2 a −2 1 ≤ 1− 1− < 1. β β −1 (cid:18) (cid:19) Therefore, item 1 of Proposition 7 gives that Property (−F) holds. NowwecanclassifythecubicPisot unitswithProperty(−F).Thefollowingdescription of cubic Pisot numbers in terms of the coefficients of the minimal polynomial is due to Akiyama [1, Lemma 1]. Lemma 8. A number β > 1 with minimal polynomial x3 − ax2 + bx − c is Pisot if and only if |b+1| < a+c and b+c2 < sgn(c)(1+ac). Proof of Theorem 2. Let β > 1 be a cubic Pisot unit with minimal polynomial x3−ax2+ bx − c. If c = −1, then β has a negative conjugate, which contradicts Property (−F) by [20]. Therefore, we assume in the following that c = 1. Then from Lemma 8 we have that −a−1 ≤ b < a. By Proposition 5, Property (−F) holds if and only if (r ,r ) ∈ D0 , 0 1 2,α with (r ,r ) = (1, b − 1 ) and α = β . We distinguish five cases for the value of b. 0 1 β β β2 β+1 1. b = 0: If a ≥ 2, then we have |r |+|r | = 1 + 1 < α and r +r > 0 > α −1, so 0 1 β β2 0 1 we apply item 3 of Proposition 7. If a = 1, then we have T 1(0) = {0} as β < 1+√5, −β 2 thus Fin(−β) = {0}. − 2. b = −1: If a ≥ 1, then r +r = − 1 > 1 = α −1. If a ≥ 3, then we also have 0 1 β2 β−+1 |r |+|r | < α and use item 3 of Proposition 7. If a = 2, then r ≈ 0.39, r ≈ −0.55, 0 1 0 1 α ≈ 0.72, {−1,0,1}2 is a set of witnesses, and Property (−F) holds because τr,α acts on this set in the following way: (−1,1) 7→ (1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,−1) 7→ (−1,0) 7→ (0,0), (0,1) 7→ (1,0), (1,−1) 7→ (−1,−1). For a = 1, we refer to Theorem 3, which is proved below. If a = 0, then β < 1+√5 2 and thus Fin(−β) = {0}. 7 3. 1 ≤ b ≤ a−2: For b ≥ 2, we have 0 < r < r < α and thus (r ,r ) ∈ D0 by item 2 0 1 0 1 2,α of Proposition 7. If b = 1, then we can use item 1 of Proposition 7 because r ,r > 0 0 1 and r +r < α. 0 1 4. 1 ≤ b = a−1: We have β = b+ 1 . For b ≥ 3, we have 0 < r < α < r < 1, the β(β 1) 0 1 set {−1,0,1}2 \{(1,1),(−1,−1)} is−a set of witnesses, and τr,α acts on this set by (1,0) 7→ (0,−1) 7→ (−1,1) 7→ (1,−1) 7→ (−1,0) 7→ (0,0), (0,1) 7→ (1,−1), thus Property (−F) holds. If b = 2, then 0 < r < r < α and we can use item 2 of 0 1 Proposition 7. If b = 1, then r ≈ 0.57, r ≈ 0.25, α ≈ 0.64, thus {−1,0,1}2 is a set 0 1 of witnesses, with (−1,−1) 7→ (−1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,0) 7→ (0,0), (0,1) 7→ (1,0), (1,1) 7→ (1,−1) 7→ (−1,0). 5. −a−1 ≤ b ≤ −2: We have −r0−r1+α = −bβ−1 + β12 + β+β1 > 1, thus τr,α(−1,−1) = (−1,−1), hence (r ,r ) ∈/ D0 . 0 1 2,α Therefore, β has Property (−F) if and only if −1 ≤ b < a, |a|+|b| ≥ 2. Finally, we study generalized d-bonacci numbers. Proof of Theorem 3. Let β > 1 be a root of xd −mxd 1 −···−mx−m with d,m ∈ N. − If d = 1 (and m ≥ 2), then β is an integer, and Property (−F) follows from Z = Z; β − see e.g. [20]. If d = 3, then r = (mβ,−mβ −βm2), 0 < r0 < α < −r1 < 1, with α = β+β1, and τr,α satisfies (0,1) 7→ (1,1) 7→ (1,0) 7→ (0,−1) 7→ (−1,−1) 7→ (−1,0) 7→ (0,0), with {−1,0,1}2 \{(1,−1),(−1,1)} being a set of witnesses. If d = 5, then r = (m,−m − m, m + m + m,−m − m − m − m), which gives 0 < r < β β β2 β β2 β3 β β2 β3 β4 0 α < −r1 < r2 < −r3 < 1 and the τr,α-transitions (0,1,0,0) 7→ (1,0,0,1) 7→ (0,0,1,0) 7→ (0,1,0,−1) 7→ (1,0,−1,0) 7→ (0,−1,0,0) 7→ (−1,0,0,−1)7→(0,0,−1,−1)7→(0,−1,−1,0)7→(−1,−1,0,0)7→(−1,0,0,0)7→(0,0,0,0), (0,0,−1,0) 7→ (0,−1,0,1) 7→ (−1,0,1,0) 7→ (0,1,0,−1), (0,1,1,1) 7→ (1,1,1,1) 7→ (1,1,1,0) 7→ (1,1,0,−1) 7→ (1,0,−1,−1) 7→ (0,−1,−1,−1) 7→ (−1,−1,−1,−1) 7→ (−1,−1,−1,0) 7→ (−1,−1,0,0), (0,0,0,1)7→(0,0,1,1)7→(0,1,1,0)7→(1,1,0,0)7→(1,0,0,0)7→(0,0,0,−1)7→(0,0,−1,−1), (−1,−1,0,1) 7→ (−1,0,1,1) 7→ (0,1,1,0). 8 Let V be the set of these states. We have ±e ∈ V, z ∈ V if and only if −z ∈ V and i τr,0(z) ∈ V for all z ∈ V, thus V is a set of witnesses. As τr1,1α(z) = (0,0,0,0) for all z ∈ V, β has Property (−F). Foroddd ≥ 7,Property(−F)doesnotholdsinceTd 1(m+m+m−1) = m+m+m−1, −β β2 β3 β4 β2 β3 β4 − i.e., τd 1(−1,0,0,−1,0,0,...,0) = (−1,0,0,−1,0,0,...,0). For even d ≥ 2, we use the r,−α second condition of Theorem 1, or that τr,α(−1,...,−1) = (−1,...,−1). Therefore, β has Property (−F) if and only if d ∈ {1,3,5}. 4 Addition and subtraction In this section, we consider the lengths of fractional parts arising in the addition and subtraction of (−β)-integers; we prove the following theorem. Theorem 9. Let β > 1 be a root of x3 −mβ2 −mβ −m, m ≥ 1. We have 3 if m = 1 or m is even, max{fr(x±y) : x,y ∈ Z } = 3m+ β − (4 if m ≥ 3 is odd. Throughout the section, let β be as in Theorem 9, r = (r ,r ) = (m,−m − m) and 0 1 β β β2 α = β . Recallthatx,y ∈ Z meansthatTk ( x ) = 0 = Tk ( y ), andfr(x±y) = n β+1 β β ( β)k β ( β)k is the minimal n ≥ 0 such tha−t Tk+n( x y ) =−0, w−ith k ≥ 0 suc−h th−at x , y , x y ∈ β ( ±β)k ( β)k ( β)k ( ±β)k − − − − − (ℓ ,ℓ +1). To determine fr(x−y), set β β s = Tj ( x y )+Tj ( y )−Tj ( x ) j β ( −β)k β ( β)k β ( β)k − − − − − − for j ≥ 0. Then we have s = Tj ( x y ) for j ≥ k, and, for all j ≥ 0, j β ( −β)k − − s ∈ −βs +B with B = −A−A+A = {−2m,−2m+1,...,m}, j+1 j s ∈ [ℓ ,ℓ +1)+[ℓ ,ℓ +1)−[ℓ ,ℓ +1) = (ℓ −1,ℓ +2). j β β β β β β β β As s = 0, we have s ∈ Z[β] for j ≥ 0. Therefore, we extend the bijection φ : Z2 → 0 j Z[β]∩[ℓ ,ℓ +1) to β β Φ : Z2 ×{−1,0,1} → Z[β]∩[ℓ −1,ℓ +2), (z,h) 7→ rz−⌊rz+α⌋+h. β β Note that Φ(z,0) = φ(z). Lemma 10. Let z = (z ,z ) ∈ Z2, h ∈ {−1,0,1} and b ∈ B. Then 0 1 −βΦ(z,h)+b = Φ z ,h−⌊rz+α⌋, z −z −h+⌊rz+α⌋ m+ r z +r h−r ⌊rz+α⌋+α +b . 1 1 0 0 1 1 1 (cid:0) (cid:0) (cid:1) (cid:4) (cid:5) (cid:1) 9 Proof. We have z m −βΦ(z,h)+b = −z m+z m+ 1 +⌊rz+α⌋β −hβ +b 0 1 β = r z +r h−⌊rz+α⌋ + z −z −h+⌊rz+α⌋ m+b. 0 1 1 1 0 (cid:0) (cid:1) (cid:0) (cid:1) Hence, we have sj ∈ Φ(τ˜rj,α(0,0)), where τ˜r,α extends τr,α to a set-valued function by τ˜r,α : Z2 ×{−1,0,1} → P(Z2 ×{−1,0,1}), (z,h) 7→ (z1,h−⌊rz+α⌋,h′) : h ∈ {−1,0,1}∩ z −z −h+⌊rz+α⌋ m+ r z +r h−r ⌊rz+α⌋+α +B . ′ 1 0 0 1 (cid:8) 1 1 To give a bound for(cid:0)t(cid:0)he sets τ˜j (0,0), let (cid:1) (cid:4) (cid:5) (cid:1)(cid:9) r,α A = {(j,k) : −1 ≤ j < k}, B = {(k,j) : 1 ≤ j ≤ k}, C = {(j,j −k) : 1 ≤ j ≤ k}, k k k D = {(−j,−k) : 0 ≤ j < k}, E = {(−k,−j) : 2 ≤ j ≤ k}, k k F = {(−j,k −j) : 2 ≤ j ≤ k +1}. k Then {A ,B ,C ,D ,E ,F } forms a partition of Z2 \ {(0,0),(−1,−1)}, with the k 0 k k k k k k sets B , C≥ , D , E , F , and E being empty, see Figure 1. If m ≥ 2, then let 0 0 0 0 0 1 S V = A ∪B ∪C ∪D ∪E ∪F ×{−1,0,1} \{(−1,m,1),(0,m,1)} k k k k k k (cid:18)0 k m (cid:19) ≤[≤ (cid:0) (cid:1) ∪ C \{(m+1,0)} ×{1} ∪ D ×{1} ∪ D \{(0,−m−1)} ×{0} m+1 m+1 m+1 ∪(cid:16)D(cid:0) \{(0,−m−1),(cid:1)(−1,−m(cid:17) −(cid:16)1),(−2,−m−(cid:17) 1)(cid:16)}(cid:0)×{−1} (cid:1) (cid:17) m+1 ∪(cid:0) {(0,0),(−1,−1)}∪E \{(−m−1,−m−1)}(cid:1)×{−1,0,1} m+1 (cid:16) (cid:17) ∪ (cid:0)F \{(−m−2,−1),(−(cid:1)m−1,0)} ×{−1,0} . m+1 (cid:16) (cid:17) (cid:0) (cid:1) If m = 1, then we add the point (−2,0,−1) to this set, i.e., V = (0,0),(1,1),(1,0),(0,−1),(−1,−1),(−1,0),(−2,−1) ×{−1,0,1} ∪ (−1,1),(0,1) ×{−1,0} ∪ {(−1,−2)}×{0,1} ∪ (1,−1,1),(0,−2,1),(−2,0,−1) . (cid:0)(cid:8) (cid:9) (cid:1) W(cid:0)e(cid:8)call a point z(cid:9)∈ Z2 full i(cid:1)f {(cid:0)z}×{−1,0,1} ⊂ V(cid:1). (cid:8) (cid:9) The following result is the key lemma of this section. Lemma 11. Let x,y ∈ [ℓ ,ℓ + 1) such that x − y ∈ [ℓ ,ℓ + 1). Then Tj (x − y) + β β β β β Tj (y)−Tj (x) ∈ Φ(V) for all j ≥ 0. − β β − − To prove Lemma 11, we first determine the value of ⌊rz+α⌋ for (z,h) ∈ V. 10

Finite beta-expansions with negative bases

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