On the Delone property of ( β )-integers

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1 On the Delone property of -integers Wolfgang Steiner LIAFA, CNRS, Université Paris Diderot Paris 7, Case 7014, 7505 Paris Cedex 13, France The-integers are natural generalisations of the -integers, and thus of the integers, for negative real bases. They can be described by infinite words which are fixed points of anti-morphisms. We show that they are not necessarily uniformly discrete and relatively dense in the real numbers. 1 Introduction We study the set of -integers for a real number > 1. This set is defined by Z = n T n 0, n 0 where T is the -transformation, defined by Ito and Sadahiro [5] as T : [ +1, 1 +1, x x +1 x. Equivalently, a-integer is a real number of the form n 1 d, with =0 m d m < =0 for all 1 m n, where d 0,d 1,...,d 1 are integers. Examples of -transformations are depicted in Figure 1. Recall that the set of -integers is defined by Z =Z + Z+ with Z+ = n T n 0, n 0 where T is the -transformation, T : [0,1 [0,1, x x x. These sets were introduced in the domain of quasicrystallography, see e.g. []. It is not difficult to see thatz =Z when Z, and thatz ={0} when < For 1+ 5, Ambrož et al. [1] showed thatz can be described by the fixed point of an anti-morphism on a possibly infinite alphabet. They also calculated explicitely the set of distances between consecutive-integers when T n +1 0 and T n for all n 1. It seems to be difficult to extend their methods to the general case. For the case when is an Yrrap number, i.e., when { T n } +1 n 0 is a finite set, a different approach can be found in [8]. The approach in Section resembles that in [8], but it is simpler and wors for general. In Section 3, we discuss the Delone property for sets Z. P. Ambrož, Š. Holub and Z. Masáová Eds.: 8th International Conference WORDS 011 EPTCS 63, 011, pp , doi:10.404/eptcs This wor is licensed under the Creative Commons Attribution License.

2 48 On the Delone property of -integers Fixed point of an anti-morphism By Lemma 1, we can consider the set of -integers, > 1, as a special instance of the preimage of a point in [ +1, 1 +1 of the map ι : R [ +1, 1 +1, x T n n x, with n 0 such that n x +1, Since T 1 x = x for all x +1, 1 +1, the map ι is well defined. Lemma 1. For any > 1, we have Z = ι 1 0. Proof. If x ι 1 0, then T n n x = 0 for some n 0, thus x n T n 0, i.e., x Z. On the other hand, x Z implies that x n T n 0 for some n 0. If x n +1, 1 +1, this immediately implies that ι x=0. If n x= +1, then T n+ thus ι x= 0 as well. n+ x = T n+ 1/ +1 = T n+1 +1 = T n+1 n x = T 0= 0, Note that ι x=x for all x +1, For other x, we use the following relation. Lemma. For any > 1, x R, we have ι x=t ι x. Proof. Let x Rwith n x +1, 1 +1, n 0. Then we have ι x=t n+1 n+1 x = T T n n x = T ι x. An important role in the study of the -transformation is played by the orbit of the left endpoint. In the following, fix > 1, and let +1 t n = T n +1 n 0, a n = t 0 t n 1 n 1. As usual, x denotes the largest integer x, and x denotes the smallest integer x. Then a 1 a is the -expansion of +1, i.e., t n = a n+ for all n 0, see [5]. Setting t 1 = 1 +1, t = 0, N ={0,1,,...} {}, we consider open intervals J i, j =t i,t j 1 with i, j N, 0 t i < t j 1 or t i < t j 1 0 where =and 1=. We also set a 0 = a = 0, and A ={i, j i, j N, 0 t i < t j 1 or t i < t j 1 0}.

3 W. Steiner 49 Here, i, j is a pair of elements inn, and not an open interval. Let be the length of the interval J i, j, and set L i, j = t j 1 t i i, j A L v 1 v =L v 1 + +L v, v 1 v =, for any word v 1 v A, where A denotes the free monoid over A. Let ψ : A A be an anti-morphism, which is defined on i, j A by ψ i, j = j,i+1 if ai+1 = a j, t i+1 t j 0, and otherwise by j,,00, a i+1 a j,i+1 if t i+1 > 0, t j < 0, j,0 0,,0 a i+1 a j 1 0,,i+1 if ti+1 > 0, t j 0, j,0 0,,0 a i+1 a j 1 0,i+1 if t0 < t i+1 0, t j 0, ψ i, j = j,,0 0,,0 a i+1 a j 1 0,i+1 if t0 < t i+1 0, t j < 0, j,0 0,,0 a i+1 a j 1 j,,0 0,,0 a i+1 a j 1 if t i+1 = t 0, t j 0, if t i+1 = t 0, t j < 0. Here, anti-morphism means that ψ vw = ψ wψ v for all v,w A. The anti-morphism ψ is naturally extended to infinite words over A. Right infinite words are mapped to left infinite words and vice versa. Lemma 3. Let > 1. For any u A,we have L ψ u = L u. Moreover, for any 1 l ψ u, 0<x<L v l, with ψ u=v 1 v ψ u, we have T t j 1 1 L v1 v l 1 1 x = t i + x, where u=i, j, v l =i, j. Proof. This follows from the definitions of T and ψ. Let u 1 u 0 u 1 A Z be the fixed point of ψ such that u 0 =,0, u 0 u 1 is a fixed point of ψ and u u 1 = ψ u 0 u 1, in particular u 1 =0,. Let Y ={y Z} with y = { L u 0 u 1 if 0, L u u 1 if <0. Proposition 1. Let > 1. On every interval y,y +1, Z, the map ι is a translation, with ι y,y +1 = J u.

4 50 On the Delone property of -integers Proof. We have ι x = x on the intervals y 0,y 1 = t,t 1 = 0, 1 +1 and y 1,y 0 = t0,t = +1,0, thus the statement of the proposition holds for y,y +1 +1, Assume that the statement holds for Z. By Lemma 3, we have y,y +1 = y,y + ψ u and ψ u = u u + ψ u 1, with = ψ u 0 u if 0, = ψ u +1 u 1 if < 0. Then Lemma, the assumption ι y,y +1 = J u, and Lemma 3 yield that ι y +l,y +l+1 = T ι 1 y +l,y +l+1 = T ι y+1 1 L u u +l, y +1 1 L u u +l 1 = T t j 1 1 L u u +l, t j 1 1 L u u +l 1 = J u +l for all 0 l< ψ u, where u =i, j. By induction on n, we obtain for every n 0 that the statement of the proposition holds for all Z with y,y +1 n +1, 1 +1, thus it holds for all Z. Now we describe the set Y ={y Z}, which is left out by the intervals y,y +1. Lemma 4. For any > 1, we have Proof. First note that m,n 0 Y = Z m+n T n m,n 0 +1 m+n T n +1. = m ι 1 m 0 +1, similarly to Lemma 1. Indeed, x ι 1 t 0 is equivalent to n x T n In the remaining case n x= t 0 T n t 0, we have x ι 1 t 0 since T n+1 T n t 0=t 0. Note also that m ι 1 t 0 ι 1 t m. t 0 t 0,t 1 for some n 0. n+ x = Since t 0 J u and 0 J u for all Z by the definition of A, Proposition 1 implies that ι 1 0 ι 1 t 0 Y, thusz Y by Lemma 1. SinceY Y by Lemma 3, we obtain m ι 1 t 0 Y for every m 0 as well. Let now x Y \{0}. Then there exists some m 0 such that m x Y and m 1 x Y, i.e., m 1 x y,y +1 for some Z. By Lemma and the definition of ψ, we obtain that ι m x {t 0,0}, i.e., x m ι 1 t 0 or x m Z Z. Since 0 Z, this proves the lemma. Theorem 1. Let > 1, x R. Then x Z if and only if x=y for some Z with t j 1 = 0 or t i = 0, where u 1 =i, j, u =i, j. Proof. SinceZ Y by Lemma 4, it is sufficient to consider x=y, Z. As in the proof of Lemma 4, let m 0 be such that m x Y and m 1 x Y. Then we have ι m x {t 0,0}. If ι m x = 0, then ι x=t m 0=0. Moreover, ι is continuous at m x in this case. Together with the continuity of T m at 0, this implies that ι is continuous at x, i.e., u 1 =i, j with t j 1 = 0 and u =i, j with t i = 0. If ι m x = t 0, then ι is right continuous at m x, and its limit from the left is t 1. We obtain that ι x = t m, u 1 = i, m/ and u = m/, j for some i, j N. Let i = m/,

5 W. Steiner 51 j = m/. Since i = m if m is even and j 1 = m if m is odd, x Z implies that t i = 0 or t j 1 = 0. On the other hand, t i = 0 or t j 1 = 0 yields that t m 1 = 0 or t m = 0. Since t m 1 = 0 implies t m = 0, we must have x Z. By Theorem 1, the study ofz is reduced to the study of the fixed point of ψ. Note that i, j and i, j can be identified when t i = t i and t j 1 = t j 1. After identification, A is finite if and only if {t n n 0} is a finite set, i.e., if is an Yrrap number. With the help of ψ, we can construct an anti-morphism describing the structure of Z for 1+ 5, similarly to [8]. First note that 1+ 5 implies a 1 = 1, t 1 0, or a 1, thus ψ,0 = ψ 0, starts with,00,. Therefore, L,00, = t 1 t 0 = 1 is the smallest positive element of Z. Using Theorem 1 and Z Z, the word ψ,00, determines the set Z [,0]. Splitting up ψ,00, according to Theorem 1 and applying ψ on each of the factors yields the set Z [0, ], etc. Consider all these factors, i.e., all words u u 1 between consecutive elements y,y Z, as letters. Using the described strategy, we define an anti-morphism ϕ on words consisting of these letters. Then the fixed point of ϕ codes the distances between the elements of Z, see the two simple examples below. For more complicated examples, we refer to [8]. By [1, 8], the alphabet is finite if and only if is an Yrrap number. 1/ t 1 0 t t 1 1/ 1/ 1/ 3 0 1/ t 0 t 0 t 1 t t 1 Figure 1: The-transformation for = 1+ 5 left and = 3+ 5 right. Example 1. Let = , i.e., = 3 1. Then we have t 1 = +1 = 1/ +1 and t = =t 0. Therefore, we can identify 0, and 1,, and obtain ψ :,0 0,, 0,,00,,00,1, 0,1 0,,00,1. The two-sided fixed point u u 1 u 0 u 1 of ψ is equal to 0,,00,10,,00,,00,10,,00,,00,1. Applying ψ to,00, and to the factors described by Theorem 1 yields,00,,00,,00,10,,,00,10,,00,,00,10,,00,10,.

6 5 On the Delone property of -integers Therefore, setting A=,00, and B=,00,10,, the fixed point ABBABABBABBAB ABBABABBABBAB of the anti-morphism ϕ : A AB, B ABB, describes the set of-integers, see Figure. The distances between consecutive elements ofz are L A=1 and L B= 1>1. A B B A B A B B A B B A B A B B A B Figure : The set Z [ 3, ], =3+ 5/. Example. Let = , i.e., = + 1. Then we have t 1 = +1 1=0= t. Identifying 0, and 0, 1, we obtain ψ :,0 0,, 0,,00,. The two-sided fixed point u u 1 u 0 u 1 of ψ is equal to,00,0,,00,,00,0,. The words u u 1 between consecutive elements y,y Z are A=,00, and B=0,, since,00,,00, 0,, 0,,00,. Note that B does not start with a letter i, j with t i = 0, thus ι is discontinuous at the corresponding points y Z. The fixed point AABAABABAABAB AABAABABAABAABABAABAB of the anti-morphism ϕ : A AB, B A, describes the set of -integers, with L A = 1 and L B = 1 < 1, see Figure 3. Note that n can also be represented as n+ + n+1. A A B A B A A B A A B A B Figure 3: The-integers in[ 3, 4 ], =1+ 5/.

7 W. Steiner 53 3 Delone property A set S R is called Delone set or Delaunay set if it is uniformly discrete and relatively dense; i.e., if there are numbers R>r > 0, such that each interval of length r contains at most one point of S, and every interval of length R contains at least one point of S. If is an Yrrap number, then the set of distances between consecutive -integers is finite by [1, 8], thus Z is a Delone set. For general, we show in this section that Z need neither be uniformly discrete nor relatively dense. 3.1 Uniform discreteness Since every point x Y is separated from Y \{x} by an interval around x, Y is discrete, and the same holds forz. It is well nown thatz is uniformly discrete if and only if 0 is not an accumulation point of {T n1 n 0} where T 1=. For -integers, the situation is more complicated. Proposition. Let > 1. If 0 is not an accumulation point of { T n 1 Z is uniformly discrete. +1 > 0 n 1 }, then the set Proof. When is an Yrrap number, then {L u Z} is finite, thus Z is uniformly discrete. Therefore, assume that is not an Yrrap number, in particular that t n 0 for all finite n 0, with the notation of Section. Then ι is continuous at every point y Z, Z, thus u =i, j with t i = 0. Since L u inf{t n 1 > 0 n 0}, the setz is uniformly discrete if inf{t n 1 > 0 n 1}>0. In order to give examples of where Z is not uniformly discrete or not relatively dense, we need to now which sequences are possible-expansions of +1. The corresponding problem for - expansions was solved by Parry [7]. Góra [4, Theorem 5] gave an answer to a more general question, but his theorem is incorrect, as noticed in [3]. However, Góra proved the following result, where alt denotes the alternate order on words, i.e., x 1 x < alt y 1 y if 1 n x n+1 y n+1 <0, x 1 x n = y 1 y n for some n 0. Lemma 5. Let a 1 a be a sequence of non-negative integers satisfying a n+1 a n+ alt a 1 a for all n 1, with a 1. Then there exists a unique a 1 a such that a = + 1, and + 1 a n for all n 1. Proof. As the generalised -transformations in [4] with E =1,...,1 are intimately related to our T see the introduction of [6], the lemma follows from the proof of Theorem 5 in [4]. The flaw in [4, Theorem 5] is that for a 1 a to be the -expansion of a n+ = 1 +1 a + = 1 3 cannot be excluded, which would be necessary +1. Indeed, the sequence a 1a = 10 ω satisfies the ; the -expansion of 3 is ω. In order conditions of Lemma 5 and yields =, but to avoid this problem, we define a relation < alt by x 1 x < alt y 1y if 1 n x n+1 y n+1 < 1, x 1 x n = y 1 y n for some n 0. Lemma 6. Let a 1 a and be as in Lemma 5. If a n+1 a n+ < alt a 1a for all n such that a n = 0, then a 1 a is the -expansion of +1.

8 54 On the Delone property of -integers Proof. We have to show that a n 1+ < 1 +1 for all n. If a n > 0, then this inequality follows from a n < 0. If a n= 0, then we have some m 0 such that 1 m a n+m+1 a m+1, a n+1 a n+m = a 1 a m. This implies that thus a m a n+ = + 1 m m a n 1+ = 1 a n+m+ +1 a + a m+1 + m+1 m+1 > a n+ < m+1 a + +1 m+1 a = + 1, Proposition 3. Let a 1 a = Then a 1 a is the-expansion of +1 for some > 1, and Z is not uniformly discrete. +1. Proof. By Lemmas 5 and 6, there exists a > 1 such that a 1 a is the -expansion of Since a n a n+1 starts with an odd number of zeros for all n 1, we have t n 1 > 0 for all n 0. Therefore, induction on n yields that ψ n,0 ends with,n, and ψ n+1,0 starts with n, for all n 0. This implies that ψ n+1,00, contains the factor,nn, for all n 0, and L,nn, = tn 1 t n is a distance between consecutive -integers. For any 1, we have a 1+ a +1+1 = 0 1 1, thus 0< t 1+1 < +1 and 3 +1 < t 1+< 0. Therefore, the distance between consecutive elements of Z can be arbitrarily small. The following proposition shows that the converse of Proposition is not true. Proposition 4. Let a 1 a = be a fixed point of the morphism σ 1 : ,, Then a 1 a is the -expansion of uniformly discrete. +1 for a > 1, inf{ T n 1 +1 > 0 n 1 } = 0 and Z is Proof. When a n+1 = 3, n 1, then a n+1 a n+ starts with σ1 3 for some 0. Since σ 1 3 is odd for all 0, we have σ1 3 < alt σ 1 30, thus a n+1a n+ < alt a 1a. Since a n = 0 implies a n+1 {0,3}, the conditions of Lemmas 5 and 6 are satisfied, i.e., there exists a > 1 such that a 1 a is the-expansion of +1. Let n = σ σ1 3 +1, 1. Then a n +1 a n + = 0 1 3, thus we have t n > 0 and lim t n = 0. Since n is odd, this yields that inf{t n 1 > 0 n 1}=0. Let n 1 with t n 1 > 0. Then a n a n+1 starts with an odd number of zeros, thus a 1 a n 1 ends with σ1 30 j 1 for some 0, 1 j 1. Let m = σ1 3 +j 1. Recall that u 1u 0 u 1 is a fixed point of ψ. Any letter u = i,n, i N, occurs only in ψ m i,n m, i N with t 0 t i < t n m 1. Since a 1 a m = a n m a n 1, we obtain that t m t i < t n 1. Moreover, a m+1 a m+ starts with 0 j+1 3, thus t m > 0. Now, the continuity of ι at every point inz yields that y Z if u =i,n, n 1. Therefore, y Z implies that u =,0, hence Z is uniformly discrete.

9 W. Steiner Relative denseness Since the distance between consecutive -integers is at most 1, Z is always relatively dense. We show that this is not always true for Z. Proposition 5. Let a 1 a = be a fixed point of the morphism σ : 3 313,, 1 1. Then a 1 a is the -expansion of +1 for a > 1, and Z is not relatively dense. Proof. Since σ 3 is odd for all 0, we have σ 3< alt σ 31, thus a 1a satisfies the conditions of Lemma 5. Since a n > 0 for all n, the condition of Lemma 6 holds trivially. Therefore, there exists a > 1 such that a 1 a is the -expansion of Next we show that, for any 0, ψ σ 3 σ 3 /, σ 1 3 / σ3 / 1, σ3 / 0 0,. 1 We have ψ,0 =0, and +1.,0 starts with ψ 4 0, = ψ 3 0,1 = ψ 1, = ψ,00, =,10,, where we have used that a 1 > a, a 4 = a 1, and a n > 0 for all n 1, which implies t n < 0 for all n 1. This yields the statement for =0and = 1. Supose that ψ σ 3,0 starts with 1 for some 1. Then we have ψ +1 σ 3 σ,0 = ψ 3 +3 σ 3 /, σ 1 0, σ 3 / = ψ σ 3 + = ψ σ 3 +1 = ψ σ 3 3 / σ 3 /,,0 0, σ 3 / + 1 = σ 3 /, σ 3 / σ 3 /, σ 1 3 / σ 1 3 /, σ 0 3 / 0,, where we have used the relations a σ 3 = > 1 = a σ , a 1 > a σ 3 +1, and a 1 a σ 3 = a σ 3 +3 a σ 3 +. Since σ 3 / = σ 3 +1 = σ +1 3 /, we obtain inductively that ψ σ 3,0 starts with 1 for all 0. For any 0, we have a σ 3, a σ 3 +1 = 1, a σ 3 + =, and t < 0. For any 1, this yields that L σ 3 /, σ 1 3 / = t σ 1 3 t σ 3 1 > t > 1. Let = ψ σ 3. Then we have u + j = σ j 3 /, j 1 σ 3 / for 0 j <, thus y,y + Z = /0 and y + y > /. Since 1 was chosen arbitrary, the distances between consecutive -integers are unbounded.

10 56 On the Delone property of -integers Many other examples of sets Z which are not relatively dense can be found by setting a 1 a = σ 3 σ 3 with a morphism σ 3 such that σ 3 =, σ 3 1=1, and σ 3 3 is a suitable word of odd length. We conclude the paper by stating the following three open problems, for which partial solutions are given in this section. Note that all the corresponding problems for positive bases have well-nown, simple solutions, as mentioned above. 1. Characterise the sequences a 1 a which are possible -expansions of +1.. Characterise the numbers > 1 such that Z is uniformly discrete. 3. Characterise the numbers > 1 such that Z is relatively dense. References [1] P. Ambrož, D. Dombe, Z. Masáová & E. Pelantová: Numbers with integer expansion in the numeration system with negative base. ArXiv: v3 [math.nt]. [] Č. Burdí, C. Frougny, J. P. Gazeau & R. Krejcar 1998: Beta-integers as natural counting systems for quasicrystals. J. Phys. A 3130, pp , doi: / /31/30/011. [3] D. Dombe, Z. Masáová & E. Pelantová: Number representation using generalized -transformation. ArXiv: v1 [cs.dm]. [4] P. Góra 007: Invariant densities for generalized -maps. Ergodic Theory Dynam. Systems 75, pp , doi: /s [5] S. Ito & T. Sadahiro 009: Beta-expansions with negative bases. Integers 9, pp. A, 39 59, doi: /INTEG [6] L. Liao & W. Steiner: Dynamical properties of the negative beta-transformation. ArXiv: v [math.ds]. [7] W. Parry 1960: On the -expansions of real numbers. Acta Math. Acad. Sci. Hungar. 11, pp , doi: /bf [8] W. Steiner: On the structure of -integers. ArXiv: v [math.nt].