Integer Semigroups Associated with Dumont-Thomas Numeration Systems

Given a primitive substitution, we define different binary operations on infinite subsets of the nonnegative integers. These binary operations are defined with the help of the Dumont-Thomas numeration system; that is, a numeration system associated with the substitution. We give conditions for these semigroups to have an identity element. We show that they are not finitely generated. These semigroups define actions on the set of positive integers. We describe the orbits of these actions. We also estimate the density of these sets as subsets of the positive integers.


Introduction
Integer semigroups have been studied in different contexts (cf.[1][2][3][4]).One of the well known binary operations is the Fibonacci multiplication, introduced by Knuth [1].This operation is related to the dynamical system obtained by the Fibonacci substitution (cf.[3]).It has been generalized in different ways to the tribonacci substitution; see [4][5][6].The author studied the usage of binary operations on the set of nonnegative integers, in order to describe the self-similar structure of the -bonacci substitutions [4] and on the socalled flipped tribonacci substitution [7].
In the present article, we introduce binary operations on infinite subsets of the nonnegative integers so that they are integer semigroups.These binary operations are associated with a substitution, in particular to the numeration system defined by the substitution, the so-called Dumont-Thomas numeration system.These numeration systems play an important role in the study of Rauzy fractals; for details see [8,9] and references within.The integer semigroups defined in the paper, in general, do not have identity elements.We explore when they have identity elements.In Theorem 3, we show that these semigroups are not finitely generated.
Moreover, we show that these semigroups define actions on the set of positive integers, and we describe the orbits of these actions.
In Section 4, we study the density of these semigroups as subsets of the positive integers.

Substitutions and Automata
A substitution on a finite alphabet A = {1, . . ., } is a map  from A to the set of finite words on A; that is, A * = ∪ ≥0 A  .This map extends to A * by concatenation by (0) = 0 and () = ()(), for all ,  ∈ A * .Let  N denote the set of one-sided infinite sequences in A. The map  is extended to A N in the obvious way.We call u ∈ A N a fixed point of  if (u) = u and periodic if there exists  > 0 so that it is fixed for   .There is a dynamical system associate to the sequence u, called substitution dynamical system; for details see [8][9][10].
The incidence matrix of the substitution  is defined as the matrix  = (  ) whose entry   is the number of occurrences of the symbol  in the word (), for 1 ≤ ,  ≤ .We say the substitution is primitive if its incidence matrix is primitive; that is, all the entries of   are positive for some  > 0. We shall assume throughout the paper that the substitutions are primitive.
For a primitive substitution there are a finite number of periodic points.So, we can assume without loss of generality that a primitive substitution has a fixed point; say u =  1  2 ⋅ ⋅ ⋅ and  1 = 1; that is, the first symbol of ( 1) is 1.Let  ⊂ A * = ∪ ≥0 A  be a set of finite words on the alphabet A = {1, . . ., }.An automaton over A, A = (, , , ) is a direct graph labelled by elements of A.  is the set of states,  ⊂  is the set of initial states,  is the set of labels, and  ⊂  ×  ×  is the set of labelled edges or transitions.If (, , ) ∈ , we say that  is a transition between  and .
The prefix automaton of the substitution  (cf.[11]) is the automaton A = (, , , ) so that one has the following.
(ii)  = Pref, the set of proper prefixes of the words (), for  ∈ A. We will denote by  the empty prefix.
(iii) (, , ) is in  if ,  are elements of A,  ∈  and  is a prefix of ().
The automaton reads words from left to right.A finite path in the automaton, A, is word in , the set of transitions: such that   =  −1 for 1 ≤  ≤  and   ∈ ; however, we usually denote the paths using only the labels, that is,   ⋅ ⋅ ⋅  0 .We say that the path   ⋅ ⋅ ⋅  0 terminates at the state  0 and we say that the path passes through the state, if there is  ∈ {0, . . ., } such that   = , in (1).See [12] for more details on automata theory.

Semigroups
We consider the following subset of positive integers: We define "⋆", a binary operation in P, in the following way: let ,  ∈ P such that This binary operation is well defined since the paths in the prefix-automaton, associated with the elements of P, terminate at the initial state, so we can concatenate them.It is clear that the operation ⋆ is associative.Hence, (P, ⋆) is a semigroup.However, the binary operation ⋆ is not always commutative; for example, in the Fibonacci case, we consider (2) = 1 and (3) = 1, so (2 ⋆ 3) = 11 =  (10) and (3 ⋆ 2) = 11 = (9), so 2 ⋆ 3 = 10 and 3 ⋆ 2 = 9.
We shall change the definition of the binary operation so that we have a semigroup with identity.Without loss of generality, we can suppose that the substitution  is of the form (1) = 12  , where   is a finite word on the alphabet A and  =  or  = 1 ⋅ ⋅ ⋅ 1; that is,  is the empty word or a word consisting only of a concatenation of the symbol 1.Let P + be the set of natural numbers  whose representation is () =   ⋅ ⋅ ⋅  1  0 , where the   ⋅ ⋅ ⋅  0 is a path in A that starts in the initial state and terminates in the state 2 and the path   ⋅ ⋅ ⋅  1 terminates at the initial state.
We can think of the set P + as the set of numbers whose representation is a path in the prefix-automaton that starts in the state 1 and terminates in the state 2, such that it passes through 1 in the step before terminating.
We define a binary operation "⬦" on P + as follows.Let ,  ∈ P + , whose representations are () =   ⋅ ⋅ ⋅  1  0 and () =   ⋅ ⋅ ⋅  1  0 ; so This binary operation is well defined since it is an element of P + .The operation ⬦ is associative.Let , ,  ∈ P + such that The semigroup (P + , ⬦) has a left identity; however, there might not be more than one left identity element.Let (1, , 2) be any transition in the prefix automaton A from the state 1 to the state 2, and  ∈ N such that () = .By definition  ∈ P + and  ⬦  = , for any  ∈ P + .In general (P + , ⬦) does not have a right identity.Proof.Let  = 1 if  ̸ = , or  = 1, if  = .We remark that  is a proper prefix of (1), so  ∈ Pref, and (1, , 2) ∈  in the prefix automaton; that is,  is a transition from the state 1 to the state 2. Since   does not have the symbol 2,  is the only transition in the prefix automaton from the state 1 to 2. The transition in the automaton A from the initial state to itself and from the state 1 to the state 2 are shown in Figure 2; in this figure we suppose that the word   does not contain the symbol 1.
Let  ∈ P + with () =   ⋅ ⋅ ⋅  1  0 .We have ( ⬦ ) =   ⋅ ⋅ ⋅  1 .Since  ∈ P + ,  0 is a transition from the state 1 to 2 and there is only one transition from state 1 to state 2, so  0 = .Hence,  ⬦  = .On the other hand,  is a left identity, as it was pointed out before.
We remark that  = 1 if and only if  is the empty word.For example, this is the situation in the -bonacci substitutions.
We remark that in the Fibonacci case, the set (P, ⋆) is not finitely generated because we have the paths 1, 1, 1, . . .that correspond to elements in P but the word  is not in P. Similarly the semigroup (P + , ⬦) is not finitely generated; considering the paths 1 ⋅ ⋅ ⋅  ⏟⏟⏟⏟⏟⏟⏟⏟⏟  1, with  ≥ 1, the numbers associated with these paths cannot be finitely generated since () is not in P + .The following theorem shows that this is a general situation.
Proof.We suppose, as before, that (1) = 12  1) and (1, , 2) is a transition in the prefix automaton.Let   ⋅ ⋅ ⋅  1 be a path in the prefix automaton such that it starts and ends at the state 1 so that   ⋅ ⋅ ⋅   does not terminate at the state 1, for 1 <  ≤ .Let  be the number in P + , such that () =   ⋅ ⋅ ⋅  1  0 , with  0 = .We consider the numbers   = (  ⋅ ⋅ ⋅  1  ⋅ ⋅ ⋅  ⏟⏟⏟⏟⏟⏟⏟⏟⏟  ), with  ≥ 1; by definition they belong to P + .These numbers cannot be finitely generated, in fact.Let us consider  1 ; if we want to write it as  1 =  ⬦  1 , then  1 has to be of the form ( 1 ) = , but this is not a representation given by the Dumont-Thomas numeration system, since the first symbol is .There is no other possible "factorization" of  1 , since (  ⋅ ⋅ ⋅   ) ∉ P + for 1 ≤  ≤ .Therefore,  1 cannot be written as product of P + , different from itself, and the identities.Similar argument holds for   for  > 1.Therefore, (P + , ⬦) is not finitely generated.
The proof that (P, ⋆) is not finitely generated follows in a similar manner.
We generalize the classical concepts of prime and composite numbers into this context.We say that an element  of P + is P + -composite if there exist ,   ∈ P + different of an identity element such that  =  ⬦   .We say that  is P + -prime if it is not composite.In the proof of Theorem 3, we have shown that the numbers   are P + -primes.

Corollary 4.
The number  ∈ P + is P + -prime if and only if () =   ⋅ ⋅ ⋅  0 where   ⋅ ⋅ ⋅  0 is a path in the automaton A with the property that   ⋅ ⋅ ⋅   does not terminate at the state 1 for 1 <  < .
Proof.Let  ∈ P + whose Dumont-Thomas representation is () =   ⋅ ⋅ ⋅  0 where   ⋅ ⋅ ⋅  0 is a path in A such that   ⋅ ⋅ ⋅   does not terminate at the state 1 for 1 <  < .So the numbers   given by (  ) =   ⋅ ⋅ ⋅   are not in P. Since the representation is unique, there are not ,   ∈ P + \ {} such that  =  ⬦   .Hence  is P + -prime.
We point out that the semigroup P + acts on the set of the positive integers.Let  ∈ P + , and  ∈ N; we suppose that  ∉ P + .If () =   ⋅ ⋅ ⋅  1  0 and () =   ⋅ ⋅ ⋅  0 , where   ⋅ ⋅ ⋅  0 is a path in the prefix automaton, A, that terminates at the state , then   ⋅ ⋅ ⋅  1   ⋅ ⋅ ⋅  0 is a path in A, which starts at the initial state and terminates at the state .So  ⬦  is defined and ( ⬦ ) =   ⋅ ⋅ ⋅  1   ⋅ ⋅ ⋅  0 ; however,  ⬦  is not in P + , if  ̸ = 2. Let   : N → N be the map defined by   () :=  ⬦ .This map defines a left action of P + on N, since   is the identity, where  is a left identity of P + , and The dynamics of this action can be studied.Let  ∈ N; the orbit of  under P + is the set {  () :  ∈ P + }, which is denoted by P + ⬦ .Similarly we define the orbit of  under (P, ⋆).The following proposition shows how the orbits of both actions are related.Proposition 5. Let  be a positive integer.Then P + ⬦  = (P ⋆ ) ∪ {}.
A further step in the study of the dynamics of this action is the characterization of its invariant sets.The set  ⊂ N is (P + , ⬦)-invariant if ∪ ∈ P + ⬦  = .

Density of the Semigroups
In this section, we will study the density of the sets P and P + in the set of positive integers.Let  be the Perron-Frobenius eigenvalue of  and  and  the right and left nonnegative eigenvectors associated with ; that is,  = ,    =   .If we choose them so that   V = ∑  =1     = 1, then the matrix   is called the Perron projection of  (cf.[20,21]).

Proposition 6. An estimation of the density of P in the set of positive integers is given by
where  = ( 1 , . . .,   ) is the left positive eigenvector, associated with Perron-Frobenius eigenvalue, of the transition matrix .
Proof.Let   () be the (, ) entry of the matrix   .The number of paths of length  in A that start in the state  and end in the state  is given by   ().
Let () be the number of paths of length  in the automaton A that start at the initial state; that is, () = ∑  =1  1 ().By the Perron-Frobenius Theorem for primitive matrices lim  → ∞    − = V  (cf.[21]).So the asymptotical growth of () is given by lim  → ∞ () − = ∑  =1 (  ) 1 .Let () be the number of paths of length  in A, that start in the state 1 and terminate in the state 1, so () =  11 ().So its asymptotical growth is given by lim  → ∞ () − = (  ) 11 .Therefore This expression simplifies in the following way: We prove in a similar manner the corresponding result for P + .
Example 8.In the tribonacci case, the transition matrix is The Perron-Frobenius eigenvalue is , the real root of the polynomial  3 −  2 −  − 1.The left eigenvector of  is (,  2 ,  3 ), where  =  −1 and the right eigenvector of  is ( 2 ,  3 +  4 ,  3 ), without normalization.According to Proposition 6 and Corollary 7, the upper bound of the density of P is  ≈ 0.5436 and that of P + is  2 ≈ 0.2955.

Figure 2 :
Figure 2: Part of the automaton described in the proof of Proposition 2.

Proposition 2 .
Let  be a primitive substitution on A such that (1) = 12  where  = 1 ⋅ ⋅ ⋅ 1 or  =  and   is a finite word on the alphabet.If the symbol 2 does not belong to the word   , then the identity of the semigroup (P + , ⬦) is the element  of P + whose representation is () = , with  = 1 if  ̸ = , or  = 1, if  = .