A Rauzy fractal unbounded in all directions of the plane

We construct an Arnoux-Rauzy word for which the set of all differences of two abelianized factors is equal to Z. In particular, the imbalance of this word is infinite and its Rauzy fractal is unbounded in all directions of the plane.


Introduction (short English version)
Until 2000, it was believed that, as for Sturmian words, the imbalance of Arnoux-Rauzy words was bounded -or at least finite. Cassaigne, Ferenczi and Zamboni disproved this conjecture by constructing an Arnoux-Rauzy word with infinite imbalance, i.e. a word whose broken line deviates regularly and further and further from its average direction [4]. Today, we know virtually nothing about the geometrical and topological properties of these unbalanced Rauzy fractals. The Oseledets theorem suggests that these fractals are contained in a strip of the plane: indeed, if the Lyapunov exponents of the matricial product associated with the word exist, one of these exponents at least is nonpositive since their sum equals zero. This article aims at disproving this belief.

Theorem 1.
There exists an Arnoux-Rauzy word whose Rauzy fractal is unbounded in all directions of the plane.
Theorem 1 also holds, on one hand, for C-adic words, which are the infinite words over {1, 2, 3} associated with the Cassaigne-Selmer multidimensional continued fraction algorithm introduced in [5] and, on the other hand, for strict episturmian words, which are the generalization of Arnoux-Rauzy words. We recall that a strict episturmian word is a word whose language is close by mirror and which admits, for each length, a unique right-special factor -which is moreover prolonged by each letter in the alphabet.
Theorem 1'. There exists w ∞ a C-adic word whose Rauzy fractal is unbounded is all directions of the plane.
There exists w ∞ a strict episturmian word over the alphabet {1, ..., d} such that for any hyperplane H in R d , the distance between H and the broken line (ab(p n (w ∞ ))) n∈N is unbounded.
The proofs of Theorems 1' and 1" are based on techniques similar to those of Theorem 1; they can be found in [1].
Besides, we propose an elementary proof of: The vector of letter frequencies of an Arnoux-Rauzy word has rationally independent entries.
This theorem completes the works of Arnoux and Starosta, who conjectured it in 2013, to prove that the Arnoux-Rauzy continued fraction algorithm detects all kind of rational dependencies [3]. Note that it has been recently proved by Dynnikov, Hubert and Skripchenko using quadratic forms [6].
Again, with a similar proof (see [1]), this result holds in arbitrary dimension: The vector of letter frequencies of a strict episturmian word over {1, ..., d} has rationally independent entries.

Preliminaries
We denote by A * the set of all finite words over an alphabet A.
where u[k] denotes the (k + 1)-th letter of u, is a factor of length n of a (finite or infinite) word w if there exists a nonnegative integer i such that for all k ∈ {0, ..., n − 1}, w[i + k] = u[k]; in the particular case i = 0, we say that u is the prefix of length n of w, and denote it by u = p n (w). We denote by F n (w) the set of factors of w of length n and by F(w) its set of factors of all lengths. A substitution is an application mapping letters to finite words: A → A * , that we extend into a morphism on the free monoid for the concatenation operation A * on one hand, and on the set of infinite words A N on the other hand. Three substitutions will be of high interest in this paper: σ 1 , σ 2 and σ 3 defined over A = {1, 2, 3} by: They are called Arnoux-Rauzy substitutions; we denote AR = {σ 1 , σ 2 , σ 3 }. The set AR can be seen as a three letter alphabet -it should not be confused with A = {1, 2, 3} over which the substitutions are defined. As much as we can, we refer to the elements of AR * or AR N as "sequences" instead of "words"; nonetheless, some tools like the notions of factor and prefix will turn out to be useful for this second alphabet as well, especially in Section 4.
The set A N of infinite words over A is endowed with the distance δ: If (s n ) n∈N ∈ AR N is a sequence containing infinitely many occurrences of each Arnoux-Rauzy substitution σ 1 , σ 2 and σ 3 , then the sequence of finite words (s 0 •...•s n−1 (α)), with α ∈ A, converges to an infinite word w 0 which does not depend on α. The infinite words w 0 obtained this way are called standard Arnoux-Rauzy words. An infinite word w is an Arnoux-Rauzy word if it has the same set of factors than a standard Arnoux-Rauzy word w 0 . One can show that the standard Arnoux-Rauzy word w 0 and the directive sequence (s n ) n∈N associated with w are unique. This definition of Arnoux-Rauzy words is equivalent to the more usual one: an infinite word is an Arnoux-Rauzy word if it has complexity 2n + 1 and admits exactly one right and one left special factor of each length.
Given a finite word u ∈ A * and a letter α ∈ A, we denote by |u| α the number of occurrences of α in u. The abelianized vector of u, sometimes called Parikh vector of u, is the vector ab(u) = (|u| α ) α∈A , which counts the number of times that each letter occurs in the finite word u. At this point, it is useful to order the alphabet. For the convenience of typing, we choose to represent abelianized words as line vectors. Observe that the sum of the entries of ab(u) is equal to the length of the word u, that we denote by |u|. Now, given a substitution s : A → A * , the incidence matrix of s is the matrix M s whose i − th row is the abelianized of the image by s of the i − th letter in the alphabet. Abelianized words and incidence matrices are made to satisfy: ab(s(u)) = ab(u)M s for any substitution s : A → A * and any finite word u ∈ A * .
If w ∈ A N is an infinite word and α ∈ A is a letter, the frequency of α in w is the limit, if it exists, of the proportion of α in the sequence of growing prefixes of w: f w (α) = lim n→∞ |pn(w)|α n . We denote by f w = (f w (α)) α∈A the vector of letter frequencies of w, if it exists. When the vector of letter frequencies exists, as it is the case for any Arnoux-Rauzy word, it is natural to study the difference between the predicted frequencies of letters and their observed occurrences. Given an infinite word w ∈ A N for which the vector of letter frequencies is defined, we consider the discrepancy function: The discrepancy is linked to a combinatorial property: the imbalance. The imbalance of an infinite word w is the quantity (possibly infinite) : The imbalance of an infinite word w is finite if and only if its discrepancy function is bounded. Geometrically, the discrepancy is linked to the diameter of the Rauzy fractal. Let ∆ 0 denotes the plane of R 3 with equation x + y + z = 0. For w an Arnoux-Rauzy word, denote by f w its letter frequencies vector and by π w the (oblique) projection onto ∆ 0 associated with the direct sum: Rf w ⊕ ∆ 0 = R 3 . The Rauzy fractal of w, denoted by R w , is the closure of the image of the set of abelianized prefixes of w (the broken line of w) by the projection π w : R w = ∪ k∈N {π w (ab(p k (w)))} ⊂ ∆ 0 . Note that the statement of our main result (Theorem 1) does not depend on the choice of the plane we project onto.
Remark 1 (Abuse of notation). If s = s 0 · ... · s n−1 ∈ AR * , and if w ∈ A * ∪ A N , then s(w) denotes the image of the word w by the substitution s 0 • ... • s n−1 .
Proof. Section 5 is devoted to the proof of Lemma 1.
Therefore, all standard Arnoux-Rauzy words -and thereby all Arnoux-Rauzy words-whose directive sequence starts with the prefix s will admit (a, b, c) as difference of abelianized factors.
Proof. Let p ∈ AR * and (a, b, c) ∈ Z 3 . Denote by M p the incidence matrix of the substitution associated with p (following Remark 1), which is a product of the Arnoux-Rauzy matrices M σ 1 , M σ 2 and M σ 3 , and thus belongs to GL 3 (Z). By Lemma 1, there exists s ∈ AR * and there exist u and v ∈ F(s(1)) such that ab(u)−ab(v) = (a, b, c) M −1 p . But then, p(u) and p(v) are factors of F(p·s(1)) and satisfy ab(p(u)) − ab(p(v)) = (ab(u) − ab(v))M p = (a, b, c).
We now construct a standard Arnoux-Rauzy word for which all triplets of integers can be obtained as a difference of two of its abelianized factors. Proposition 1. There exists an Arnoux-Rauzy word w ∞ such that for all (a, b, c) ∈ Z 3 , there exist u and v ∈ F(w ∞ ) satisfying ab(u) − ab(v) = (a, b, c).
Proof. Let ϕ : N → Z 3 a bijection (that can be chosen explicitly). We construct an infinite word d ∈ AR N as the limit of the sequence of finite words (p k ) k∈N ∈ (AR * ) N that we define by recurrence as follows. We first set p 0 as the prefix given by Lemma 1 for (a, b, c) = ϕ(0). Now, for k ∈ N, we set p k+1 = p k .σ 1 .σ 2 .σ 3 .s, where s ∈ AR * is given by applying Lemma 2 to the word p k .σ 1 .σ 2 .σ 3 ∈ AR * and the vector ϕ(k + 1) ∈ Z 3 . By construction, the sequence of finite words (p k ) k∈N converges to an infinite sequence d which contains infinitely many occurrences of σ 1 , σ 2 and σ 3 . This guarantees that the sequence of finite words (d 0 • ... • d n−1 (1)) n∈N converges to an Arnoux-Rauzy word, that we denote by w ∞ . Finally, for any k ∈ N, since the directive sequence of w ∞ starts with the prefix p k , there exist u k , v k ∈ F(w ∞ ) such that ab(u k ) − ab(v k ) = ϕ(k). Corollary 1. The imbalance of the word w ∞ is infinite.
Proof. For any n ∈ N, there exist u n and v n ∈ F(w ∞ ) such that ab(u n ) − ab(v n ) = (n, 0, −n); this implies both |u n | = |v n | and |u n | 1 − |v n | 1 = n. The imbalance of w ∞ is thus infinite.
The imbalance of a word, which is a combinatorial quantity, is linked to the geometrical shape of its associated broken line. More precisely: a word w admitting frequencies has an infinite imbalance if and only if its Rauzy fractal is unbounded. We now propose to show that the word w ∞ actually satisfies a stronger property: its Rauzy fractal is unbounded in all directions of the plane. This relies on the following proposition.
where ∆ 0 denotes the plane of R 3 with equation x + y + z = 0, there exist u and v ∈ F(w) such that ab(u) − ab(v) = d, then, for any plane Π and for any D ∈ R + , there exists k ∈ N such that the euclidean distance between the point ab(p k (w)) and the plane Π is larger than D.
Remark 2. Proposition 2 and its proof remain valid by replacing ∆ 0 by any other plane whose intersection with Z 3 is not trapped between two parallel lines. Theorem 1. There exists an Arnoux-Rauzy word whose Rauzy fractal is unbounded in all directions of the plane.
Proof. We obtain, by applying Proposition 2 to the word w ∞ described in Proposition 1 and to planes spanned by f w and a vector of ∆ 0 , that the Rauzy fractal associated with w ∞ cannot be trapped between two parallels lines.

Proof of Lemma 1
We consider the infinite oriented graph whose vertices are the elements of Z 3 and whose edges map triplets to their images by one the 15 following applications. For δ ∈ {−2, −1, 0, 1, 2} and i ∈ {1, 2, 3}, consider: Our aim is to show that all vertices can be reached from the triplet O = (0, 0, 0) ∈ Z 3 , moving through a finite number of edges (see Definition 1 and Proposition 3 below.) The motivation lies in the following lemma.
The steps marked with ( * ) (removal of the initial i l ) are always possible since, for all l ∈ {0, ..., n−1}, the words u l and v l (and thusũ l+1 andṽ l+1 ) are nonempty.
In the sequel, it is convenient to introduce some vocabulary from graph theory.   Proof. For the first assertion, change δ l into −δ l in the finite sequence of edges going from O to (a, b, c). For the second assertion, change i l into s(i l ) in the finite sequence of edges going from O to (x j ) j∈{1,2,3} .
6 The vector of letter frequencies of w ∞ has rationally independent entries We sketch an elementary proof of the much wider result: Theorem 2. The vector of letter frequencies of an Arnoux-Rauzy word has rationally independent entries.
The proof is inspired from a similar result that holds for C-adic words [5].
Proof. Let w an Arnoux-Rauzy word; denote by (s n ) n∈N its directive sequence and by f its letter frequencies vector. We recall that for all nonnegative integer n, s n = σ i if and only if the i − th entry of F n AR (f ) (F AR is defined in Section 1) is greater than the sum of the two others. By contradiction, assume that the entries of f are not rationally independent.
First, observe that if for some r ∈ N, the i − th entry of F r AR (f ) is zero, then it will remain zero; and from this point on the directive sequence will not contain the substitution σ i , which is conflicting with the definition of Arnoux-Rauzy words and the uniqueness of the directive sequence. Thus, for all n ∈ N, all entries of Denote l m = (a, b, c) t and consider D m = max(|b − a|, |c − b|, |c − a|) ∈ N the difference between the maximum and the minimum entry of l m , that we call spread of l m . We claim that the sequence of nonnegative integers (D m ) m∈N is non-increasing and that it furthermore decreases infinitely oftenand here will be the contradiction. Indeed, the vector l m+1 is of the form M l m , where M is one the the three Arnoux-Rauzy matrices M σ 1 , M σ 2 or M σ 3 , which give respectively: l m+1 = (a, a + b, a + c) t , l m+1 = (a + b, b, c + b) t and l m+1 = (a + c, b + c, c) t . One can easily show, observing that the extreme entries of l m have opposite signs, that in all cases D m ≥ D m+1 . Similarly, we write l m+2 = M s m+1 M sm l m . A quick argument show that as soon as s m+1 = s m , which happens infinitely many times by definition of Arnoux-Rauzy words, we have D m > D m+2 .