Sturmian words. Lecture notes

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1 Sturmian words Lecture notes Sturmian words are a challenging topic, which is a bridge between combinatorics on words, number theory and dynamical systems. Sturmian words have been widely studied for their theoretical importance and applications to various fields of science. They admit several equivalent definitions and can even be described explicitly in arithmetic form. They also have some remarkable characterizations of geometrical nature (mechanical words, rotations), as well as characterizations via the monoid of Sturmian morphisms. In the course we consider some basics on Sturmian words and related topics. In particular, we will consider various characterizations of Sturmian words, and their generalizations to larger alphabets (episturmian and Arnaux-Rauzy words). 1 Complexity and Sturmian words Definition 1. An infinite word w is ultimately periodic if there exist N,T N such that w n+t = w n for each n N. Otherwise w is called aperiodic. Definition 2. The factor complexity of an infinite word w is the function p w (n) counting the number of its factors of length n. Properties: non-decreasing function, i.e., p w (n+1) p w (n) p(1) = #Σ for a periodic word w there exists C N such that p w (n) C for all n N Denote by F n (w) the set of factors of a word w of length n. Definition 3. The Rauzy graph G n (w) of order n is the labelled graph with vertex set F n (w) and the edge set {(bv,a,va) a,b Σ,bva F n+1 (w)}. A word y is a label of path from u to v in G n (w) if and only if F n+1 (uy) F n+1 (w) and v is a suffix of length n of uy. Example. Consider the Fibonacci word f = , 1

2 Figure 1: The Rauzy graph G 2 (f) of the Fibonacci word. defined as a fixed point of a morphism ϕ : 0 01, 1 0. On Fig. 1 one can see the Rauzy graph of the Fibonacci word of order 2. Exercise. One has also define the Fibonacci word via recurrence relations. Namely, f = lim n f n, where the sequence of words f n is defined by f 0 = 0,f 1 = 01, Prove the equivalence of the two definitions. f n+1 = f n f n 1 Remark that the sequence of lengths of the words f n is the traditional sequence of Fibonacci numbers, i. e., f n = F n, where F 0 = 1, F 1 = 2,..., F n+1 = F n + F n 1. Besides that, f n 1 = F n 2 for n 2. Here and in the further text we denote by v a the number of occurrences of the letter a in v. Theorem 1 (Morse and Hedlund, 1940). If p w (n) n for some n, then w is ultimately periodic. Proof. First notice that it is enough to prove the theorem for the case #Σ 2, since any unary word is clearly periodic. Now notice that p w (n) n implies that p w (n) = p w (n+1) for some n. Consider a Rauzy graph G n (w). Since every factor of length n is a prefix of a factor of length n+1, there is at least one edge leaving each vertex. Since p w (n) = p w (n+1), there is exactly one edge leaving each vertex. This implies that the strongly connected components of the graph are simple circuits. Thus every infinite path will loop through a fixed circuit after a while, and hence its label is eventually periodic. Since w is a label of a path, the theorem is proved. Corollary 1. For each n and each aperiodic word w it holds p w (n) n+1. Definition 4. An infinite word s is called Sturmian if p s (n) = n+1 for every n. Thus Sturmian words are aperiodic infinite words with minimal complexity. Since p s (1) = 2, these words are binary. A right special factor of a word w is a word u such that there exist letters a and b such that ua and ub are factors of w. Thus, s is Sturmian word if and only if it has exactly one right special factor of each length. 2

3 Example. The Fibonacci word is Sturmian. Proof. Since f = ϕ(f), it is a product of words 01 and 0. Therefore, 11 is not a factor of f and hence p f (2) = 3. The word 000 is not a factor of ϕ(f), since otherwise it is a prefix of some ϕ(v) for a factor v of f, and v has to start with 11. To show that f is Sturmian, we prove that f has exactly one right special factor of each length. Claim: For no word v, both 0v0 and 1v1 are factors of f. This is clear is v is the empty word and if v is a letter. Arguing by induction on v, assume that 0v0 and 1v1 are factors of f. Then v starts and ends with 0, i.e. v = 0u0 for some u. Since 00u00 and 10u01 are factors of f, there exists a factor z of f such that ϕ(z) = 0u. Moreover, 00u0 = ϕ(1z1) and 010u01 = ϕ(0z0). Hence 1z1 and 0z0 are factors of f. This is a contradiction because z ϕ(z) < v. So, the Claim is proved. We show now that f has at most one right special factor of each length. Assume that u and v are right special factors of the same length, and let z be the longest common suffix of u and v. Then the four words 0z0,0z1,1z0,1z1 are factors of f, which contradicts the Claim. To show that f has at least one right special factor of each length, we use the relation where g 2 = ε and for n 3 f n+2 = g n f R n fr n t n (n 2) (1) g n = f n 3...f 1 f 0, t n = { 01, if n is odd 10, otherwise. Here and in the further text v R denotes the reversal of v, i.e., for v = v 1...v n we have v R = v n...v 1. Observe that the first letter of fn R is opposite from the first letter of t n. This proves that fn R is a right special factor for each n 2. Since a suffix of a right special factor is right special, this proves that right special factors of any length exist. Equation(1)isprovedbyinduction. Indeed, f 4 = ε(010)(010)10andf 5 = 0(10010)(10010)01. Next, it is easily checked by induction that ϕ(u R )0 = 0(ϕ(u)) R for any word u. It follows that ϕ(f R n t n) = 0f R n+1 t n+1 and since ϕ(g n )0 = g n+1, one gets (1). 2 Balance Denote by x the length of the word x, and by x a the number of occurrences of the letter a in X. A set of words X is balanced if for every x,y X with x = y one has x a y a 1. A finite or infinite word is balanced if the set of its factors is balanced. A set of words is called factorial, if together with each word in it it contains all factors of this word. Proposition 1. Let X be a factorial set of binary words. If X is balanced, then for all n 0 #(X Σ n ) n+1. 3

4 Proof. Clearly, the statement holds for n = 0,1, and it holds for n = 2 since X cannot contain both 00 and 11. Assume the converse, and let n 3 be the smallest number for which the statement does not hold. Set Y = X Σ n 1 and Z = X Σ n. Then #X n and #Z n+2. For each z Z, its suffix of length n 1 is in Y. By the pigeon-hole principle, there exist two distinct words y,y Y such that for all four words 0y,1y,0y,1y are in Z. Since y y, there exists a word x such that x0 and x1 are prefixes of y and y. Therefore, both 0x0 and 1x1 are in X, and thus X is unbalanced. A word u = u 1...u n is a palindrome if u R = u n...u 1 is equal to u. For example, 0110 is palindrome. Moreover, the following holds: Proposition 2. Let X be a factorial set of binary words. Then X is unbalanced if and only if there exists a palindrome p such that both 0p0 and 1p1 are in X. Proof. The condition is clearly sufficient. Conversely, assume that X is unbalanced. By definition, consider two words u,v X, u = v, such that u 1 v 1 2, and choose them with minimal length. The first letters of u and v are distinct, and so are the last letters. Assuming that u starts with 0 and v with 1, there are factorizations u = 0pau and v = 1pbv for some words p, u, v and letters a b. In fact a = 0 and b = 1, since otherwise u and v are shorter words satisfying the condition. Thus, again by minimality, u = 0p0 and v = 1p1. Assume next that p is not a palindrome. Then there exists a prefix z of p and a letter a such that za is a prefix of p, z R is a suffix of p, but az R is not a suffix of p. Then bz R is a suffix of p with b a. This gives a proper prefix 0za of u and a proper suffix bz R of p. If a = 0 and b = 1, then 1z R 1 and 0z0 are shorter words than u and v satisfying 0z0 1 1z R 1 2, contradicting the minimality of length. Then u = 0z1u and v = v 1z R 0 for two words u and v with u 1 v 1 2, contradicting again the minimality. Thus p is a palindrome. Remark. When we proved that the Fibonacci word is Sturmian we in fact showed that it is balanced. Theorem 2. Let w be an infinite word. The following conditions are equivalent: (i) w is Sturmian, (ii) w is balanced and aperiodic. Proof. (ii) (i) Since w is aperiodic, then p w (n) n+1 for all n by Morse and Hedlund theorem. Since w is balanced, by Proposition 1 we have p w (n) n+1 for all n. Therefore, p w (n) = n+1 and w is Sturmian by definition. (i) (ii) Assume that w is Sturmian and unbalanced, we will show that w is eventually periodic. Since w is unbalanced, there exists a palindrome p such that 0p0, 1p1 are factors of w by Proposition 2. Hence p is right special. Since w is Sturmian, there is a unique right special factor of length n = p +1, which is either 0p or 1p. Assume 0p is right special, and 1p is not, and so, 0p1 is a factor, and 1p0 is not. Any occurrence of 1p in w is followed by 1. Let v be a word of length n 1 such that u = 1p1v is a factor of w. The word u has length 2n. We will prove the following claim: Claim. All factors of length n of u are not right special. 4

5 1 p 1 v 0 p 1 x 0 t 1 y z To prove the claim, it is enough to show that the only right special factor 0p is not a factor of u. Assume the contrary. Then there exist factorizations p = x0t, v = yz, p = t1y. Since p = x0t is palindrome, the first factorization implies p = t R 0x R, and hence the letter following the prefix t in p is both 0 and 1. A contradiction. The claim follows. The condition u = 2n implies that there are n+1 non-special factors of length n in u. But there are n non-special factors in w. So, there are two factors of length n in u which coincide. It follows that w is ultimately periodic. The theorem is proved. The slope of a nonempty binary word x is the number π(x) = x 1. In other words, this is x a frequency on the letter 1 in the word. It is easy to see that π(xy) = x y π(x)+ xy xy π(y). Proposition 3. A factorial set of words X is balanced if and only if, for all x,y X, x,y ǫ, π(x) π(y) < 1 x + 1 y. (2) Proof. Assume first that (2) holds. For x,y X of the same length, the inequality gives x 1 y 1 < 2, showing that X is balanced. Conversely, assume that X is balanced, and let x,y be in X. If x = y, then (2) holds. Assume x > y, and set x = zt, with z = y. Arguing by induction on x + y, we have π(t) π(y) < 1 t + 1 y and since X is factorial, z 1 y 1 1, whence π(z) π(y) < 1 y. Next, thus π(x) π(y) = z t π(z)+ x x π(t) π(y) = z t (π(z) π(y))+ x x (π(t) π(y)), π(x) π(y) < 1 x + t ( 1 x y + 1 ) = 1 t x + 1 y. The inequality (2) implies that for an infinite balanced binary word w the sequence (π(pref n (w))) n 1 is a Cauchy sequence and hence it converges as n. The limit 5

6 α = lim n π(pref n (w)) is called the slope of the infinite balanced word w. Exercise. Prove that π(f) = 1 τ 2, where τ = (1+ 5)/2 is the golden ratio. Proposition 4. Let w be an infinite balanced binary word with slope α. For every nonempty factor u of w, one has, π(u) α 1 u. (3) More precisely, one of the following two inequalities holds: either or α u 1 < u 1 α u +1 for all u F(w) (4) α u 1 u 1 < α u +1 for all u F(w) (5) Proof. Given some ε, consider n 0 such that for all n n 0, Then, using (2), π(pref n w) α ε. π(u) α π(u) π(pref n w) + π(pref n w) α < 1 u + 1 n +ε. Take n and then ε 0, the inequality (3) follows. Now assume that neither (4) nor (5) holds. Then there exist u,v F(w) such that α u 1 = u 1 and α v +1 = v 1. Hence π(u) π(v) = 1 + 1, a contradiction with (2). u v The proposition is proved. Proposition 5. Let w be an infinite binary balanced word. The slope α of w is a rational number if and only if w is eventually periodic. Proof. If w is eventually periodic, then by definition there exist finite words u and v such that w = uv ω. Then π(uv n ) = u 1 +n v 1 π(v) u +n v as n, so the slope is rational. Now suppose α is rational, α = q/p with q and p relatively prime integers. Without loss of generality assume that (4) holds (the case of (5) is symmetric). So, for any factor u of w of length p either u 1 = q or u 1 = q+1. There are only finitely many occurrences of factors of length p and with q+1 many 1 s, since otherwise there is a factor uzv of w with u = v = p and u 1 = v 1 = q +1. In view of (4) 2+2q + z 1 = uzv 1 1+α(p+ z +p) = 1+2q +α z, whence z 1 α z 1 in contradiction with (4). By the preceding observation, there is a factorization w = ty such that every word in F p (y) has the same number of 1 s. Consider now an occurrence azb of a factor in y of length p + 1, with a and b being letters. Since az 1 = zb 1, one has a = b. This means that y is periodic with period p. Consequently, w is eventually periodic. 6

7 y = αx+ρ Figure 2: Mechanical words corresponding to the line y = αx+ρ. 3 Geometrical characterization: mechanical and rotational words Given two real numbers α (0,1), ρ [0,1). We define two binary infinite words s α,ρ and s α,ρ by s α,ρ (n) = α(n+1)+ρ αn+ρ s α,ρ(n) = α(n+1)+ρ αn+ρ The word s α,ρ is upper mechanical word and s α,ρ is lower mechanical word of slope α and intercept ρ. On Fig. 2 one can see a graphical interpretation of the above definition. Consider the straight line with equation y = αx+ρ. The points with integer coordinates just below this line are P n = (n, αn+ρ ). Two consecutive points P n and P n+1 are joined by a straight line segment that is horizontal if s α,ρ (n) = 0 and diagonal if s α,ρ (n) = 1. The same observation holds for the points P n = (n, αn+ρ ) located just above the line. Since 1 + αn + ρ = αn + ρ whenever αn + ρ is not an integer, one has s α,ρ = s α,ρ except when αn+ρ is an integer for some n 0. In this case and, if n > 0, s α,ρ (n) = 0,s α,ρ (n) = 1 s α,ρ (n 1) = 1,s α,ρ (n 1) = 0. Thus, if α is irrational, s α,ρ and s α,ρ differ by at most one factor of length 2. A mechanical word is irrational or rational according to its slope is rational or irrational. An important special case is given ρ = 0 and α irrational. In this case s α,0 (0) = α = 0, s α,0 (0) = α = 1, and s α,0 = 0c α, s α,0 = 0c α, where the infinite word c α is called the characteristic word of slope α. 7

8 Theorem 3. Let s be an infinite binary word. The following are equivalent: (1) s is Sturmian (2) s is irrational mechanical. The proof will be a simple consequence of two lemmas. In the proofs, we will use several times the formula x x 1 < x x < x x+1. Lemma 1. Let s be a mechanical word with slope α. Then s is balanced of slope α. If α is rational, then s is purely periodic. If α is irrational, then s is aperiodic. Proof. Let s = s α,ρ be a lower mechanical word. The proof is similar for upper mechanical words. The number of 1 s in a factor u = s(n)...s(n+p 1) is the number u 1 = α(n+ p)+ρ αn+ρ, thus α u 1 < u 1 < α u +1 (6) This implies α u u 1 1+ α u, and shows that u 1 takes only two consecutive values, when u ranges over the factors of a fixed length of s. Thus, s is balanced. Moreover, by (6) π(u) α < 1 u Thus π(u) α as u, and α is the slope of s as it was defined for balanced words. This proves the first statement. If α is irrational, the word s is aperiodic by Proposition 5. If α = q/p is rational, then α(n+p)+ρ = q + αn+ρ, for all n > 0. Thus s(n+p) = s(n) for all n, showing that s is purely periodic. Lemma 2. Let s be a balanced infinite word. If s is aperiodic, then s is irrational mechanical. If s is purely periodic, then s is rational mechanical. Proof. Denote the slope of the balanced word s by α. For every real number τ, at least one of the following holds: pref n s 1 αn+τ for all n, pref n s 1 αn+τ for all n. Indeed, otherwise there exists a real number τ and two integers n, n + k such that pref n s 1 < αn + τ and pref n+k s 1 > αn+k + τ (or the symmetric relation). This implies that pref n+k s 1 pref n s 1 2+ αn+k +τ αn+τ > 1+αk, a contradiction with (3). Set ρ = inf{τ prefn s 1 αn+τ for all n}. By Proposition 4 one has ρ 1 and ρ < 1 if α is irrational. Observe that for all n 0 pref n s 1 αn+ρ pref n s 1 +1, (7) 8

9 since otherwise there is an integer n such that pref n s < αn + ρ, and setting σ = pref n s 1 +1 αn, one has σ < ρ and αn+σ > pref n s 1, in contradiction with the definition of ρ. If s is aperiodic, then α is irrational by Proposition 5, and αn+ρ is an integer for at most one n. By (7), either pref n s 1 = αn+ρ for all n, and then s = s α,ρ, or pref n s 1 = αn+ρ for all but one n 0, and pref n0 s 1 +1 = αn 0 +ρ. In this case, one has pref n s 1 = αn+ρ 1 for all n and s = s α,ρ 1. If s = u ω is purely periodic with period u = p, then α = q/p with q = u 1. Again pref n s 1 = αn+ρ if αn+ρ is never an integer (this depends on ρ). If pref n s 1 = αn+ρ for some n, we claim that pref n s 1 = αn+ρ for all n. Assume the converse. Then by (7), pref m s 1 +1 = αm+ρ for some m and we may assume n < m < n+p. Consider the words y = s(n+1)...s(m) and z = s(m+1)...s(n+p). Then π(y) = pref ms 1 pref n s 1 m n = α 1 y and π(z) = pref n+ps 1 pref m s 1 n+p m = α+ 1 z, whence π(y) π(z) = 1/ y +1/ z, in contradiction with Proposition 2. Similarly, if 1 + pref n s 1 = αn+ρ for some n, then pref n s 1 = αn+ρ for all n. Proof of Theorem 3. Follows from the previous two lemmas and Theorem 2. Remark. Remark that a balanced infinite word is not always mechanical when the slope is rational. For example, consider the infinite balanced word 10. It is not a mechanical word. Indeed, it has slope 0, and the only mechanical word of slope 0 is 0. Mechanical words can also be generated by rotations. The rotation by angle α is the mapping R α from [0,1) (identified with the unit circle) to itself defined by R α (x) = {x+α}, where {x} = x x is the fractional part of x. Iterating R, one gets A straightforward computation shows that Rα n (x) = {x+nα}. (n+1)α+ρ = 1+ nα+ρ {nα+ρ} 1 α. Considering a partition of [0,1) into I 0 = [0,1 α), I 1 = [1 α,1), one gets { 0, if Rα n s α,ρ (n) = (ρ) = {ρ+nα} I 0, 1, if Rα(ρ) n = {ρ+nα} I 1. (8) One can also define I 0 = (0,1 α], I 1 = (1 α,1], the corresponding word is s α,ρ. 9

10 We will identify the interval [0,1) with the unit circle. For 0 a < b < 1, the sets [a,b) and [0,a) [b,1) are intervals, and the latter one is denoted [b,a). Then, for any subinterval I of [0,1), the sets R(I) and R 1 (I) are always intervals. Now consider a binary word w = b 0 b 1...b m 1, we want to know whether w is a factor of some s α,ρ = a 0 a 1... By (8), a n+k = b i if and only if R n+i (ρ) I bi, or equivalently, if and only if R n (ρ) R i (I bi ). Thus, for n 0, where I w is the interval w = a n a n+1...a n+m 1 R n (ρ) I w, I w = I b0 R 1 (I b1 ) R m+1 (I bm 1 ). The interval I w is non empty if and only if w is a factor of s α,ρ. An infinite word w is uniformly recurrent, if for any its factor u there exist an integer N such that for every i, u F(w i...w i+n ). Exercise. Prove that Sturmian words are uniformly recurrent. 4 Properties of factors of a Sturmian word Proposition 6. Let s and t be Sturmian words. 1. If s and t have same slope, then F(s) = F(t). 2. If s and t have distinct slopes, then F(s) F(t) is finite. Proof. 1. Let α be the common slope of s and t. By Proposition 4, every factor u of s satisfies π(u) α < 1 u. Equality is impossible because α is irrational. Next, for every factor v of t, π(v) α < 1 v Let X = F(s) F(t). The set X is factorial. It is also balanced since In view of Proposition 1 π(u) π(v) π(u) α + π(v) α < 1 u + 1 v. #(X Σ n ) n+1 for every n. Thus F(s) = X = F(t). Let now α be the slope of s and β be the slope of t. We may suppose that β > α. For any factor u of s such that β α > 2 1, one has α π(u) > by Proposition 4, whence u u β π(u) = (β α)+(α π(u)) > 1 showing that u is not a factor of t. u Proposition 7. The set F(s) of factors of a Sturmian word s is closed under reversal. 10

11 Proof. Set F(s) R = {x R x F(s)}. The set X = F(s) F(s) R is balanced. In view of Proposition 1, #(X Σ n ) n + 1, for each n, and since #(F(s) Σ n ) = n + 1, one has X = F(s). Thus F(s) R = F(s). Let an alphabet (Σ, ) be a linearly ordered set. Then induces an order on the set of infinitewordsoverσ. Giventwoinfinitewordsxandy overσ,wesaythatxislexicographically less than y, denoted x < y if there is an integer n such that x i = y i for i < n, and x n < y n. We write x y if either x < y or x = y. Proposition 8. Let 0 < α < 1 be an irrational number and let 0 ρ,ρ < 1 be real numbers. Then s α,ρ < s α,ρ ρ < ρ Proof. Since α is irrational, the set of fractional parts {αn} for n 0 is dense in the interval [0,1). Thus ρ < ρ if and only if there exists an integer n 1 such that 1 ρ {αn} < 1 ρ, and this is equivalent to αn + ρ = 1 + αn + ρ. If n is the smallest integer for which this equality holds, then s α,ρ (n 1) = 0 and s α,ρ (n 1) = 1 and s α,ρ (k) = s α,ρ (k) for k < n 1. Lemma 3. Let 0 < α,α < 1 be irrational numbers and let 0 ρ,ρ < 1 be real numbers. Any of the equalities s α,ρ = s α,ρ, s α,ρ = s α,ρ, s α,ρ = s α,ρ implies α = α and ρ = ρ. Proof. Any of the equalities implies that α = α because equal words have the same slope. Next, s α,ρ = s α,ρ implies ρ = ρ by the previous proposition. Finally, consider the equality s α,ρ = s α,ρ. In the case αn+ρ is never an integer, then s α,ρ = s α,ρ and the conclusion holds. In the case αn 0 + ρ is an integer for some n 0 we have s α,ρ +(n 0 +1)α = s α,ρ +(n 0 +1)α showing ρ = ρ. Sturmian words with intercept 0 have many interesting properties and are in some sense the easiest to investigate. Remind that for an irrational number 0 < α < 1, the words s α,0 and s α,1 differ only by their first letter, and that s α,0 = 0c α,s α,0 = 1c α, where c α is the characteristic word of slope α. Equivalently, c α = s α,α = s α,α. Proposition 9. For every Sturmian word s, either 0s or 1s is Sturmian. A Sturmian word s is characteristic if and only if 0s and 1s are both Sturmian. Proof. The first claim follows from the fact that s α,ρ α = as α,ρ for some a {0,1}. If s = s α,α = s α,α is the characteristic word of slope α, then 0s = s α,0 and 1s = s α,0 are Sturmian. Conversely, thesturmianwords 0sand1shave sameslope, sayα. Denoteby ρandρ their intercepts. Then their common shift s has intercept α+ρ = α+ρ, and by Lemma 3, ρ = ρ. Thus 0s = s α,ρ, 1s = s α,ρ. Assume ρ > 0. The first letter 0s(0) = 0 = α+ρ ρ = α+ρ, 1s(0) = 1 = α+ρ ρ = α+ρ 1, which gives α+ρ = 2, a contradiction. Remind that we call a factor v of a word w a right (resp., left) special if both va and vb (resp., av and bv) are factors of w for a b Σ. A factor is bispecial, if it is both right and left special. 11

12 Proposition 10. The set of right special factors of a Sturmian word is the set of reversals of the prefixes of the characteristic word of same slope. Clearly, v is left special if and only if v R is right special. In other words, the proposition states that the set of left special factors of a Sturmian word is the set of prefixes of the characteristic word of same slope. So, bispecial factors of Sturmian words a palindromes. Proof. Let s be a Sturmian word of slope α. By Proposition 9, the infinite words 0s and 1s are Sturmian and clearly have slope α. Thus F(s) = F(c α ) = F(0c α ) = F(1c α ) by Proposition 6. Consequently, for each prefix v of c α, both 0v and 1v are factors of s. Since F(s) is closed under reversal, this shows that v R is right special. Thus v R is the unique right special factor of length v. 5 Iterated palindromic closure We denote with w (+) the right palindromic closure of the word w, i.e., the shortest palindrome which has w as a prefix. The iterated palindromic operator ψ is defined inductively as follows: ψ(ε) = ε, For any word w and any letter a, ψ(wa) = (ψ(w)a) (+). For example, ψ(aaba) = aabaaabaa. It follows immediately from the definition that if u is a prefix of v, then ψ(u) is a prefix of ψ(v). Thus, given an infinite word = on an alphabet Σ we can define ψ( ) = lim n ψ( n ). The following lemma summarizes the properties of ψ. Lemma 4. Let be a right infinite word over the (finite or infinite) alphabet Σ and let w = ψ( ). Then the following statements hold: 1. The word w is closed under reversal, i.e., if v = v 1 v 2...v k is a factor of w, then so is its mirror image v k...v 2 v The word w is uniformly recurrent. 3. If each letter a Σ appears in an infinite number of times, then for each prefix u of w and each a Σ, we have au is a factor of w. Proof. Since any factor of w is contained in some ψ(u) for a sufficiently long prefix u of, and ψ(u) is by definition a palindrome (and hence closed under reversal), the first statement is proved. The second statement is easily derived from the fact that for any finite prefix va of (a being a letter), we have that ψ(va) 2 ψ(v) +1 and moreover ψ(va) begins and ends 12

13 in ψ(v). It follows that any factor of length (for example) 3 ψ(v) contains an occurrence of ψ(v). Finally suppose each a Σ appears infinitely many times in. Thus for any letter a and any prefix v of there exists a prefix of of the form vv a. From the definition of ψ we then have that ψ(vv )a is a prefix of w and ψ(vv ) ends in ψ(v), so ψ(v)a is a factor of w. Since ψ(v) is a palindrome and w is closed under reversal, we obtain that for any prefix v of and for any letter a, the word aψ(v) is a factor of w and the third statement easily follows. In fact every characteristic Sturmian word can be obtained as an iterated palindromic closure of an infinite word containing infinitely many occurrences of 0 s and 1 s and vise versa. 6 Standard sequences Let (d 1,d 2,...d n,...) be a sequence of integers, with d 1 0 and d n > 0 for n > 1. To such a sequence, we associate a sequence (s n ) n 1 of words by s 1 = 1, s 0 = 0, s n = s dn n 1s n 2 (n 1). The sequence (s n ) n 1 is a standard sequence, and the sequence (d 1,d 2,...) is its directive sequence. Observe that if d 1 > 0, then any s n (n 0) starts with 0; on the contrary, if d 1 = 0, then s 1 = s 1 = 1, and s n starts with 1 for n 0. Every s 2n ends with 0, every s 2n+1 ends with 1. This sequence defines a limit: s = lim n s n. Example. We saw that the Fibonacci word can e defined via the recurrence relation f n = f n 1 f n 2. So,itsdirectivesequenceis(1,1,...). Observethatthedirectivesequence(0,1,1,...) results in the word obtained from Fibonacci word by exchanging 0 and 1. Now we will show the connection between infinite standard words and infinite palindromic closures. Let x {0,1} ω be an infinite word containing infinitely many occurrences of 0 s and 1 s. Then x can be uniquely expressed as x = 0 d 1 1 d 2..., with d 1 0 and d i > 0 for i > 1. We call the infinite sequence d l,d 2,...,d n,... the integral representation of x. Theorem 4. Let be an infinite binary word containing infinitely many occurrences of 0 s and 1 s, and d l,d 2,...,d n,... its integral representation. Then ψ( ) is the infinite standard word with the directive sequence d l,d 2,...,d n,... We refer for the proof to; A. de Luca: Sturmian words, structure, combinatorics and their arithmetics, Theoret. Comput. Sci.,

14 Now we are going to relate standard words to characteristic words and to continued fraction expansion of the slope of a characteristic word. For these we need the following morphisms: E : , ϕ : , ϕ : From these, we get other morphisms, denoted G, G, D, D and defined by G = ϕ E : , G = ϕ E : , D = E ϕ : D = E ϕ : Clearly, ϕ = G E = E D and ϕ = G E = E D. Now we are going to prove that for each Sturmian word s, he infinite words E(s), G(s), G(s), ϕ(s), ϕ(s), D(s), D(s) are Sturmian. Lemma 5. For any real number ρ, the following relations hold: Proof. For n 0, E(s α,ρ ) = s 1 α,1 ρ and E(s α,ρ ) = s 1 α,1 ρ. s 1 α,1 ρ = (1 α)(n+1)+1 ρ (1 α)n+1 ρ = 1 ( (αn ρ α(n+1) ρ ) = 1 s α,ρ (n) because r = r for every real number r. This proves the first equality; the second one is symmetric. Lemma 6. Let 0 < α < 1. For any real number 0 ρ < 1, the following relations hold: G(s α,ρ ) = s α 1+α, ρ 1+α, G(sα,ρ ) = s α 1+α,ρ+α 1+α For any real number 0 < ρ 1, the following relations hold: G(s α,ρ ) = s α 1+α, ρ, G(s 1+α α,ρ ) = s α 1+α,ρ+α 1+α, ϕ(s α,ρ ) = s 1 α, ϕ(s α,ρ ) = s1 α 2 α, 1 ρ 2 α 2 α, 1 ρ 2 α Proof. Let s = a 0 a 1...a n... be an infinite word, the a i being letters. An integer n is the index of the k-th occurrence of the letter 1 in s if a 0...a n contains k letters 1 and a n = 1. If s = s α,ρ and 0 ρ < 1, this means that α(n+1)+ρ = k, αn+ρ = k 1, which implies αn+ρ < k α(n+1)+ρ, that is k ρ n = α

15 Similarly, if s = s α,ρ and 0 < ρ 1, then α(n+1)+ρ = k +1, αn+ρ = k, which implies n = k ρ α. Set G(s α,ρ ) = b 0 b 1...b i..., with b i {0,1}. Since every letter 1 in s α,ρ is mapped to 01 in G(s α,ρ ), the prefix a 0...a n of s α,ρ (where n is the index of the k-th letter 1) is mapped onto the prefix b 0 b 1...b n+k of G(s α,ρ ). Thus the index of the k-th letter 1 in G(s α,ρ ) is k ρ 1+α n+k = α 1. 1+α This proves the first formula. Next, we observe that, for any infinite word x, one has G(x) = 0 G(x). Indeed, the formula G(w) = 0 G(w) is easily shown to hold for finite words w by induction. Furthermore, if a Sturmian word s α,ρ starts with 0 and setting s α,ρ = 0t, one gets t = s α,α+ρ. Altogether G(s α,ρ ) = s α 1+α,ρ+α 1+α for 0 ρ < 1. The proof of the other formula is similar. Finally, since ϕ = G E, ϕ(s α,ρ ) = G(s 1 α,1 ρ ) = s 1 α. 2 α, 1 ρ 2 α Exercise. Find similar formulas for D, D and ϕ. Corollary 2. For any irrational α, 0 < α < 1, one has For m 1, define a morphism θ m by E(c α ) = c 1 α G(c α ) = c α/(1+α). It is easily checked that θ m : 0 0m m θ m := G m 1 E G. Corollary 3. For m 1, one has θ m (c α ) = c 1. m+α Proof. Since E G(c α ) = c 1/(1+α), the formula holds for m = 1. Next, G(c 1/(k+α) ) = c 1/(1+k+α), so the claim is true by induction. We use this corollary for connecting continued fractions to characteristic words. Every irrational number γ admits a unique expansion as a continued fraction γ = m m m m ,

16 where m 0,m 1,... are integers, m 0 0, m i > 0 for i 1. We write it as γ = [m 0,m 1,m 2,...]. The integers m i are called the partial quotients of γ. If the sequence (m i ) is eventually periodic, and m i = m k+i for i h, we write this by overlining the purely periodic part: γ = [m 0,m 1,m 2,...,m h 1,m h,...,m h+k 1 ]. Let α = [0,m 1,m 2,...] be the continued fraction expansion of an irrational α with 0 < α < 1. If, for some β with 0 < β < 1, we agree to write β = [0,m i+1,m i+2,...] α = [0,m 1,m 2,...,m i +β] Corollary 4. If α = [0,m 1,m 2,...,m i +β] for some irrational α, 0 < α,β < 1, one has c α = θ m1 θ m2...θ m1 (c β ). Proposition 11. Let α = [0,1 + d 1,d 2,...] be the continued fraction expansion of some irrational α with 0 < α < 1, and let (s n ) be the standard sequence associated to (d 1,d 2,...). Then every s n is a prefix of c α and c α = lim n s n. Proof. By definition, s n = s dn n 1s n 2 for n 1. Define morphisms h n by We claim that h n = θ 1+d1 θ d2 θ dn. s n = h n (0),s n s n 1 = h n (1), n 1. This holds for n = 1 since h 1 (0) = 0 d 1 1 = s 1 and h 1 (1) = 0 d 1 10 = s 1 s 0. Next, for n 2, and h n (0) = h n 1 (θ dn (0)) = h n 1 (0 dn 1 1) = s dn 1 n 1 s n 1s n 2 = s n h n (1) = h n 1 (0 dn 1 10) = s n s n 1 For any infinite word x, the infinite word h n (x) starts with s n because both h n (0) and h n (1) start with s n. Thus, setting β n = [0,d n+1,d n+2,...], one has c α = h n (c β ) by Corollary 4 and thus c α starts with s n. This proves the first claim. The second is an immediate consequence. Example. The directive sequence for the Fibonacci word is (1,1,...). The corresponding irrational is 1/τ 2 = [0,2,1,1..., and indeed the infinite Fibonacci word is the characteristic word of slope 1/τ 2. Exercise. Find a directive sequence for the characteristic Sturmian word of slope α = Proposition 11 has several interesting consequences, in particular the relation to fixed points and the powers that may appear in a Sturmian word. Let x be an infinite word. For w F(x), the index of w in x is the greatest integer d such that w d F(x), if such an integer exists. Otherwise, w is said to have infinite index. 16

17 Proposition 12. Every nonempty factor of a Sturmian word s has finite index in s. Proof. Assume the contrary. There exists a Sturmian word s and a nonempty factor u of s such that u n is a factor of s for every n 1. For long enough factor u n, the slope of this word differs from α by more than 1/n u, a contradiction with Proposition 4. An infinite word x has bounded index if there exists an integer d such that every nonempty factor of x has an index less than or equal to d. Theorem 5. A Sturmian word has bounded index if and only if the continued fraction expansion of its slope has bounded partial quotients. Proposition 13. A finite standard word s n is primitive. Proof. First we prove that We prove that by induction, and using that So, by the definition of s n, s n s n 1 1 s n 1 s n 1 = 1. s 1 s 0 1 s 1 1 s 0 = (d 1 +1) 0 1 = 1. s n s n 1 1 s n 1 s n 1 = (d n s n 1 + s n 2 ) s n 1 1 (d n s n s n 2 1 ) s n 1 = s n 2 s n 1 1 s n 2 1 s n 1 = = 1. Lemma 7. Let (s n ) n 1 be the standard sequence of the characteristic word c α, with α = [0,1+d 1,d 2,...]. For n 3, the word s 1+d n+1 n is a prefix of c α, and s 2+d n+1 n is not a prefix. If d 1 1, this holds also for n = 2. Proof. We show that for n 3, (and for n 2, if d 1 1), one has Indeed, s n 1 s n = s n t n 1 ; with t n = s dn 1 n 1 s n 2s n 1. s n 1 s n = s n 1 s dn n 1 s n 2 = s dn n 1 sd n 1 n 2 s n 3s n 2 = s dn n 1s n 2 s d n 1 1 n 2 s n 3 s n 2 = s n t n 1 provided d n 1 1. Observe that t n 1 is not a prefix of s n, since otherwise s n = t n 1 u for some word u, and s n 1 s n u = s 2 n and s n is not primitive (see Ex.3 in Combinatorics on Words lecture notes). Clearly, s n+1 s n is a prefix of the characteristic word c α. Since s n+1 s n = s d n+1 n s n 1 s n = s 1+d n+1 n t n 1, the word s 1+d n+1 n is a prefix of c α, and since t n 1 is not a prefix of s n, the word s 2+d n+1 n a prefix of c α. is not 17

18 Proof of Theorem 5. Since a Sturmian word has the same factors as the characteristic word of same slope, it suffices to prove the result for characteristic words. Let c α be the characteristic word of slope α = [0,1+d 1,d 2,...]. Let (s n ) n 1 be the associated standard sequence. To prove that the condition is necessary, observe that s d n+1 n is a prefix of c α for each n 1. Consequently, if the sequence (d n ) of partial quotients is unbounded, the infinite word has factors of arbitrarily large exponent. Conversely, assume that the partial quotients (d n ) are bounded by some D and arguing by contradiction, suppose that c α has unbounded index. Let r be some integer such that F(c α ) contains a primitive word of length r with index greater than D+4. Among those words, let w be a word of length r of maximal index. Let d+1 be the index of w. Then d D+3. For the proof we need two claims. Claim (1). The characteristic word has prefixes of the form w d, with d D +3. Proof of Claim (1): Indeed, if w d+1 is a prefix of c α, we are done. Otherwise, consider an occurrence ofw d+1. Set w = za withaaletter, andlet bbetheletter preceding theoccurrence of w d+1. If b = a, replace w by az and proceed. The process will stop after at most w 1 steps because either a prefix of c α is obtained, or because otherwise w would occur in c α at the power d+2. Thus, we may assume b a. Thus b(za) d+1 is a factor of c α. This implies that a(za) d and b(za) d are factors, so w d is a right special factor, and therefore it is a prefix of c α. Claim (1) is proved. Claim (2). If w d is a prefix of the characteristic word c α, then w is one of the standard words s n. Proof of Claim (2): Indeed, set e = d 2, so that e D +1. Let n be the greatest integer such that s n is a prefix of w e+1. Then w e+1 is a prefix of s n+1 = s d n+1 n s n 1, thus also of s 1+d n+1 n. This shows that (1+D) w (1+e) w (1+d n+1 ) s n (1+D) s n, whence w s n. Now, since both w e+2 and s 1+d n+1 n are prefixes of c α, one is a prefix of the other. If w e+2 is the shorter one, then w e+2 = w e+1 + w s n + w. Thus, w e+2 and s 1+d n+1 n share a common prefix of length s n + w. Consequently, w and s n are powers of the same word, and since they are primitive, they are equal. If s 1+d n+1 n is the shorter one then, since (1+e) w (1+d n+1 ) s n, s 1+d n+1 n = s n +d n+1 s n s n + d n+1 1+d n+1 (1+e) w s n + w, and the same conclusion holds. Claim (2) is proved. If follows that s 1+e n is a prefix of c α and, since e D+1 d n+1 +1, also s 2+d n+1 n is a prefix of c α, contradicting Lemma 7. We say that an infinite word w Σ ω is critical if there exists a real number Ω > 1, called the critical exponent of w, such that the follows conditions hold: (i) if u Σ +, t > 1 and u t F(w), then t Ω, (ii) for all ε > 0 there exists a v Σ + and a rational number r > Ω ε, such that v r F(w). The following theorem establishes critical exponents for fixed points of morphisms: 18

19 Theorem 6. Let 0 < α < 1 with α = [0,a 0,a 1,...,a m ], where a 0,...a m are positive integers and a m > a 0. Then a Sturmian word of slope α has a critical exponent Ω, where Ω = max 1 t m [2+a t,a t 1,...,a 1,a m,...,a 1 ] Example. Consider τ = ( 5 1)/2, then 1/τ 2 = [0,2,1,1,...] and so by Theorem 6 the Fibonacci infinite word f has a critical exponent Ω where Ω = 2+[1,1,1,1,...] = = 3, For irrational α (0,1), α = [0,a 1,a 2,...], the associated rational approximants p n /q n are defined by p 0 = 0, p 1 = 1, p n = a n p n 1 +p n 2, q 0 = 1, q 1 = a 1, q n = a n q n 1 +q n 2. Theorem 7. The critical exponent of a Sturmian sequence s of slope α is given by 7 Sturmian morphisms Ω = max{a 1,2+sup{a n+1 +(q n 1 2)/q n }}. n N In this section the identity morphism Id and the morphism E that exchanges the letters 0 and 1 will be called trivial morphisms. A morphism f is Sturmian if f(s) is a Sturmian word for every Sturmian word s. Since an erasing morphism can never be Sturmian, all morphisms considered here are assumed to be nonerasing. The trivial morphisms Id and E are Sturmian. The set of Sturmian morphisms is closed under composition, and consequently is a submonoid of the monoid of endomorphisms of {0,1}. The main result of this section is the characterization of Sturmian morphisms (Theorem 8). Consider the morphisms ϕ : , ϕ : Proposition 14. The morphisms E, ϕ and ϕ are Sturmian. Proof. This follows from Lemma 6. We shall see below that in fact every Sturmian morphism is a composition of these three morphisms. The following proposition gives a converse of Proposition 14. Proposition 15. Let x be an infinite word. (i) If ϕ(x) is Sturmian then x is Sturmian. (ii) If ϕ(x) is Sturmian and x starts with the letter 0, then x is Sturmian. 19

20 Proof. Let x be an infinite word. If ϕ(x) or ϕ(x) is Sturmian, then x is clearly aperiodic. Arguing by contradiction, let us suppose that x is not balanced and suppose that 0v0 and 1v1 are both factors of x. Clearly, ϕ(0v0) = 01ϕ(v)01, ϕ(1v1) = 0ϕ(v)0 and every occurrence of ϕ(1v1) in ϕ(x) is followed by the letter 0. Consequently 1ϕ(v)01 and 0ϕ(v)00 are both factors of ϕ(x) which is not balanced. Next, if x does not start with 1, then either 01v1 or 11v1 is a factor of x. But ϕ(0v0) contains the factor 10 ϕ(v)1, and ϕ(01v1) and ϕ(11v1) both contain the factor 00 ϕ(v)0. Consequently, ϕ(x) is not balanced. Corollary 5. Let x be an infinite word and let f be a morphism that is a composition of E and ϕ. If f(x) is Sturmian then x is Sturmian. Example. We give an example of a non-sturmian word x starting with 1 and such that ϕ(x) is Sturmian. Let f be the Fibonacci word. The infinite word 11f is not Sturmian because it contains both 00 and 11 as factors. However, since f is a characteristic word, the infinite word 0f is Sturmian. Consequently ϕ(ϕ(0f)) = ϕ(01f) = 100 ϕ(f) is Sturmian. Thus 00 ϕ(f) also is Sturmian and, since 00 = ϕ(11), ϕ(11f) is Sturmian. Let us denote St the submonoid of the monoid of endomorphisms obtained by composition of E, ϕ and ϕ in any number and order. St is called the monoid of Sturm and by Proposition 14 all its elements are Sturmian. A first step to the converse is the following. Lemma 8. Let f and g be two morphisms and let x be a Sturmian word. If f St and f g(x) is a Sturmian word, then g(x) is a Sturmian word. Proof. Let x be a Sturmian word and g a morphism. It suffices to prove the conclusion for f = E, f = ϕ and f = ϕ. Set y = g(x). If E(y) is a Sturmian word then y is also a Sturmian word too and, by Proposition 15, this also holds if ϕ(y) is a Sturmian word. It remains to prove that if ϕ(y) is a Sturmian word then so is y. Suppose that y is not a Sturmian word. Observe that y is aperiodic, since otherwise ϕ(y) is eventually periodic thus it is not Sturmian. Thus y = g(x) is not balanced and contains two factors 0v0 and 1v1 which are factors of images of some factors of x. The Sturmian word x is recurrent, thus 1v1 occurs infinitely often in y, which implies that 01v1 or 11v1 is a factor of y. Since ϕ(0v0) = 10 ϕ(v)10 and ϕ(1v1) = 0 ϕ(v)0, both 10 ϕ(v)1 and 00 ϕ(v)0 are factors of ϕ(y) and thus ϕ(y) is not balanced. A contradiction. Corollary 6. Let f St and g be a morphism. The morphism f g is Sturmian if and only if g is Sturmian. Proof. Assume first that g is Sturmian. Since f isacomposition ofe, ϕ and ϕ, themorphism f g is Sturmian by Proposition 14. Conversely, if f g is Sturmian, then for every Sturmian word x, the infinite word f g(x) is Sturmian and, by Lemma 8, the infinite word g(x) is Sturmian. This means that g is Sturmian. A morphism f is locally Sturmian if there exists at least one Sturmian word x such that f(x) is a Sturmian word. 20

21 Theorem 8. Let f be a morphism. The following three conditions are equivalent: (i) f St, (ii) f is Sturmian, (iii) f is locally Sturmian. The equivalence of (i) and (ii) means that the monoid of Sturm is exactly the monoid of Sturmian morphisms. The length of a morphism f is the number f = f(0) + f(1). The proof of Theorem 8 is based on the following fundamental lemma. Lemma 9. Let f be a non trivial morphism. If f is locally Sturmian then f(0) and f(1) both start or end with the same letter. Proof. Let f be a non trivial morphism and suppose that f(0) and f(1) do not start nor end with the same letter. Suppose f(0) starts with the letter 0. Then f(1) starts with the letter 1. If f(0) ends with 1 then f(1) ends with 0. But in this case f(01) contains a factor 11 and f(10) contains a factor 00. Thus the image of any Sturmian word contains the two factors 00 and 11 which means that f is not locally Sturmian. Otherwise f(0) 0{0,1} 0 {0} and f(1) 1{0,1} 1 {1}, and we prove the result by induction on f. If f = 3, then f(a) = cc and f(b) = d for letters a,b,c,d, a b, and since any Sturmian word x contains the two factors a n+1 and ba n b for some integer n, f(x) contains (cc) n+1 and d(cc) n d and thus is not Sturmian. Arguing by contradiction, suppose that f 4 and f is locally Sturmian. Let x be a Sturmian word such that f(x) is Sturmian (such a word exists because f is locally Sturmian) and suppose that x contains the factor 00 (the case where x contains 11 is clearly the same). Since f(0) starts and ends with 0, f(x) contains also 00. Consequently, since the infinite word f(x) is balanced, neither f(0) nor f(1) contains the factor 11. Since x is Sturmian, x does not contain 11 and there is an integer m 1 such that every block of 0 between two consecutive occurrences of 1 is either 0 m or 0 m+1. The word f(0) does not contain the factor 00. Indeed, otherwise f(0) = u00v and f(1) = r1 = 1s for some words u,v,r,s. Since 0 m+1 and 10 m 1 are factors of x, the words f(0 m+1 ) and f(10 m 1) are factors of f(x). But f(0 m+1 ) = u00vf(0 m 1 )u00v = uw 1 v, f(10 m 1) = r1f(0 m 1 )u00v1s = rw 2 s for suitable w 1,w 2, and one has w 1 = w 2 and w 1 1 w 2 1 = 2, a contradiction. Consequently f(0) = (01) n 0 for some integer n 0. Since 10 m 1 and 10 m+1 1 are factors of x, the infinite word f(x) contains the two factors 10 m 1 and 10 m+1 1 if n = 0, and the two factors 101 and 1001 if n 0. Set p = m if n = 0, and p = 1 if n 0. Then in both cases, f(x) contains the factors 10 p 1 and 10 p+1 1, and in both cases 1 p m. Since f(1) does not contain the factor 11, there exist an integer k 0, and integers m 1,...,m k {0,1} such that f(1) = 10 p+m 1 10 p+m p+m k 1. Consider a new alphabet B = {a,b} and two morphisms ρ, η : B {0,1} 21

22 ρ : a 01 b 0 p 1, η : a (01)n 0 b 0 p 1. We show that there exists a word u over B such that f(ρ(b)) = η(bub). (i) If n = 0, set u = a m 1 ba m 2 b...ba m k. Since f(1) 1, one has f(1) = 1η(u)0 p 1. Thus f(ρ(b)) = f(0 p 1) = η(bub). (ii) If n 0 and m 1 = = m k = 0, set u = b k+n 1. Since f(1) = (10) k 1, one gets η(u) = (01) k+n 1 and f(ρ(b)) = f(01) = η(bub). (iii) Otherwise n 0 and m i = 1 for at least one integer i, 1 i k. Thus there exist integers t 2,n 1,...,n t such that f(1) = 1(01) n 1 0(01) n (01) n t 1 0(01) nt. Since f(01) starts with (01) n+1, one has n 1 0, n i n for 2 i t 1 and n t 1. Set u = b n 1 ab n 2 n a...b n t 1 n a b n t 1. Then, again, f(ρ(b)) = f(01) = η(bub). Define a morphism g : B B by g : a a b bub. Then f ρ = η g. Since m p, by deleting if necessary some letters at the beginning of x, one may suppose that x starts with 0p1. It follows that there exists a (unique) infinite word x over B such that ρ(x ) = x. Thus there exists a (unique) infinite word y over B such that f x ρ x g f(x) y η Identifying a with 0 and b with 1, one has ρ = (ϕ E) p. If n = 0 then η = ρ. If n 0, then p = 1, so η = ϕ E (E ϕ) n. Thus since x and f(x) are Sturmian, the words x and y are Sturmian by Corollary 5. Consequently the morphism g is locally Sturmian. However, the words g(0) and g(1) do not start nor end with the same letter and 3 g < f. By induction, g is not locally Sturmian, a contradiction. The lemma is proved. Proof of Theorem 8. It is easily seen that (i) = (ii) and (ii) = (iii). So let us suppose that f is a locally Sturmian morphism. The property is straightforward if f = Id or f = E. Thus we assume f 3. Let x be a Sturmian word such that f(x) is also a Sturmian word. Since f(x) is balanced, it contains only one of the two words 00 or 11. Suppose that f(x) contains 00. From Lemma 9, the words f(0) and f(1) both start or end with 0. Consider first the case where f(0) and f(1) both start with 0. Then f(0),f(1) {0,01} + and there exists two words u and v such that f(0) = ϕ(u) and f(1) = ϕ(v). Define g a morphism by g(0) = u and g(1) = v. Then f = ϕ g and, by Lemma 8, g(x) is a Sturmian 22

23 word. Next, f = g + uv 0 and uv 0 > 0. Otherwise, f(0) = ϕ(u) and f(1) = ϕ(v) would contain only 0 and f(x) = 0 ω would not be Sturmian. Thus g < f and the result follows by induction. If f(0) and f(1) both end with 0, the same argument holds with ϕ instead of ϕ, and if f(x) contains 11 then E f is of the same length and contains 00. Now we are going to describe those characteristic words that are fixed points of standard morphisms. As an example, we know that the morphism ϕ fixes the infinite Fibonacci word f. We say that a morphism h fixes an infinite word x if h(x) = x. In this case, x is a fixed point of h. For the description of characteristic words which are fixed points of morphisms, we introduce a special set of irrational numbers. A Sturm number is a number α that has a continued fraction expansion of one of the following kinds: (i) α = [0,1,a 0,a 1,...,a k ], with a k a 0, (ii) α = [0,1+a 0,a 1,...,a k ] with a k a 0 1. Observer that (i) implies α > 1/2 and and (ii) implies α < 1/2. More precisely, α has an expansion of type (i) if and only if 1 α has an expansion of type (ii). Consequently, α is a Sturm number if and only 1 α is a Sturm number. As an example, 1/τ = [0,1] is covered by the first case (for k = 1 and a k = a 0 = 1), and 1 = τ 2 = [0,2,1] is covered by the second case. Theorem 9. Let 0 < α < 1 be an irrational number. The characteristic word c α is a fixed point of some nontrivial morphism if and only if α is a Sturm number. Theorem 10. Let x {0,1} ω be a characteristic Sturmian word. If y is a pure morphic word in the orbit of x, then y {x,0x,1x,01x,10x}. 23