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 theirtheoretical importance and applications to various fields of science. They admit several equiv-alent definitions and can even be described explicitly in arithmetic form. They also have someremarkable characterizations of geometrical nature (mechanical words, rotations), as well ascharacterizations via the monoid of Sturmian morphisms. In the course we consider somebasics on Sturmian words and related topics. In particular, we will consider various charac-terizations of Sturmian words, and their generalizations to larger alphabets (episturmian andArnaux-Rauzy words).

1 Complexity and Sturmian words

Definition 1. An infinite word w is ultimately periodic if there exist N, T ∈ N such thatwn+T = wn for each n ≥ N . Otherwise w is called aperiodic.

Definition 2. The factor complexity of an infinite word w is the function pw(n) counting thenumber of its factors of length n.

Properties:

• non-decreasing function, i.e., pw(n + 1) ≥ pw(n)

• p(1) = #Σ

• for a periodic word w there exists C ∈ N such that pw(n) ≤ C for all n ∈ N

Denote by Fn(w) the set of factors of a word w of length n.

Definition 3. The Rauzy graph Gn(w) of order n is the labelled graph with vertex set Fn(w)and the edge set

(bv, a, va)|a, b ∈ Σ, bva ∈ Fn+1(w).

A word y is a label of path from u to v in Gn(w) if and only if Fn+1(uy) ⊆ Fn+1(w) andv is a suffix of length n of uy.

Example. Consider the Fibonacci word

f = 01001010010010100101001001010010 . . . ,

1

01 10

00

1

01

0

Figure 1: The Rauzy graph G2(f) of the Fibonacci word.

defined as a fixed point of a morphism ϕ : 0 7→ 01, 1 7→ 0. On Fig. 1 one can see the Rauzygraph of the Fibonacci word of order 2.

Exercise. One has also define the Fibonacci word via recurrence relations. Namely, f =limn→∞ fn, where the sequence of words fn is defined by

f0 = 0, f1 = 01,

fn+1 = fnfn−1

Prove the equivalence of the two definitions.

Remark that the sequence of lengths of the words fn is the traditional sequence of Fi-bonacci numbers, i. e., |fn| = Fn, where F0 = 1, F1 = 2, . . . , Fn+1 = Fn + Fn−1. Besidesthat, |fn|1 = Fn−2 for n ≥ 2. Here and in the further text we denote by |v|a the number ofoccurrences of the letter a in v.

Theorem 1 (Morse and Hedlund, 1940). If pw(n) ≤ n for some n, then w is ultimatelyperiodic.

Proof. First notice that it is enough to prove the theorem for the case #Σ ≥ 2, since anyunary word is clearly periodic. Now notice that pw(n) ≤ n implies that pw(n) = pw(n+1) forsome n. Consider a Rauzy graph Gn(w). Since every factor of length n is a prefix of a factorof length n+1, there is at least one edge leaving each vertex. Since pw(n) = pw(n+1), there isexactly one edge leaving each vertex. This implies that the strongly connected components ofthe graph are simple circuits. Thus every infinite path will loop through a fixed circuit aftera while, and hence its label is eventually periodic. Since w is a label of a path, the theoremis proved.

Corollary 1. For each n and each aperiodic word w it holds pw(n) ≥ n+ 1.

Definition 4. An infinite word s is called Sturmian if ps(n) = n+ 1 for every n.

Thus Sturmian words are aperiodic infinite words with minimal complexity. Since ps(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 suchthat ua and ub are factors of w. Thus, s is Sturmian word if and only if it has exactly oneright special factor of each length.

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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 fand hence pf(2) = 3. The word 000 is not a factor of ϕ(f), since otherwise it is a prefix ofsome 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 eachlength.

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|, assumethat 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 isa 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 uand v are right special factors of the same length, and let z be the longest common suffix of uand 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

fn+2 = gnfRn f

Rn tn (n ≥ 2) (1)

where g2 = ε and for n ≥ 3

gn = fn−3 . . . f1f0, tn =

01, if n is odd

10, otherwise.

Here and in the further text vR denotes the reversal of v, i.e., for v = v1 . . . vn we havevR = vn . . . v1.

Observe that the first letter of fRn is opposite from the first letter of tn. This proves that

fRn is a right special factor for each n ≥ 2. Since a suffix of a right special factor is rightspecial, this proves that right special factors of any length exist.

Equation (1) is proved by induction. Indeed, f4 = ε(010)(010)10 and f5 = 0(10010)(10010)01.Next, it is easily checked by induction that

ϕ(uR)0 = 0(ϕ(u))R

for any word u. It follows that ϕ(fRn tn) = 0fR

n+1tn+1 and since ϕ(gn)0 = gn+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 ain 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 iscalled 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 ∩ Σn) ≤ n+ 1.

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Proof. Clearly, the statement holds for n = 0, 1, and it holds for n = 2 since X cannot containboth 00 and 11. Assume the converse, and let n ≥ 3 be the smallest number for which thestatement does not hold. Set Y = X∩Σn−1 and Z = X∩Σn. 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 existtwo distinct words y, y′ ∈ Y such that for all four words 0y, 1y, 0y′, 1y′ are in Z. Since y 6= y′,there exists a word x such that x0 and x1 are prefixes of y and y′. Therefore, both 0x0 and1x1 are in X , and thus X is unbalanced.

A word u = u1 . . . un is a palindrome if uR = un . . . u1 is equal to u. For example, 0110 ispalindrome.

Moreover, the following holds:

Proposition 2. Let X be a factorial set of binary words. Then X is unbalanced if and onlyif 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. Bydefinition, consider two words u, v ∈ X , |u| = |v|, such that ||u|1−|v|1| ≥ 2, and choose themwith 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 6= 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, zR is a suffix of p, but azR is not a suffix of p. Then bzR is a suffixof p with b 6= a. This gives a proper prefix 0za of u and a proper suffix bzR of p. If a = 0and b = 1, then 1zR1 and 0z0 are shorter words than u and v satisfying ||0z0|1− |1zR|1| ≥ 2,contradicting the minimality of length. Then u = 0z1u′′ and v = v′′1zR0 for two words u′′ andv′′ 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 isbalanced.

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 pw(n) ≥ n + 1 for all n by Morse and Hedlundtheorem. Since w is balanced, by Proposition 1 we have pw(n) ≤ n + 1 for all n. Therefore,pw(n) = n+ 1 and w is Sturmian by definition.

(i) ⇒ (ii) Assume that w is Sturmian and unbalanced, we will show that w is eventuallyperiodic. Since w is unbalanced, there exists a palindrome p such that 0p0, 1p1 are factorsof w by Proposition 2. Hence p is right special. Since w is Sturmian, there is a unique rightspecial factor of length n = |p|+ 1, which is either 0p or 1p. Assume 0p is right special, and1p 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 thatu = 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

1 p 1 v0 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 factorof u. Assume the contrary. Then there exist factorizations p = x0t, v = yz, p = t1y.

Since p = x0t is palindrome, the first factorization implies p = tR0xR, and hence the letterfollowing 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 whichcoincide. 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|x| . In other words, this is

a frequency on the letter 1 in the word.It is easy to see that

π(xy) =|x||xy|π(x) +

|y||xy|π(y).

Proposition 3. A factorial set of words X is balanced if and only if, for all x, y ∈ X, x, y 6= 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,

π(x)− π(y) =|z||x|π(z) +

|t||x|π(t)− π(y)

=|z||x|(π(z)− π(y)) +

|t||x|(π(t)− π(y)),

thus

|π(x)− π(y)| < 1

|x| +|t||x|

(

1

|y| +1

|t|

)

=1

|x| +1

|y| .

The inequality (2) implies that for an infinite balanced binary word w the sequence(π(prefn(w)))n≥1 is a Cauchy sequence and hence it converges as n→∞. The limit

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α = limn→∞

π(prefn(w))

is called the slope of the infinite balanced word w.

Exercise. Prove that π(f) = 1τ2, where τ = (1 +

√5)/2 is the golden ratio.

Proposition 4. Let w be an infinite balanced binary word with slope α. For every nonemptyfactor u of w, one has,

|π(u)− α| ≤ 1

|u| . (3)

More precisely, one of the following two inequalities holds: either

α|u| − 1 < |u|1 ≤ α|u|+ 1 for all u ∈ F (w) (4)

or

α|u| − 1 ≤ |u|1 < α|u|+ 1 for all u ∈ F (w) (5)

Proof. Given some ε, consider n0 such that for all n ≥ n0,

|π(prefnw)− α| ≤ ε.

Then, using (2),

|π(u)− α| ≤ |π(u)− π(prefnw)|+ |π(prefnw)− 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|u| +

1|v| , a contradiction with (2).

The proposition is proved.

Proposition 5. Let w be an infinite binary balanced word. The slope α of w is a rationalnumber if and only if w is eventually periodic.

Proof. If w is eventually periodic, then by definition there exist finite words u and v suchthat w = uvω. Then

π(uvn) =|u|1 + n|v|1|u|+ n|v| → π(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 oflength p either |u|1 = q or |u|1 = q+1. There are only finitely many occurrences of factors oflength p and with q+1 many 1’s, since otherwise there is a factor uzv of w with |u| = |v| = pand |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 inFp(y) has the same number of 1’s. Consider now an occurrence azb of a factor in y of lengthp + 1, with a and b being letters. Since |az|1 = |zb|1, one has a = b. This means that y isperiodic with period p. Consequently, w is eventually periodic.

6

-

6

t!!!!!!!!!!!!!!!

r

r r r

r r

r r r

r

r r r

r r

r r r

1 0 0 1 0 1 0 0 1

y = αx+ ρ

Figure 2: Mechanical words corresponding to the line y = αx+ ρ.

3 Geometrical characterization: mechanical and rota-

tional words

Given two real numbers α ∈ (0, 1), ρ ∈ [0, 1). We define two binary infinite words sα,ρ ands′α,ρ by

sα,ρ(n) = ⌊α(n+ 1) + ρ⌋ − ⌊αn+ ρ⌋

s′α,ρ(n) = ⌈α(n+ 1) + ρ⌉ − ⌈αn+ ρ⌉The word sα,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. Considerthe straight line with equation y = αx+ρ. The points with integer coordinates just below thisline are Pn = (n, ⌊αn+ ρ⌋). Two consecutive points Pn and Pn+1 are joined by a straight linesegment that is horizontal if sα,ρ(n) = 0 and diagonal if sα,ρ(n) = 1. The same observationholds 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 sα,ρ = s′α,ρexcept when αn+ ρ is an integer for some n ≥ 0. In this case

sα,ρ(n) = 0, sα,ρ(n) = 1

and, if n > 0,sα,ρ(n− 1) = 1, sα,ρ(n− 1) = 0.

Thus, if α is irrational, sα,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 sα,0(0) = ⌊α⌋ = 0,

sα,0(0) = ⌈α⌉ = 1, and

sα,0 = 0cα, s′α,0 = 0cα,

where the infinite word cα is called the characteristic word of slope α.

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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 severaltimes 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 isrational, then s is purely periodic. If α is irrational, then s is aperiodic.

Proof. Let s = sα,ρ be a lower mechanical word. The proof is similar for upper mechanicalwords. 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 sis 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 oneof the following holds:

• |prefns|1 ≤ ⌊αn + τ⌋ for all n,

• |prefns|1 ≥ ⌊αn + τ⌋ for all n.

Indeed, otherwise there exists a real number τ and two integers n, n + k such that|prefns|1 < ⌊αn + τ⌋ and |prefn+ks|1 > ⌊αn+ k + τ⌋ (or the symmetric relation). Thisimplies that |prefn+ks|1 − |prefns|1 ≥ 2 + ⌊αn+ k + τ⌋ − ⌊αn + τ⌋ > 1 + αk, a contradictionwith (3).

Set

ρ = infτ∣

∣ |prefns|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

|prefns|1 ≤ αn + ρ ≤ |prefns|1 + 1, (7)

8

since otherwise there is an integer n such that |prefns|1 + 1 < αn + ρ, and setting σ =|prefns|1 + 1− αn, one has σ < ρ and αn+ σ > |prefns|1, in contradiction with the definitionof ρ.

If s is aperiodic, then α is irrational by Proposition 5, and αn+ρ is an integer for at mostone n. By (7), either |prefns|1 = ⌊αn+ ρ⌋ for all n, and then s = sα,ρ, or |prefns|1 = ⌊αn+ ρ⌋for all but one n0, and |prefn0

s|1 +1 = αn0+ ρ. In this case, one has |prefns|1 = ⌈αn+ ρ− 1⌉for all n and s = s′α,ρ−1.

If s = uω is purely periodic with period |u| = p, then α = q/p with q = |u|1. Again|prefns|1 = ⌊αn+ ρ⌋ if αn+ ρ is never an integer (this depends on ρ).

If |prefns|1 = αn+ ρ for some n, we claim that |prefns|1 = ⌊αn+ ρ⌋ for all n. Assume theconverse. Then by (7), |prefms|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) =|prefms|1 − |prefns|1

m− n = α− 1

|y|

and

π(z) =|prefn+ps|1 − |prefms|1

n+ p−m = α +1

|z| ,

whence |π(y) − π(z)| = 1/|y| + 1/|z|, in contradiction with Proposition 2. Similarly, if 1 +|prefns|1 = αn+ ρ for some n, then |prefns|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 isrational. 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 themapping Rα from [0, 1) (identified with the unit circle) to itself defined by

Rα(x) = x+ α,

where x = x− ⌊x⌋ is the fractional part of x. Iterating R, one gets

Rnα(x) = x+ nα.

A straightforward computation shows that

⌊(n + 1)α+ ρ⌋ = 1 + ⌊nα + ρ⌋ ⇔ nα + ρ ≥ 1− α.

Considering a partition of [0, 1) into I0 = [0, 1− α), I1 = [1− α, 1), one gets

sα,ρ(n) =

0, if Rnα(ρ) = ρ+ nα ∈ I0,

1, if Rnα(ρ) = ρ+ nα ∈ I1.

(8)

One can also define I ′0 = (0, 1− α], I ′1 = (1− α, 1], the corresponding word is s′α,ρ.

9

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 subintervalI of [0, 1), the sets R(I) and R−1(I) are always intervals.

Now consider a binary word w = b0b1 . . . bm−1, we want to know whether w is a factor ofsome sα,ρ = a0a1 . . . . By (8), an+k = bi if and only if Rn+i(ρ) ∈ Ibi , or equivalently, if andonly if Rn(ρ) ∈ R−i(Ibi). Thus, for n ≥ 0,

w = anan+1 . . . an+m−1 ⇐⇒ Rn(ρ) ∈ Iw,

where Iw is the interval

Iw = Ib0 ∩R−1(Ib1) ∩ · · · ∩R−m+1(Ibm−1).

The interval Iw is non empty if and only if w is a factor of sα,ρ.

An infinite word w is uniformly recurrent, if for any its factor u there exist an integer Nsuch that for every i, u ∈ F (wi . . . wi+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

|π(u)− π(v)| ≤ |π(u)− α|+ |π(v)− α| < 1

|u| +1

|v| .

In view of Proposition 1#(X ∩ Σn) ≤ 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|u| , one has α − π(u) > − 1

|u| by Proposition 4, whence

β − π(u) = (β − α) + (α− π(u)) > 1|u| showing that u is not a factor of t.

Proposition 7. The set F (s) of factors of a Sturmian word s is closed under reversal.

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Proof. Set F (s)R = xR|x ∈ F (s). The set X = F (s) ∩ F (s)R is balanced. In view ofProposition 1, #(X ∩ Σn) ≤ n + 1, for each n, and since #(F (s) ∩ Σn) = n + 1, one hasX = F (s). Thus F (s)R = F (s).

Let an alphabet (Σ,≤) be a linearly ordered set. Then ≤ induces an order on the set ofinfinite words over Σ. Given two infinite words x and y over Σ, we say that x is lexicographicallyless than y, denoted x < y if there is an integer n such that xi = yi for i < n, and xn < yn.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

sα,ρ < sα,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 whichthis equality holds, then sα,ρ(n − 1) = 0 and sα,ρ′(n − 1) = 1 and sα,ρ(k) = sα,ρ′(k) fork < 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 sα,ρ = sα′,ρ′, sα,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, sα,ρ = sα′,ρ′ implies ρ = ρ′ by the previous proposition. Finally, consider the equalitysα,ρ = s′α′,ρ′. In the case αn+ρ′ is never an integer, then s′α,ρ′ = sα,ρ′ and the conclusion holds.In the case αn0 + ρ′ is an integer for some n0 we have sα,ρ′+(n0+1)α = s′α,ρ′+(n0+1)α showingρ = ρ′.

Sturmian words with intercept 0 have many interesting properties and are in some sensethe easiest to investigate. Remind that for an irrational number 0 < α < 1, the words sα,0and sα,1 differ only by their first letter, and that

sα,0 = 0cα, s′α,0 = 1cα,

where cα is the characteristic word of slope α. Equivalently, cα = sα,α = s′α,α.

Proposition 9. For every Sturmian word s, either 0s or 1s is Sturmian. A Sturmian words is characteristic if and only if 0s and 1s are both Sturmian.

Proof. The first claim follows from the fact that sα,ρ−α = asα,ρ for some a ∈ 0, 1.If s = sα,α = s′α,α is the characteristic word of slope α, then 0s = sα,0 and 1s = s′α,0 are

Sturmian.Conversely, the Sturmian words 0s and 1s have same slope, say α. Denote by ρ and ρ′ their

intercepts. Then their common shift s has intercept α+ ρ = α+ ρ′, and by Lemma 3, ρ = ρ′.Thus 0s = sα,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 6= b ∈ Σ. A factor is bispecial, if it is both right andleft special.

11

Proposition 10. The set of right special factors of a Sturmian word is the set of reversalsof the prefixes of the characteristic word of same slope.

Clearly, v is left special if and only if vR is right special. In other words, the propositionstates that the set of left special factors of a Sturmian word is the set of prefixes of thecharacteristic 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 1sare Sturmian and clearly have slope α. Thus

F (s) = F (cα) = F (0cα) = F (1cα)

by Proposition 6. Consequently, for each prefix v of cα, both 0v and 1v are factors of s. SinceF (s) is closed under reversal, this shows that vR is right special. Thus vR is the unique rightspecial factor of length |v|.

5 Iterated palindromic closure

We denote with w(+) the right palindromic closure of the word w, i.e., the shortest palindromewhich 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 isa prefix of v, then ψ(u) is a prefix of ψ(v). Thus, given an infinite word ∆ = ∆0∆1∆2 . . . onan alphabet Σ we can define

ψ(∆) = limn→∞

ψ(∆0∆1∆2 . . .∆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 letw = ψ(∆). Then the following statements hold:

1. The word w is closed under reversal, i.e., if v = v1v2 . . . vk is a factor of w, then so isits mirror image vk . . . v2v1.

2. The word w is uniformly recurrent.

3. If each letter a ∈ Σ appears in ∆ an infinite number of times, then for each prefix u ofw 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 statementis 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

in ψ(v). It follows that any factor of length (for example) 3|ψ(v)| contains an occurrence ofψ(v).

Finally suppose each a ∈ Σ appears infinitely many times in ∆. Thus for any letter a andany prefix v of ∆ there exists a prefix of ∆ of the form vv′a. From the definition of ψ we thenhave 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. Sinceψ(v) is a palindrome and w is closed under reversal, we obtain that for any prefix v of ∆ andfor 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 palindromicclosure of an infinite word containing infinitely many occurrences of 0’s and 1’s and viseversa.

6 Standard sequences

Let (d1, d2, . . . dn, . . . ) be a sequence of integers, with d1 ≥ 0 and dn > 0 for n > 1. To such asequence, we associate a sequence (sn)n≥1 of words by

s−1 = 1, s0 = 0, sn = sdnn−1sn−2(n ≥ 1).

The sequence (sn)n≥−1 is a standard sequence, and the sequence (d1, d2, . . . ) is its directivesequence. Observe that if d1 > 0, then any sn (n ≥ 0) starts with 0; on the contrary, if d1 = 0,then s1 = s−1 = 1, and sn starts with 1 for n 6= 0. Every s2n ends with 0, every s2n+1 endswith 1. This sequence defines a limit:

s = limn→∞

sn.

Example. We saw that the Fibonacci word can e defined via the recurrence relation fn =fn−1fn−2. So, its directive sequence is (1, 1, . . . ). Observe that the directive sequence (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 palindromicclosures.

Let x ∈ 0, 1ω be an infinite word containing infinitely many occurrences of 0’s and 1’s.Then x can be uniquely expressed as

x = 0d11d2 . . . ,

with d1 ≥ 0 and di > 0 for i > 1. We call the infinite sequence dl, d2, . . . , dn, . . . the integralrepresentation of x.

Theorem 4. Let ∆ be an infinite binary word containing infinitely many occurrences of 0’sand 1’s, and dl, d2, . . . , dn, . . . its integral representation. Then ψ(∆) is the infinite standardword with the directive sequence dl, d2, . . . , dn, . . . .

We refer for the proof to; A. de Luca: Sturmian words, structure, combinatorics and theirarithmetics, Theoret. Comput. Sci., 1997

13

Now we are going to relate standard words to characteristic words and to continued fractionexpansion of the slope of a characteristic word. For these we need the following morphisms:

E :0→ 11→ 0

, ϕ :0→ 011→ 0

, ϕ :0→ 101→ 0

.

From these, we get other morphisms, denoted G, G, D, D and defined by

G = ϕ E :0→ 01→ 01

, G = ϕ E :0→ 01→ 10

,

D = E ϕ :0→ 101→ 1

D = E ϕ :0→ 011→ 1

.

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:

E(sα,ρ) = s′1−α,1−ρ and E(s′α,ρ) = s1−α,1−ρ.

Proof. For n ≥ 0,

s′1−α,1−ρ = ⌈(1 − α)(n+ 1) + 1− ρ⌉ − ⌈(1− α)n+ 1− ρ⌉= 1− (⌈(αn− ρ⌉ − ⌈−α(n + 1)− ρ⌉) = 1− sα,ρ(n)

because −⌈r⌉ = ⌊r⌋ for every real number r. This proves the first equality; the second one issymmetric.

Lemma 6. Let 0 < α < 1. For any real number 0 ≤ ρ < 1, the following relations hold:

G(sα,ρ) = s α1+α

, ρ

1+α, G(sα,ρ) = s α

1+α, ρ+α1+α

, ϕ(sα,ρ) = s′1−α2−α

, 1−ρ

2−α

.

For any real number 0 < ρ ≤ 1, the following relations hold:

G(s′α,ρ) = s′ α1+α

, ρ

1+α, G(s′α,ρ) = s′ α

1+α, ρ+α1+α

, ϕ(s′α,ρ) = s 1−α2−α

, 1−ρ

2−α.

Proof. Let s = a0a1 . . . an . . . be an infinite word, the ai being letters. An integer n is theindex of the k-th occurrence of the letter 1 in s if a0 . . . an contains k letters 1 and an = 1. Ifs = sα,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

n =

⌈

k − ρα− 1

⌉

.

14

Similarly, if s = s′α,ρ and 0 < ρ ≤ 1, then

⌈α(n+ 1) + ρ⌉ = k + 1, ⌈αn+ ρ⌉ = k,

which implies n =⌊

k−ρα

⌋

.Set G(sα,ρ) = b0b1 . . . bi . . . , with bi ∈ 0, 1. Since every letter 1 in sα,ρ is mapped to 01

in G(sα,ρ), the prefix a0 . . . an of sα,ρ (where n is the index of the k-th letter 1) is mappedonto the prefix b0b1 . . . bn+k of G(sα,ρ). Thus the index of the k-th letter 1 in G(sα,ρ) is

n+ k =

⌈

k − ρ1+α

α1+α

− 1

⌉

.

This proves the first formula.Next, we observe that, for any infinite word x, one has

G(x) = 0G(x).

Indeed, the formula G(w) = 0G(w) is easily shown to hold for finite words w by induction.Furthermore, if a Sturmian word sα,ρ starts with 0 and setting sα,ρ = 0t, one gets t = sα,α+ρ.Altogether

G(sα,ρ) = s α1+α

, ρ+α1+α

for 0 ≤ ρ < 1. The proof of the other formula is similar. Finally, since ϕ = G E,ϕ(sα,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

E(cα) = c1−α G(cα) = cα/(1+α).

For m ≥ 1, define a morphism θm by

θm :0→ 0m−111→ 0m−110

.

It is easily checked that

θm := Gm−1 E G.

Corollary 3. For m ≥ 1, one has θm(cα) = c 1

m+α.

Proof. Since EG(cα) = c1/(1+α), the formula holds form = 1. Next, G(c1/(k+α)) = c1/(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 = m0 +1

m1 +1

m2+1

m3+1...

,

15

where m0, m1, . . . are integers, m0 ≥ 0, mi > 0 for i ≥ 1. We write it as

γ = [m0, m1, m2, . . . ].

The integers mi are called the partial quotients of γ. If the sequence (mi) is eventuallyperiodic, and mi = mk+i for i ≥ h, we write this by overlining the purely periodic part:

γ = [m0, m1, m2, . . . , mh−1, mh, . . . , mh+k−1].

Let α = [0, m1, m2, . . . ] be the continued fraction expansion of an irrational α with 0 <α < 1. If, for some β with 0 < β < 1,

β = [0, mi+1, mi+2, . . . ]

we agree to writeα = [0, m1, m2, . . . , mi + β]

Corollary 4. If α = [0, m1, m2, . . . , mi + β] for some irrational α, 0 < α, β < 1, one has

cα = θm1 θm2

. . . θm1(cβ).

Proposition 11. Let α = [0, 1 + d1, d2, . . . ] be the continued fraction expansion of someirrational α with 0 < α < 1, and let (sn) be the standard sequence associated to (d1, d2, . . . ).Then every sn is a prefix of cα and cα = limn→∞ sn.

Proof. By definition, sn = sdnn−1sn−2 for n ≥ 1. Define morphisms hn by

hn = θ1+d1 θd2 · · · θdn .We claim that

sn = hn(0), snsn−1 = hn(1), n ≥ 1.

This holds for n = 1 since h1(0) = 0d11 = s1 and h1(1) = 0d110 = s1s0. Next, for n ≥ 2,

hn(0) = hn−1(θdn(0)) = hn−1(0dn−11) = sdn−1

n−1 sn−1sn−2 = sn

andhn(1) = hn−1(0

dn−110) = snsn−1

For any infinite word x, the infinite word hn(x) starts with sn because both hn(0) and hn(1)start with sn. Thus, setting βn = [0, dn+1, dn+2, . . . ], one has cα = hn(cβ) by Corollary4 and thus cα starts with sn. This proves the first claim. The second is an immediateconsequence.

Example. The directive sequence for the Fibonacci word is (1, 1, . . . ). The correspondingirrational is 1/τ 2 = [0, 2, 1, 1 . . . , and indeed the infinite Fibonacci word is the characteristicword of slope 1/τ 2.

Exercise. Find a directive sequence for the characteristic Sturmian word of slope α =√3−12

.

Proposition 11 has several interesting consequences, in particular the relation to fixedpoints and the powers that may appear in a Sturmian word. Let x be an infinite word. Forw ∈ F (x), the index of w in x is the greatest integer d such that wd ∈ F (x), if such an integerexists. Otherwise, w is said to have infinite index.

16

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 ssuch that un is a factor of s for every n ≥ 1. For long enough factor un, the slope of this worddiffers 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 nonemptyfactor 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 ex-pansion of its slope has bounded partial quotients.

Proposition 13. A finite standard word sn is primitive.

Proof. First we prove that||sn||sn−1|1 − |sn|1|sn−1|| = 1.

We prove that by induction, and using that

||s1||s0|1 − |s1|1|s0|| = |(d1 + 1) · 0− 1| = 1.

So, by the definition of sn,

||sn||sn−1|1 − |sn|1|sn−1|| = |(dn|sn−1|+ |sn−2|)|sn−1|1 − (dn|sn−1|1 + |sn−2|1)|sn−1||= ||sn−2||sn−1|1 − |sn−2|1|sn−1|| = · · · = 1.

Lemma 7. Let (sn)n≥1 be the standard sequence of the characteristic word cα, with α =[0, 1+ d1, d2, . . . ]. For n ≥ 3, the word s1+dn+1

n is a prefix of cα, and s2+dn+1

n is not a prefix. Ifd1 ≥ 1, this holds also for n = 2.

Proof. We show that for n ≥ 3, (and for n ≥ 2, if d1 ≥ 1), one has

sn−1sn = sntn−1; with tn = sdn−1n−1 sn−2sn−1.

Indeed,

sn−1sn = sn−1sdnn−1sn−2 = sdnn−1s

dn−1

n−2 sn−3sn−2

= sdnn−1sn−2sdn−1−1n−2 sn−3sn−2 = sntn−1

provided dn−1 ≥ 1. Observe that tn−1 is not a prefix of sn, since otherwise sn = tn−1u forsome word u, and sn−1snu = s2n and sn is not primitive (see Ex.3 in Combinatorics on Wordslecture notes). Clearly, sn+1sn is a prefix of the characteristic word cα. Since

sn+1sn = sdn+1

n sn−1sn = s1+dn+1

n tn−1,

the word s1+dn+1

n is a prefix of cα, and since tn−1 is not a prefix of sn, the word s2+dn+1

n is nota prefix of cα.

17

Proof of Theorem 5. Since a Sturmian word has the same factors as the characteristic word ofsame slope, it suffices to prove the result for characteristic words. Let cα be the characteristicword of slope α = [0, 1 + d1, d2, . . . ]. Let (sn)n≥1 be the associated standard sequence.

To prove that the condition is necessary, observe that sdn+1

n is a prefix of cα for each n ≥ 1.Consequently, if the sequence (dn) of partial quotients is unbounded, the infinite word hasfactors of arbitrarily large exponent.

Conversely, assume that the partial quotients (dn) are bounded by some D and arguing bycontradiction, suppose that cα has unbounded index. Let r be some integer such that F (cα)contains a primitive word of length r with index greater than D+4. Among those words, letw be a word of length r of maximal index. Let d+1 be the index of w. Then d ≥ D+3. Forthe proof we need two claims.

Claim (1). The characteristic word has prefixes of the form wd, with d ≥ D + 3.

Proof of Claim (1): Indeed, if wd+1 is a prefix of cα, we are done. Otherwise, consider anoccurrence of wd+1. Set w = za with a a letter, and let b be the letter preceding the occurrenceof wd+1. If b = a, replace w by az and proceed. The process will stop after at most |w| − 1steps because either a prefix of cα is obtained, or because otherwise w would occur in cα atthe power d + 2. Thus, we may assume b 6= a. Thus b(za)d+1 is a factor of cα. This impliesthat a(za)d and b(za)d are factors, so wd is a right special factor, and therefore it is a prefixof cα. Claim (1) is proved.

Claim (2). If wd is a prefix of the characteristic word cα, then w is one of the standard wordssn.

Proof of Claim (2): Indeed, set e = d − 2, so that e ≥ D + 1. Let n be the greatest integersuch that sn is a prefix of we+1. Then we+1 is a prefix of sn+1 = sdn+1

n sn−1, thus also of s1+dn+1

n .This shows that

(1 +D)|w| ≤ (1 + e)|w| ≤ (1 + dn+1)|sn| ≤ (1 +D)|sn|,

whence |w| ≤ |sn|. Now, since both we+2 and s1+dn+1

n are prefixes of cα, one is a prefix of theother. If we+2 is the shorter one, then |we+2| = |we+1| + |w| ≥ |sn| + |w|. Thus, we+2 ands1+dn+1

n share a common prefix of length |sn|+ |w|. Consequently, w and sn are powers of thesame word, and since they are primitive, they are equal. If s1+dn+1

n is the shorter one then,since (1 + e)|w| ≤ (1 + dn+1)|sn|,

|s1+dn+1

n | = |sn|+ dn+1|sn| ≥ |sn|+dn+1

1 + dn+1(1 + e)|w| ≥ |sn|+ |w|,

and the same conclusion holds. Claim (2) is proved.

If follows that s1+en is a prefix of cα and, since e ≥ D+1 ≥ dn+1+1, also s2+dn+1

n is a prefixof cα, contradicting Lemma 7.

We say that an infinite word w ∈ Σω is critical if there exists a real number Ω > 1, calledthe critical exponent of w, such that the follows conditions hold:

(i) if u ∈ Σ+, t > 1 and ut ∈ F (w), then t ≤ Ω,(ii) for all ε > 0 there exists a v ∈ Σ+ and a rational number r > Ω − ε, such that

vr ∈ F (w).The following theorem establishes critical exponents for fixed points of morphisms:

18

Theorem 6. Let 0 < α < 1 with α = [0, a0, a1, . . . , am], where a0, . . . am are positive integersand am > a0. Then a Sturmian word of slope α has a critical exponent Ω, where

Ω = max1≤t≤m

[2 + at, at−1, . . . , a1, am, . . . , a1]

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 whereΩ = 2 + [1, 1, 1, 1, . . . ] = 3 +

√5−12

= 3, 618034 . . .

For irrational α ∈ (0, 1), α = [0, a1, a2, . . . ], the associated rational approximants pn/qnare defined by

p0 = 0, p1 = 1, pn = anpn−1 + pn−2,

q0 = 1, q1 = a1, qn = anqn−1 + qn−2.

Theorem 7. The critical exponent of a Sturmian sequence s of slope α is given by

Ω = maxa1, 2 + supn∈Nan+1 + (qn−1 − 2)/qn.

7 Sturmian morphisms

In this section the identity morphism Id and the morphism E that exchanges the letters 0and 1 will be called trivial morphisms.

A morphism f is Sturmian if f(s) is a Sturmian word for every Sturmian word s. Since anerasing morphism can never be Sturmian, all morphisms considered here are assumed to benonerasing. The trivial morphisms Id and E are Sturmian. The set of Sturmian morphisms isclosed under composition, and consequently is a submonoid of the monoid of endomorphismsof 0, 1.

The main result of this section is the characterization of Sturmian morphisms (Theorem8). Consider the morphisms

ϕ :0→ 011→ 0

, ϕ :0→ 101→ 0

.

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 threemorphisms. 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

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 and1v1 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) isfollowed by the letter 0. Consequently 1ϕ(v)01 and 0ϕ(v)00 are both factors of ϕ(x) whichis 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 10ϕ(v)1, and ϕ(01v1) and ϕ(11v1) both contain the factor 00ϕ(v)0. Con-sequently, ϕ(x) is not balanced.

Corollary 5. Let x be an infinite word and let f be a morphism that is a composition of Eand 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 becauseit contains both 00 and 11 as factors. However, since f is a characteristic word, the infiniteword 0f is Sturmian. Consequently ϕ(ϕ(0f)) = ϕ(01f) = 100ϕ(f) is Sturmian. Thus 00ϕ(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 compositionof E, ϕ and ϕ in any number and order. St is called the monoid of Sturm and by Proposition14 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 andf 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 forf = 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, byProposition 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 nota Sturmian word. Observe that y is aperiodic, since otherwise ϕ(y) is eventually periodicthus it is not Sturmian. Thus y = g(x) is not balanced and contains two factors 0v0 and1v1 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. Sinceϕ(0v0) = 10ϕ(v)10 and ϕ(1v1) = 0ϕ(v)0, both 10ϕ(v)1 and 00ϕ(v)0 are factors of ϕ(y) andthus 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 onlyif g is Sturmian.

Proof. Assume first that g is Sturmian. Since f is a composition of E, ϕ and ϕ, the morphismf g is Sturmian by Proposition 14. Conversely, if f g is Sturmian, then for every Sturmianword x, the infinite word f g(x) is Sturmian and, by Lemma 8, the infinite word g(x) isSturmian. This means that g is Sturmian.

A morphism f is locally Sturmian if there exists at least one Sturmian word x such thatf(x) is a Sturmian word.

20

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 ofSturmian morphisms.

The length of a morphism f is the number ‖f‖ = |f(0)| + |f(1)|. The proof of Theorem8 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) bothstart 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 endwith the same letter. Suppose f(0) starts with the letter 0. Then f(1) starts with the letter1. If f(0) ends with 1 then f(1) ends with 0. But in this case f(01) contains a factor 11 andf(10) contains a factor 00. Thus the image of any Sturmian word contains the two factors 00and 11 which means that f is not locally Sturmian.

Otherwise f(0) ∈ 00, 1∗0 ∪ 0 and f(1) ∈ 10, 1∗1 ∪ 1, and we prove the result byinduction on ‖f‖.

If ‖f‖ = 3, then f(a) = cc and f(b) = d for letters a, b, c, d, a 6= b, and since any Sturmianword x contains the two factors an+1 and banb for some integer n, f(x) contains (cc)n+1 andd(cc)nd and thus is not Sturmian.

Arguing by contradiction, suppose that ‖f‖ ≥ 4 and f is locally Sturmian. Let x be aSturmian 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 wordf(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 everyblock of 0 between two consecutive occurrences of 1 is either 0m or 0m+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 0m+1 and 10m1 are factors of x, the words f(0m+1)and f(10m1) are factors of f(x). But

f(0m+1) = u00vf(0m−1)u00v = uw1v, f(10m1) = r1f(0m−1)u00v1s = rw2s

for suitable w1, w2, and one has |w1| = |w2| and ||w1|1 − |w2|1| = 2, a contradiction.Consequently f(0) = (01)n0 for some integer n ≥ 0.Since 10m1 and 10m+11 are factors of x, the infinite word f(x) contains the two factors

10m1 and 10m+11 if n = 0, and the two factors 101 and 1001 if n 6= 0. Set p = m if n = 0,and p = 1 if n 6= 0. Then in both cases, f(x) contains the factors 10p1 and 10p+11, and inboth cases 1 ≤ p ≤ m.

Since f(1) does not contain the factor 11, there exist an integer k ≥ 0, and integersm1, . . . , mk ∈ 0, 1 such that

f(1) = 10p+m110p+m21 . . . 10p+mk1.

Consider a new alphabet B = a, b and two morphisms ρ, η : B∗ → 0, 1∗

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ρ :a→ 01b→ 0p1

, η :a→ (01)n0b→ 0p1

.

We show that there exists a word u over B such that f(ρ(b)) = η(bub).(i) If n = 0, set u = am1bam2b . . . bamk . Since f(1) 6= 1, one has f(1) = 1η(u)0p1. Thus

f(ρ(b)) = f(0p1) = η(bub).(ii) If n 6= 0 and m1 = · · · = mk = 0, set u = bk+n−1. Since f(1) = (10)k1, one gets

η(u) = (01)k+n−1 and f(ρ(b)) = f(01) = η(bub).(iii) Otherwise n 6= 0 and mi = 1 for at least one integer i, 1 ≤ i ≤ k. Thus there exist

integers t ≥ 2, n1, . . . , nt such that

f(1) = 1(01)n10(01)n20 . . . (01)nt−10(01)nt.

Since f(01) starts with (01)n+1, one has n1 ≥ 0, ni ≥ n for 2 ≤ i ≤ t − 1 and nt ≥ 1. Setu = bn1abn2−na . . . bnt−1−nabnt−1. Then, again, f(ρ(b)) = f(01) = η(bub).

Define a morphism g : B∗ → B∗ by

g :a→ ab→ 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 wordx′ over B such that ρ(x′) = x.

Thus there exists a (unique) infinite word y′ over B such that

xρ←−−− x′

f

y

y

g

f(x) ←−−−η

y′

Identifying a with 0 and b with 1, one has ρ = (ϕ E)p. If n = 0 then η = ρ. If n 6= 0,then p = 1, so η = ϕ E (E ϕ)n. Thus since x and f(x) are Sturmian, the words x′ andy′ 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 ga morphism by g(0) = u and g(1) = v. Then f = ϕ g and, by Lemma 8, g(x) is a Sturmian

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word. Next, ‖f‖ = ‖g‖+ |uv|0 and |uv|0 > 0. Otherwise, f(0) = ϕ(u) and f(1) = ϕ(v) wouldcontain only 0 and f(x) = 0ω would not be Sturmian. Thus ‖g‖ < ‖f‖ and the result followsby induction.

If f(0) and f(1) both end with 0, the same argument holds with ϕ instead of ϕ, and iff(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 standardmorphisms. As an example, we know that the morphism ϕ fixes the infinite Fibonacci wordf . We say that a morphism h fixes an infinite word x if h(x) = x. In this case, x is a fixedpoint of h.

For the description of characteristic words which are fixed points of morphisms, we intro-duce a special set of irrational numbers. A Sturm number is a number α that has a continuedfraction expansion of one of the following kinds:

(i) α = [0, 1, a0, a1, . . . , ak], with ak ≥ a0,(ii) α = [0, 1 + a0, a1, . . . , ak] with ak ≥ a0 ≥ 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 aSturm 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 ak = a0 = 1), and1 = τ 2 = [0, 2, 1] is covered by the second case.

Theorem 9. Let 0 < α < 1 be an irrational number. The characteristic word cα is a fixedpoint of some nontrivial morphism if and only if α is a Sturm number.

Theorem 10. Let x ∈ 0, 1ω be a characteristic Sturmian word. If y is a pure morphic wordin the orbit of x, then y ∈ x, 0x, 1x, 01x, 10x.

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