Right special factors in Sturmian subshifts

3.2. RIGHT SPECIAL FACTORS IN STURMIAN SUBSHIFTS 41

Definition 3.2.1. For ξ = [0;a₁ +1,a₂, . . .] ∈ (0,1) irrational, where (ai)i∈N ∈ N×N^N₊ set

R₀= R₀(ξ)≔(0), L₀ =L₀(ξ)≔ (1) R_k = R_k(ξ)≔τ^a1ρ^a2τ^a3ρ^a4· · ·τ^a2k−1ρ^a2k(0), L_k = L_k(ξ)≔τ^a1ρ^a2τ^a3ρ^a4· · ·τ^a2k−1ρ^a2k(1) for anyk∈N.

The former definition is consistent with our current understanding of a se-quence induced by an irrational number, as from Definition 2.3.2 we have (−1)^2k = 1 and hence

ω^2k₀ =τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(01)

=τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k(0)

=R_k and

ω^2k₁ =τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(10)

=τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(1)τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(0)

=τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k(1)τ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(0)

=L_kτ^a1ρ^a2τ^a3. . . τ^aj−1ρ^a2k−1(0).

As limn→∞ωⁿ_l exists for l ∈ {0,1} one can yield convergence of limk→∞R_k = limk→∞ω^2k₀ and limk→∞L_k =limk→∞ω^2k₁ . There is also a recursive representation ofR_k andL_k which will mainly be used form here on onwards.

Proposition 3.2.2. Letξ= [0;a1+1,a2, . . .]∈(0,1)be an irrational number. For all k∈N₊we have

|R_k|= q_2k,

R_k = R_k−1L_k. . .L_k

⏞ˉˉˉˉ⏟⏟ˉˉˉˉ⏞

a_2k

,

|L_k|=q_2k−1,

L_k =L_k−1R_k−1. . .R_k−1

⏞ˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉ⏞

a_2k−1

.

Proof. ForR_k this is exactly Lemma 2.3.5 for j = 2kandl = 0 by the previous

discussion, whether forL_k one has j=2k,l= 1.

Corollary 3.2.3. Letξ = [0;a₁ +1,a₂, . . .] ∈ (0,1) be irrational, k ∈ N+, n ∈ {0,1, . . . ,a2(k+1)−1}and m∈ {0,1, . . . ,a2(k+1)−1−1}. The words

R_kL_k+1. . .L_k+1

⏞ˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉ⏞

n

and L_k+1R_k. . .R_k

⏞ˉˉˉˉ⏟⏟ˉˉˉˉ⏞

m

are right special.

3.2. RIGHT SPECIAL FACTORS IN STURMIAN SUBSHIFTS 43 Proof. That they are right special is due to Proposition 3.2.2, as (Rk)₀ = 0 and (L_k)0= 1 for allk∈N. For the remaining part observe

S^{|Lk|+a2(k+1)−1|Rk|+(a2(k+1)−(n+1))|Lk+1|}(Rk+1)= R_kL_k+1. . .L_k+1

⏞ˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉ⏞

n

,

S^{|Rk−1|+a2k|Lk|+(a2(k+1)−1−(m+1))|Rk|}(Lk+1)= L_kR_k. . .R_k

⏞ˉˉˉˉ⏟⏟ˉˉˉˉ⏞

m

.

Remark 3.2.4. For a Sturmian subshiftXof slopeξthe sequencesx,ywill usually denote the unique infinite words with x|_|Rk|= R_k andy|_|Lk|= L_k, for allk ∈N+. As x = limn→∞τ^a1ρ^a2τ^a3. . . ρ²ⁿ⁻¹(01) and (S^|Lk−1|L_k)||R_k−1| = R_k−1 for allk ∈ N+, both x,y∈X and hence, by minimality,X = Xx = Xy.

With that the notions α-repetetive, α-repulsive and α-finite forα > 1 can be connected to the factors R_k and L_k, k ∈ N, of a Sturmian subshift of slope ξ and the following theorem, preceded by a remark, establishes link a if ξ well-approximable ofα-type.

Remark 3.2.5. An analogue of Theorem 3.2.6 for α = 1 of a Sturmian subshift of slope ξ is given in [47] in Lemma 4.9 via Remark 2.2.17. It states that 1-repulsive (or equivalent 1-finite) is equivalent to the continued fraction expansion ofξbeing bounded. Thus Theorem 3.2.6 can be seen to deal with the case, when the continued fraction expansion ofξis unbounded.

Theorem 3.2.6. Forα >1andξ∈[0,1]irrational, the following are equivalent.

1. The Sturmian subshift of slopeξisα-repetitive.

2. The Sturmian subshift of slopeξisα-repulsive.

3. The Sturmian subshift of slopeξisα-finite.

4. The Sturmian subshift of slopeξis well-approximable ofα-type, i.e. ξ∈Θα. The proof of Theorem 3.2.6 is divided into the following implications: 1⇒2

⇒4, 4⇒1, 4⇒3. Note that 3⇔2 is due to Theorem 2.2.16.

Proof of Theorem 3.2.6. 1⇒2: Assume that the statement is false, in which case eitherℓα = 0 orℓα = ∞. First we consider the caseℓα = 0. By definition ofℓα, there exist wordsW,w∈ L(X) such thatwis a prefix and suffix ofW,W ≠w≠ ∅ and

1≤ |W| − |w| ≤

⌊︄ |w|^1/α 2^1/αR^1/αα

⌋︄

and R(n)≤2R_αn^α, (3.3)

for alln≥ |w|. Further, for alli∈ {1,2, . . . ,|w|}, we have that

wi =Wi =Wi+|W|−|w|, (3.4)

where we recall thatwkandWk respectively denote thek-th letter ofwandW. By the property ofα-repetitive, for all wordsu∈ L(X) with

|u|=

⌊︄ |w|^1/α 2^1/αR^1/αα

⌋︄

,

we have thatuis a factor ofw. In particular, lettingξ∈Xandk∈N+, the factor (︂ξk, ξk+1, . . . , ξ_k+⌊|w|1/α2^−1/αR^−1/α_α ⌋

)︂,

ofξis a factor ofw. This together with (3.3) and (3.4) yields thatξk =ξk+|W|−|w| for allk∈N+, and thus,ξis periodic. This contradicts the aperiodicity and minimality ofX. Therefore, ifXisα-repetitive and notα-repulsive, thenℓ_α= ∞. For ease of notation set Bk = inf{Aα,n: n ≥ akqk−1}. By Proposition 3.2.2, for allk ∈ N+ we have that

W ≔L_k. . .L_k

⏞ˉˉˉˉ⏟⏟ˉˉˉˉ⏞

a2k

, w≔L_k. . .L_k

⏞ˉˉˉˉ⏟⏟ˉˉˉˉ⏞

a_2k−1

, W^′ ≔R⏞ˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉ⏞k−1. . .Rk−1 a2k−1

, w^′ ≔R⏞ˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉ⏞k−1. . .Rk−1 a_2k−1−1

(3.5) all belong to the languageL(X) and

|W| − |w|

|w|^1/α = |L_k|^1−1/α

(a2k−1)^1/α = q^1−1/α_2k−1 (a2k−1)^1/α, provided thata_2k ≠ 1. In the same manner

|W^′| − |w^′|

|w^′|^1/α = |R_k−1|^1−1/α

a_2k−1−1 = q¹_2(k−^−1/α₁₎ (a_2k−1−1)^1/α, provided thata2k−1 ≠1. Hence, fork ∈N+withak ≠1,

Bk ≤q^1−1/α_k−1 (ak−1)^−1/α. (3.6)

Thus, since by assumption ℓα = ∞, since Bk ≤ Bk+1, for all k ∈ N+, and since (qk)k∈N+ is an unbounded monotonic sequence, givenN ∈N₊there existsM ∈N₊ so thatajq^1−α_j−1 < N^−α, for all j ≥ M. For alln ∈ N+letm_(n) be the largest natural number so thatqm_(n) ≤ n. By Theorem 3.1.4, for alln∈N+, so thatm(n) ≥ M,

R(n)

n^α ≤ qm(n)+1+2qm(n)−1+qm(n)+1−qm(n)

n^α

3.2. RIGHT SPECIAL FACTORS IN STURMIAN SUBSHIFTS 45

≤ 2am(n)+1qm(n)+2qm(n)−1+qm(n)

q^α_m(n) ≤ 2

N^α + 2qm(n)−1

q^α_m(n) + qm(n)

q^α_m(n).

Hence, we have thatRα ≤ 2N^−α. However,N was chosen arbitrary and soRα =0, this contradicts the initial assumption thatXisα-repetitive.

2 ⇒ 4: Let [0;a₁ + 1,a₂, . . .] denote the continued fraction expansion of ξ. Since the Sturmian subshift X is α-repulsive and α > 1 we have that the continued fraction entries of ξ are unbounded, as it is mentioned in the latter part of Remark 3.2.5 that the continued fraction entries of ξ are bounded if and only if α = 1. In particular, infinitely often we have that an ≠ 1. Setting Bk = inf{Aα,n: n ≥ akqk−1}, as in (3.6), we have that Bk ≤ q^1−1/α_k−1 (ak − 1)^−1/α, for all k ∈ N₊ with ak ≠ 1. Since Bk ≤ Bk+1, there exists N ∈ N₊ so that, 2^α/ℓ^α_α ≥ (an − 1)q^1−α_n−1, for all n ≥ N with an ≠ 1. Hence, since the sequence (qn)n∈N₊is an unbounded monotonic sequence and since,X isα-repulsive,

A_α(ξ)=lim sup

n→∞

anq^1−α_n−1 ≤ 2^α ℓ^α_α <∞.

It remains is to show that A_α(ξ) > 0. We have observed that if the Sturmian subshiftX isα-repulsive, then the continued fraction entries ofξ are unbounded.

In particular, infinitely often we have that an ≠ 1. Thus, letting W,w,W^′,w^′ be as in (3.5), if A_α(ξ) = 0, then Bk = 0, for all k ∈ N₊, and hence ℓ_α = 0. This contradicts the assumption that Xisα-repulsive. Hence, if the Sturmian subshift X isα-repulsive, thenAα(ξ)>0.

4 ⇒ 1: Let m_(n) denotes the largest integer so that qm_(n) < n. Since A_α(ξ) <

∞, there exists a constant c > 1 so that am+1 ≤ cq^α−1_m , for all m ∈ N+. By Theorem 3.1.4 and the recursive definition of the sequence (qn)n∈N₊, we have for alln∈N₊,

R(n)≤ R(qm_(n))+am_(n)+1qm_(n)

= 2am_(n)+1qm_(n)+qm_(n)−1+2qm_(n) −1

≤ 2cq^α_m(n)+qm_(n)−1+2qm_(n)

≤ (2c+3)n^α.

In particular, ifξis well-approximable ofα-type thenRαis finite. Further, by The-orem 3.1.4, the recursive definition of the sequence (qn)n∈N+ and the assumption thatAα(ξ)>0, we have that

Rα ≥lim sup

k∈N+

R(qk)

q^α_k =lim sup

k∈N+

qk+1+2qk−1

q^α_k ≥ lim sup

k∈N+

ak+1qk

q^α_k = Aα(ξ)>0.

That is, ifξis well-approximable ofα-type, then 0<Rα.

4⇒3: By Proposition 3.2.2 and the definition ofQ(n), we haveQ(qn)≥an+1

and so

Qα= lim sup

n→∞

Q(n)

n^α−1 ≥ lim sup

n→∞

Q(qn)

q^α−1_n ≥ lim sup

n→∞

an+1

q^α−1_n = Aα(ξ)> 0.

Thus, ifξis well-approximable of α-type andX was notα-finite, then Qα would be infinite. By way of contradiction assume thatξis well-approximable ofα-type andXand thatQα =∞. This means there exists a sequence of tuples ((nk,pk))k∈N+

of natural numbers such that the sequences (nk)k∈N₊ and (pk)k∈N₊ are strictly in-creasing and lim

n→∞pkn^1−α_k = ∞and for eachk∈N+there exists a wordW(k)∈ L(X) with|W_(k)|= nkandW_(k)W_(k)· · ·W_(k)

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

pk

∈ L(X). For a fixedk∈N+, setting

W = W_(k)W_(k)· · ·W_(k)

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

pk

and w= W_(k)W_(k)· · ·W_(k)

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

p_k−1

,

we have

|W| − |w|

|w|^1/α = n^1−1/α_k (pk −1)^1/α =

(︄ pk

pk−1 n^α−1_k

pk

)︄^1/α

= (︄ pk

pk−1

(︂pkn^1−α_k )︂−1)︄1/α

.

This latter value converges to zero askincreases to infinity. Therefore,ℓ_α = 0 and soX is not α-repulsive. This is a contradiction as have we already seen that ξ is well-approximable ofα-type if and only ifXisα-repulsive.

As Sturmian words are characterised by their right special factors we will pay more attention to them in the following.

Definition 3.2.7. Letu∈X, we define the function bn(u)=

⎧

⎪⎪

⎨

⎪⎪

⎩

1 ifu|n is a right special word, 0 otherwise,

for alln∈N₊.

Corollary 3.2.8. Letξ =[0;a1+1,a2, . . .]∈[0,1/2]and let X denote a Sturmian subshift of slopeξ. If x,y ∈ X are the unique infinite words such that x||R_m|= R_m and y||Lm|= L_m, for all m∈N+, then

1. bn(x) = 1 if and only if n = jq_2k−1 + q_2k−2 for some k ∈ N+ and some j∈ {0,1, . . . ,a_2k −1}, and

2. bm(y) = 1 if and only if m = iq_2l + q_2l−1 for some l ∈ N+ and some i ∈ {0,1, . . . ,a2l+1−1}.

3.2. RIGHT SPECIAL FACTORS IN STURMIAN SUBSHIFTS 47 Proof. Corollary 3.2.3 gives the reverse implication: Ifn= jq2k−1+q2k−2, for some k∈N+and some j∈ {0,1, . . . ,a2k−1}, thenbn(x)=1, and ifm=iq2l+q2l−1, for somel∈N₊andi∈ {0,1, . . . ,a_2l+1−1}, thenbm(y)=1.

For the forward implication, we show the result for bn(x) and bm(y) where n ≤ |R₁| = q₂ and where m ≤ |L₂| = q₃ after which we proceed by induction to obtain the general result.

By Remark 2.2.8 and Corollary 3.2.3 it follows b₁(x) = 1 and, for m ∈ {1,2, . . . ,q1 − 1}, that bm(y) = 0. Consider the word R₁ = x|_|R1|= x|_q2. Let n=kq₁+(j+1)q₀for somek∈ {0,1, . . . ,a₂−1}and some j∈ {1,2, . . . ,a₁}. For k= 0,

x|n= R₁|_n=(0,1,0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

j−1

).

By Proposition 3.2.2 and Corollary 3.2.3,

L₁ =(1,0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

a1

)

is a right special word and thus, by Remark 2.2.8, the set of all right special words of length at most|L₁|=a₁+1 is

{(1,0,0, . . . ,0

⏞ˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉ⏞

a1

),(0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

a1

),(0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

a₁−1

), . . . ,(0,0),(0)}.

Since there exists a unique right special word per length, it follows thatbn(x)=0.

In the case thatk ∈ {1, . . . ,a₂−1},

S^n−|L1|(x|n)=(0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

a₁−(j−1)

,1,0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

j−1

),

where we recall that a₁− (j− 1) ≥ 1. Since there exists a unique right special word per length and since

|S^n−|L1|(x|n)|=|S^n−|L1|(0,⏞ˉˉˉˉˉˉ0, . . . ,⏟⏟ˉˉˉˉˉˉ⏞0

a₁−(j−1)

,1,0,0, . . . ,0

⏞ˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉ⏞

j−1

)|=|L₁|,

it follows that bn(x) = 0. An application of Corollary 3.2.3 completes the proof forn≤ |R₁|=q₂.

Consider the wordL₂= y||L₂|=y|q₃. Let

m=lq₂+1+(i+1)q₁ =l|R₁|+1+(i+1)|L₁|

for somel ∈ {0,1, . . . ,a₃ −1}andi ∈ {0,1, . . . ,a₂−1}. By Proposition 3.2.2 we have

S^l|R1|+1(y|_m)=S^l|R1|+1(L₂|_m)=S(L₁R₀L₁L₁. . .L₁

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

i

)=R₀R₀. . .R₀

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

a₁+1=q₁=|L₁|

L₁L₁. . .L₁

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

i

and hence|S^l|R1|+1(y|_m)| = (i+1)|L₁| = (i+ 1)q1. By Remark 2.2.8 and Corol-lary 3.2.3,

S^{1+(a2−(i+1))|L1|}(x|q₂)=S^{1+(a2−(i+1))q1}(x|q₂)=S^{1+(a2−(i+1))q1}(R₁)=L₁L₁. . .L₁

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

i+1

is a right special word of length (i+1)|L₁| = (i+1)q₁. Since there is a unique right special word per length and since

R₀R₀. . .R₀

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

a₁+1=q₁=|L₁|

L₁L₁. . .L₁

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

i

≠L₁L₁. . .L₁

⏞ˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉ⏞

i+1

it follows thatbm(y) = 0. An application of Corollary 3.2.3 yields the result for m≤ |L₂|= q₃.

Assume there isr∈N+so that the result holds for all natural numbersn<q2r

andm<q2r+1, namely,

1. bn(x) = 1 if and only if n = jq2k−1 + q2k−2 for k ∈ {1,2, . . . ,r} and j ∈ {0,1, . . . ,a2k−1}, and

2. bm(y) = 1 if and only if m = iq_2l + q_2l−1 for l ∈ {1,2, . . . ,r} and i ∈ {0,1, . . . ,a_2l+1−1}.

The proof of 1. and 2. forr+1 follows in the same manner; thus below we only provide the proof of 1. forr+1. To this end consider the word

x||R_r+1|=R_r+1 =R_rL_r+1L_r+1. . .L_r+1

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

a_2(r+1)

.

By way of contradiction, suppose there exists an integernwith|R_r|<n≤ |R_r+1|,n is not of the form stated in Corollary 3.2.8 (1) andbn(x)= 1. For if not, the result is a consequence of Corollary 3.2.3. By our hypothesis, we have,

n=|R_r|+(a_2(r+1)−1−b)|Lr+1|+|L_r|+(a_2(r+1)−1−a)|Rr|, wherea∈ {1,2, . . .a2(r+1)−1}andb∈ {0,1, . . . ,a2(r+1)−1}. Set

v=R_rL_r+1L_r+1. . .L_r+1

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

a2(r+1)−1−b

L_rR_rR_r. . .R_r

⏞ˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉ⏞

a_2(r+1)−1−a

, w=R_rR_r. . .R_r

⏞ˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉ⏞

a

L_r+1L_r+1. . .L_r+1

⏞ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏟⏟ˉˉˉˉˉˉˉˉˉˉˉˉˉˉˉ⏞

b

,

3.3. SPECTRAL METRICS ON STURMIAN SUBSHIFTS 49

Im Dokument Finite and Infinite Rotation Sequences and Beyond (Seite 47-55)