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Numeration systems as dynamicalsystems
Teturo Kamae(Osaka City University, 558-8585 Japan)
Abstract: A numeration system originally implies a digitizationof real numbers, but in this paper it rather implies a compactificationof real numbers as a result of the digitization.By definition, a numeration system with G, where G is a non-
trivial closed multiplicative subgroup of R+, is a nontrivial com-pact metrizable space Ω admitting a continuous (λω + t)-action of(λ, t) ∈ G × R to ω ∈ Ω, such that the (ω + t)-action is strictlyergodic with the unique invariant probability measure µΩ, which isthe unique G-invariant probability measure attaining the topologicalentropy | log λ| of the transformation ω 7→ λω for any λ = 1.We construct a class of numeration systems coming from weighted
substitutions, which contains those coming from substitutions or β-expansions with algebraic β. It also contains those with G = R+.We obtained the ζ-function ζΩ of the numeration systems Ω com-
ing from weighted substitutions. For α ∈ C with 0 < Re(α) < 1,there exists a nontrivial adapted α-homogeneous cocycle on Ω if andonly if ζΩ has a pole at α. These cocycles are generalizations of thefractal functions of Peano type. Moreover, in the case G = R+, thesecocycles define self-similar processes of order α under µΩ. We discussone of them with α = 1/2. For a pole α of ζΩ with Re(α) < 0,there exsists a nontrivial adapted α-homogeneous cocycle on the setof integer points I(Ω) in Ω, which is proved to be a coboundary. Theimage of I(Ω) under this coboundary function becomes a fractal set
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of Rauzy type.This paper is based on [?], changed the way of presentation and
modified on a large scale.
1 Numeration systems
By a numeration system, we mean a compact metrizable space Ωwith at least 2 elements as follows:(♯1) There exists a nontrivial closed multiplicative subgroup G of
R+ and a continuous action λω + t of (λ, t) ∈ G × R to ω ∈ Ω suchthat λ′(λω + t) + t′ = λ′λω + λ′t+ t′.(♯2) The (ω + t)-action of t ∈ R to ω ∈ Ω is strictly ergodic with
the unique invariant probability measure µΩ called the equilibriummeasure on Ω. Consequently, it is invariant under the (λω+ t)-actionof (λ, t) ∈ G× R to ω ∈ Ω as well.(♯3) For any fixed λ0 ∈ G, the transformation ω 7→ λ0ω on Ω has
the | log λ0|-topological entropy. For any probability measure ν on Ωother than µΩ which is invariant under the λω-action of λ ∈ G to ω,and 1 = λ0 ∈ G, it holds that
hν(λ0) < hµΩ(λ0) = | log λ0|.
The (ω + t)-action of t ∈ R to ω ∈ Ω is called the additive actionor R-action, while the λω-action of λ ∈ G to ω ∈ Ω is called themultiplicative action or G-action.Note that if Ω is a numeration system, then Ω is a connected space
with the continuum cardinality. Also, note that the multiplicativegroup G as above is either R+ or λn; n ∈ Z for some λ > 1.Moreover, the additive action is faithful, that is, ω + t = ω impliest = 0 for any ω ∈ Ω and t ∈ R.This is because if there exist ω1 ∈ Ω and t1 = 0 such that ω1+ t1 =
ω1, then take a sequence λn inG such that λn → 0 and λnω1 convergesas n → ∞. Let ω∞ := limn→∞ λnω1. For any t ∈ R, let an be asequence of integers such that anλnt1 → t as n→ ∞. Then we have
ω∞ + t = limn→∞
(λnω1 + λnant1)
= limn→∞
λn (ω1 + ant1) = limn→∞
λnω1 = ω∞.
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Thus, ω∞ becomes a fixed point of the (ω + t)-action of t ∈ R toω ∈ Ω. S ince this action is minimal, we have Ω = ω∞, cotradictingwith that Ω has at least 2 elements.
An example of a numeration system is the set 0, 1Z with theproduct topology divided by the closed equivalence relation ∼ suchthat
(· · ·α−2, α−1;α0, α1, α2 · · · ) ∼ (· · · β−2, β−1; β0, β1, β2 · · · )
if and only if there exists N ∈ Z ∪ ±∞ satisfying that αn =βn (∀n > N), αN = βN + 1 and αn = 0, βn = 1 (∀n < N) or thesame statement with α and β exchanged. Let Ω(2) := 0, 1Z/ ∼ andthe equivalence class containing (· · ·α−2, α−1;α0, α1α2 · · · ) ∈ 0, 1Zis denoted by
∑∞n=−∞ αn2
n ∈ Ω(2). Then, Ω(2) is an additive topo-logical group with the addition as follows:
∞∑n=−∞
αn2n +
∞∑n=−∞
βn2n =
∞∑n=−∞
γn2n
if and only if there exists (· · · η−2, η−1; η0, η1, η2 · · · ) ∈ 0, 1Z satisfy-ing that
2ηn+1 + γn = αn + βn + ηn (∀n ∈ Z).This is isomorphic to the 2-adic solenoidal group which is by def-
inition the projective limit of the projective system θ : R/Z → R/Zwith θ(α) = 2α (α ∈ R/Z).Moreover, R is imbedded in Ω(2) continuously as a dense additive
subgroup in the way that a nonnegative real number α is identifiedwith
∑∞n=−∞ αn2
n such that α =∑N
n=−∞ αn2n and αn = 0 (∀n > N)
for some N ∈ Z, while a negative real number −α with α as above isidentified with
∑∞n=−∞(1−αn)2
n. Then, R acts additively to Ω(2) bythis addition. Furthermore, G := 2k; k ∈ Z acts multiplicativelyto Ω(2) by
2k∞∑
n=−∞
αn2n =
∞∑n=−∞
αn−k2n.
Thus, we have a group of actions on Ω(2) satisfying (♯1), (♯2)and (♯3)with G := 2k; k ∈ Z and the equilibrium measure (1/2, 1/2)Z.
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Theorem 1. Ω(2) is a numeration system with G = 2n; n ∈ Z.
We can express Ω(2) in the following different way. By a partitionof the upper half plane H := z = x+ iy; y > 0, we mean a disjointfamily of open sets such that the union of their closures coincideswith H. Let us consider the space Ω(2)′ of partitions ω of H by opensquares of the form (x1, x2)× (y1, y2) with x2−x1 = y2−y1 = y1 andy1 ∈ G such that (x1, x2)× (y1, y2) ∈ ω implies
(x1, (x1 + x2)/2)× (y1/2, y1) ∈ ω (type 0)
and (1)
((x1 + x2)/2, x2)× (y1/2, y1) ∈ ω (type 1).
An example of ω ∈ Ω(2)′ is shown in Figure 1. For ω ∈ Ω(2)′,let (α0, α1, · · · ) be the sequence of the types defined in (1) of thesquares in ω intersecting with the half vertical line from +0 + i to+0 + i∞ and let (α−1, α−2, · · · ) be the sequence of the types of thesquares in ω intersecting with the line segment from +0 + i to +0.Then, ω is identified with
∑∞n=−∞ αn2
n. Note that replacing +0by −0, we get
∑∞n=−∞ βn2
n such that (· · ·α−2, α−1;α0, α1, · · · ) ∼(· · · β−2, β−1; β0, β1, · · · ).The topology on Ω(2)′ is defined so that ωn ∈ Ω(2)′ converges to
ω ∈ Ω(2)′ as n → ∞ if for every R ∈ ω, there exist Rn ∈ ωn suchthat limn→∞ ρ(R,Rn) = 0, where ρ is the Hausdorff metric betweensets R, R′ ⊂ H
ρ(R,R′) := maxsupz∈R
infz′∈R′
|z − z′|, supz′∈R′
infz∈R
|z − z′|. (2)
For ω ∈ Ω(2)′, t ∈ R and λ ∈ 2n; n ∈ R, ω + t ∈ Ω(2)′ andλω ∈ Ω(2)′ are defined as the partitions
ω + t := (x1 − t, x2 − t)× (y1, y2); (x1, x2)× (y1, y2) ∈ ω
and
λω := (λx1, λx2)× (λy1, λy2); (x1, x2)× (y1, y2) ∈ ω.
Let κ : Ω(2)′ → Ω(2) be the identification mapping defined above.Then, κ is a homeomorphism between Ω(2)′ and Ω(2) such that κ(ω+
4
x
y
ui
Figure 1: the tiling corresponding to · · · 01.101 · · ·
t) = κ(ω) + t and κ(λω) = λκ(ω) for any ω ∈ Ω(2)′, t ∈ R andλ ∈ 2n; n ∈ Z. Thus, Ω(2)′ is isomorphic to Ω(2) as a numerationsystem and will be identified with Ω(2).
We generalize this construction. Let A be a nonempty finite set.An element in A is called a color. An open rectangle (x1, x2)×(y1, y2)in H is called an admissible tile if
x2 − x1 = y1
is satisfied (see Figure 2). In another word, an admissible tile is a
5
x
Figure 2: admissible tiles
rectangle (x1, x2) × (y1, y2) in H such that the lower side has thehyperbolic length 1. Let R be the set of admissible tiles in H.A colored tiling ω is a subset of R× A such that
(1) R ∩R′ = ∅ for any (R, a) and (R′, a′) in ω with (R, a) = (R′, a′),and(2) ∪(R,A)∈ωR = H.
An element in R× A is called a colored tile. We denote
dom(ω) := R; (R, a) ∈ ω for some a ∈ A.
For R ∈ dom(ω), there exists a unique a ∈ A such that (R, a) ∈ ω,which is denoted by ω(R) and is called the color of the tile R (in ω).Let R = (x1, x2)× (y1, y2). We call y2/y1 the vertical size of the tileR which is denoted by S(R).Let Ω(A) be the set of colored tilings with colors in A. A topology
is introduced on Ω(A) so that a net ωnn∈I ⊂ Ω(A) converges toω ∈ Ω(A) if for every (R, a) ∈ ω, there exists (Rn, an) ∈ ωn such that
an = a for any sufficiently large n ∈ I and limn→∞
ρ(R,Rn) = 0,
where ρ is the Hausdorff metric defined in (2).
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For an admissible tile R := (x1, x2) × (y1, y2), t ∈ R and λ ∈ R+,we denote
R + t := (x1 + t, x2 + t)× (y1, y2)
λR := (λx1, λx2)× (λy1, λy2).
Note that they are also admissible tiles.For ω ∈ Ω(A), t ∈ R and λ ∈ R+, we define ω + t ∈ Ω(A) and
λω ∈ Ω(A) as follows:
ω + t = (R− t, a); (R, a) ∈ ωλω = (λR, a); (R, a) ∈ ω.
Thus, we define a continuous group action λω + t of (λ, t) ∈ R+ ×R to ω ∈ Ω(A). We construct compact metrizable subspaces ofΩ(A) corresponding to weighted substitutions which are numerationsystems. Though ♯A ≥ 2 is assumed in [?], we consider the case♯A = 1 as well.
2 Remarks on the notations
In this paper, the notations are changed in a large scale from theprevious papers [?], [?] and [?] of the author. The main changes areas follows:(1) Here, the colored tilings are defined on the upper half planeH, noton R2 as in the previous papers. The multiplicative action here agreewith the multiplication on H, while it agree with the logarithmicversion of the multiplication at one coodinate in the previous papers.Here, the tiles are open rectangles, not half open rectangles as in theprevious papers.(2) Here, we simplified the proof in [9] for the space of colored tilingscoming from weighted substitutions to be numeration systems byomitting the arguments on the topological entropy.(3) The roles of x-axis and y-axis for colored tilings are exchangedhere and in [?] from those in [?] and [?].(4) Here and in [?], the set of colors is denoted by A instead of Σ.
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Colors are denoted by a, a′, ai (etc.) instead of σ, σ′, σi (etc.).(5) Here and in [?], the weighted substitution is denoted by (σ, τ)instead of (φ, η).(6) Here and in [?], admissible tiles are denoted by R, R′, Ri, R
i
(etc.) instead of S, S ′, Si, Si (etc.).
(7) Here and in [?], the terminology “primitive” for substitutions isused instead of “mixing” in [?] and [?].
3 Weighted substitutions
A substitution σ on a set A is a mapping A → A+, where A+ =∪∞ℓ=1 Aℓ. For ξ ∈ A+, we denote |ξ| := ℓ if ξ ∈ Aℓ, and ξ with |ξ| = ℓ
is usually denoted by ξ0ξ1 · · · ξℓ−1 with ξi ∈ A. We can extend σ tobe a homomorphism A+ → A+ as follows:
σ(ξ) := σ(ξ0)σ(ξ1) · · ·σ(ξℓ−1),
where ξ ∈ Aℓ and the right-hand side is the concatenations of σ(ξi)’s.We can define σ2, σ3, · · · as the compositions of σ : A+ → A+.A weighted substitution (σ, τ) on A is a mapping A → A+× (0, 1)+
such that |σ(a)| = |τ(a)| and∑
i<|τ(a)| τ(a)i = 1 for any a ∈ A. Notethat σ is a substitution on A. We define τn : A → (0, 1)+ (n = 2, 3, ...)(depending on σ) inductively by
τn(a)k = τ(a)iτn−1(σ(a)i)j
for any a ∈ A and i, j, k with
0 ≤ i < |σ(a)|, 0 ≤ j < |σn−1(σ(a)i)|, k =∑h<i
|σn−1(σ(a)h)|+ j.
Then, (σn, τn) is also a weighted substitution for n = 2, 3, · · · .A substitution σ on A is called primitive if there exists a positive
integer n such that for any a, a′ ∈ A, σn(a)i = a′ holds for some iwith 0 ≤ i < |σn(a)|.For a weighted substitution (σ, τ) on A, we always assume that
the substitution σ is primitive. (3)
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We define the base set B(σ, τ) as the closed, multiplicative subgroupof R+ generated by the set
τn(a)i ; a ∈ A, n = 0, 1, · · · and 0 ≤ i < |σn(a)|such that σn(a)i = a
.
x
+1/94/9 4/9: :
+ +
−
+ + + +
− −
+ + + + + + + +- - - -
− −+
Figure 3: the weighted substitution in Example 1
Example 1. Let A = +,− and (σ, τ) be a weighted substitutionsuch that
+ → (+, 4/9)(−, 1/9)(+, 4/9)− → (−, 4/9)(+, 1/9)(−, 4/9),
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where we express a weighted substitution (σ, τ) by
a→ (σ(a)0, τ(a)0)(σ(a)1, τ(a)1) · · · (a ∈ A).
Then, 4/9 ∈ B(σ, τ) since σ(+)0 = + and τ(+)0 = 4/9. More-over, 1/81 ∈ B(σ, τ) since σ2(+)4 = + and τ 2(+)4 = 1/81. Since4/9 and 1/81 do not have a common multiplicative base, we haveB(σ, τ) = R+. This weighted substitution is discussed in the follow-ing sections. The repetition of this weighted substitution starting at+ is shown in Figure 3 by colored tiles. The substituted word of acolor is represented as the sequence of colors of the connected tiles inbelow in order from left. The horizontal (additive) sizes of tiles areproportional to the weights and the vertical (multiplicative) sizes arethe inverse of the weights.
Let G := B(σ, τ). Then, there exists a function g : A → R+ suchthat
g(σ(a)i)G = g(a)τ(a)iG (4)
for any a ∈ A and 0 ≤ i < |σ(a)|. Note that if G = R+, then wecan take g ≡ 1. In the other case, we can define g by g(a0) = 1 andg(a) := τn(a0)i for some n and i such that σn(a0)i = a, where a0 isany fixed element in A.Let (σ, τ) be a weighted substitution satisfying (3). Let G =
B(σ, τ). Let g satisfy (4). Let Ω(σ, τ, g)′ be the set of all elements ωin Ω(A) such that for any ((x1, x2)× (y1, y2), a) ∈ ω, we have(I) y1 ∈ g(a)G, and(II) (Ri, σ(a)i) ∈ ω holds for i = 0, 1, · · · , |σ(a)| − 1, where
Ri := (x1 + (x2 − x1)i−1∑j=0
τ(a)j, x1 + (x2 − x1)i∑
j=0
τ(a)j)
×(τ(a)iy1, y1).
A vertical line γ := x × (−∞,∞) is called a separating line ofω ∈ Ω(σ, τ, g)′ if for any (R, a) ∈ ω, R ∩ γ = ∅. Let Ω(σ, τ, g)′′ bethe set of all ω ∈ Ω(σ, τ, g)′ which do not have a separating line andΩ(σ, τ, g) be the closure of Ω(σ, τ, g)′′. Then, the action of G × Ron Ω(σ, τ, g) satisfies (♯1). We usually denote Ω(σ, τ, 1) simply byΩ(σ, τ).
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Remark 1. [?] A nontrivial primitive substitution σ : A → A+,where “nontrivial” means
∑a∈A |σ(a)| ≥ 2, is considered as a weighted
substitution in a canonical way. Let
M := (♯0 ≤ i < |σ(a)|; σ(a)i = a′)a,a′∈A
be the associate matrix. Let λ be the maximum eigen-value of Mand ξ := (ξa)a∈A be a positive column vector such that Mξ = λξ.Define weight τ by
τ(a)i =ξσ(a)iλξa
,
which is called the natural weight of σ. Thus, we get a weighted sub-stitution (σ, τ) which admits weight 1. We modify (σ, τ) if necessaryin the following way. If there exists a ∈ A with |σ(a)| = 1, so thata → (a′, 1) is a part of (σ, τ), then we replace all the occurrences ofa in the right hand side of “→” by a′ and remove a from A togetherwith the rule a → (a′, 1) from (σ, τ). We continue this process untilno a ∈ A satisfies |σ(a)| = 1. After that if there exist a, a′ ∈ A suchthat (σ(a), τ(a)) = (σ(a′), τ(a′)), then we identify them.For example, the 2-adic expansion substitution 1 → 12, 2 → 12
corresponds to the weighted substitution 1 → (1, 1/2)(1, 1/2). TheThue-Morse substitution 1 → 12, 2 → 21 corresponds to the weightedsubstitution 1 → (1, 1/2)(2, 1/2), 2 → (2, 1/2)(1, 1/2). The Fi-bonacci substitution 1 → 12, 2 → 1 corresponds to the weightedsubstitution 1 → (1, λ−1)(1, λ−2), where λ = (1 +
√5)/2.
The weighted substitution (σ, τ) obtained in this way satisfies thatB(σ, τ) = λn; n ∈ Z and that g in (4) can be defined by g(a) =ξa (a ∈ A). Dynamical systems coming from substitutions are dis-cussed by many authors (see [?], for example). Our weighted substi-tutions are a generalization of them.
Let (σ, τ) be a weighted substitution on A satisfying (3). Let gsatisfy (4). Consider Ω(σ, τ, g). We call the tile Ri in (II) the i-thchild of the tile (x1, x2)× (y1, y2), and (x1, x2)× (y1, y2) the mother ofRi. Note that the vertical size S(Ri) of Ri coincides with the inverseof the weight τ(a)i. If Rj is a child of Rj+1 for j = 0, 1, · · · , k − 1.Then, the tile R0 is called a k-th descendant of the tile Rk. If R0 is
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the i-th tile among the set of the k-th descendants of Rk counting as0, 1, 2, · · · from left, we call R0 the (k, i)-descendant of the tile Rk.In this case, we also say that Rk is the k-th ancestor of R0.
Theorem 2. The space Ω(σ, τ, g) is a numeration system with G =B(σ, τ).
Proof. We have already proved (♯1) and (♯2) in Theorem 3 of [?].Here we prove (♯3). Let Ω := Ω(σ, τ, g) and µΩ be the equilibriummeasure. Since µΩ is the unique invariant probability measure underthe additive action, it is also invariant under the multiplicative action.By Goodman [?], it is sufficient to prove that for any λ ∈ G with
λ = 1, the transformation ω 7→ λω on Ω has the metrical entropy| log λ| under µΩ, while it has the metrical entropy less than | log λ|under any other G-invariant probability measure.
Lemma 1. Let
Σ := ω ∈ Ω; ω has a separating lineΣ0 := ω ∈ Ω; y-axis is the separating line of ω.
Then, we have(i) Σ \Σ0 is dissipative with respect to the G-action, so that ν(Σ \
Σ0) = 0 for any G-invariant probability measure ν on Ω.(ii) For any ω ∈ Σ0, ω restricted to the right quarter plane (0,∞)×
(0,∞) and to the left quarter plane (−∞, 0)× (0,∞) are cyclic indi-vidually with respect to the G-action. Hence, Gω with respect to theG-action is either cyclic or conjugate to a 2-dimensional irrationalrotation with a multiplicative time parameter.(iii) Σ0 is a finite union of minimal and equicontinuous sets with
respect to the G-action. In fact, there is a mapping from the set ofpairs a ∈ A and i with 0 ≤ i < i+1 < |σ(a)| onto the set of minimalsets in Σ0.
Proof. (i) If the line x = u is the separating line of ω ∈ Ω, thenx = λu is the separating line of λω. Hence, Σ \ Σ0 is dissipative.
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(ii) Let ω ∈ Σ0. Denote by ω+ the restriction of ω to the rightquarter plane (0,∞)× (0,∞), while by ω− the restriction of ω to theleft quarter plane (−∞, 0)× (0,∞). Let (R±
i )i∈Z be the sequence oftiles in dom(ω) such that R±
i intersects with the upper half lines ofx = ±0, and R±
i is a child of R±i+1 for any i ∈ Z (± respectively). Let
a±i := ω(R±i ) be the colors of R±
i (± respectively). Define mappingsσ± from A to A by σ+(a) = σ(a)0 and σ−(a) = σ(a)|σ(a)|−1. Sinceσ±(a
±i ) = a±i−1 (i ∈ Z) (± respectively), the sequence (a±i )i∈Z is
periodic, which also implies that the vertical sizes S(R±i ) ofR
±i , which
coincide with the inverses of the weights τ(ai+1)±, are also periodicin i ∈ Z with the period, say r± which is the minimum period of(a±i )i∈Z (± respectively). Then, λ+ := τ r
+(a+0 )
−10 is the minimum
(multiplicative) cycle of ω+, while λ− := τ r−(a−0 )
−1
|σr− (a−0 )|−1is the
minimum (multiplicative) cycle of ω−, that is, λω± = ω± holds forλ = λ± and λ± is the minimum among λ > 1 with this property (±respectively).Therefore, ω is cyclic with respect to the G-action if λ+ and λ−
have a common multiplicative base. In this case, the minimum cycleof ω is the minimum positive number x such that x = (λ+)n =(λ−)m holds for some positive integers n,m. Otherwise, the G-actionto Gω is conjugate to an 2-dimensional irrational rotation with amultiplicative time parameter.(iii) We use the notations in the proof of (ii). Take any pair
(a, i) with a ∈ A and 0 ≤ i < i + 1 < |σ(a)|. Take any ω′ ∈ Ωhaving a tile R ∈ dom(ω′) with ω′(R) = a such that the y-axispasses in between the i-th child of R and the i + 1-th child of R.Let ψ(a, i) be the set of limit points of λω′ as λ ∈ G tends to ∞.Note that this does not depend on the choice of ω′. Then, ψ(a, i)is a closed G-invariant subset of Σ0. Moreover, since the sequence(σn
−(σ(a)i), σn+(σ(a)i+1))n=0,1,2,··· enter into a cycle after some time,
ψ(a, i) is minimal and equicontinuous with respect to the G-action.To prove that the mapping ψ is onto, take any ω ∈ Σ0. There exists
ωn ∈ Ω(σ, τ, g)′′ which converges to ω as n → ∞. We may assumethat there exists a pair (a, i) such that for any n = 1, 2, · · · , thereexists R ∈ dom(ωn) with a = ωn(R) such that the y-axis separetesthe i-th child of R and the i + 1-th child of R. Then, ω ∈ ψ(a, i),
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which proves that ψ is a mapping from the set of pairs (a, i) witha ∈ A and 0 ≤ i < i + 1 < |σ(a)| onto the set of minimal sets in Σ0
with respect to the G-action.
x
y
vertical sizes
1/(1− p)
vertical sizes
1/p
Figure 4: an element in Σ0 in Example 2
Example 2. Let p with 0 < p < 1 satisfy that log p/ log(1 − p) isirrational. Let (σ, τ) be a weighted substitution on A = 1 such that1 → (1, p)(1, 1 − p). Then, B(σ, τ) = R+ holds. Let Ω = Ω(σ, τ).In this case, elements in Σ0 are not periodic, but almost periodic asshown in Figure 4. Then, the dynamical system (Σ0, λ (λ ∈ R+)) isisomorphic to ((R/Z)2, Tλ (λ ∈ R+)) with
Tλ(x, y) = (x+ log λ/ log(1/p), y + log λ/ log(1/(1− p))).
Lemma 2. It holds that hµΩ(λ) = | log λ| for any λ ∈ G. Let λ =
1 and ν be any other G-invariant probability measure on Ω, thenhν(λ) < | log λ|.
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Proof. To prove the lemma, it is sufficient to prove the statementsfor λ > 1. Take any G-invariant probability measure ν on Ω whichattains the topological entropy of the multiplication by λ1 ∈ G withλ1 > 1, that is, hν(λ1) = log λ1. We assume also that the G-action toΩ is ergodic with respect to ν. Then by Lemma 1, either ν(Σ0) = 1 orν(Ω\Σ) = 1. In the former case, hν(λ) = 0 holds for any λ ∈ G sincethe G-action on Σ0 is equicontinuous by Lemma 1, which contradictswith the assumption. Thus, we have ν(Ω \ Σ) = 1.For ω ∈ Ω, let R0(ω) ∈ dom(ω) be the tile (x1, x2) × (y1, y2) such
that x1 ≤ 0 < x2 and y1 ≤ 1 < y2.Take a0 ∈ A such that
ν(ω ∈ Ω; ω(R0(ω)) = a0) > 0.
Take b0 := maxb ≤ 1; b ∈ g(a0)G (see (4)). Let
Ω1 := ω ∈ Ω; the set λ ∈ G; λω(R0(λω)) = a0is unbounded at 0 and ∞ simultaneously
Ω0 := ω ∈ Ω1; R0(ω) = (x1, x2)× (y1, y2)
with y1 = b0 and ω(R0(ω)) = a0 .
For ω ∈ Ω1, let λ0(ω) be the smallest λ ∈ G with λ > 1 such thatλω ∈ Ω0. Define a mapping Λ : Ω0 → Ω0 by Λ(ω) := λ0(ω)ω.For k = 0, 1, 2, · · · and i = 0, 1, · · · , |σk(a0)| − 1, let
P (k, i) := ω ∈ Ω0; λ0(ω)−1R0(λ0(ω)ω)
is the (k, i)-descendant of R0(ω)
(see Figure 5) and let
P := P (k, i); k = 1, 2, · · · , 0 ≤ i < |σk(a0)|
be a measurable partition of Ω0. Note that λ0(ω) = τ k(a0)−1i if
ω ∈ P (k, i).Since ν(Ω1) = 1 by the ergodicity and
Ω1 =∪
P (k,i)∈P
∪1≤λ<τk(a0)
−1i
λ∈G
λP (k, i),
15
x
y
sb0
s(3/13)b0
a0
a0
Figure 5: ω ∈ P (3, 4) with λ0(ω) = 13/3
there exists a unique Λ-invariant probability measure ν0 on Ω0 suchthat for any Borel set B ⊂ Ω, we have
ν(B) = C(ν)−1∑
P (k,i)∈P
∫ b0τk(a0)−1i
b0
ν0(λ−1B ∩ P (k, i))dλ/λ (5)
withC(ν) :=
∑P (k,i)∈P
− log τ k(a0)i ν0(P (k, i)) <∞ (6)
if G = R+ and
ν(B) = C(ν)−1∑
P (k,i)∈P
∑λ∈G
b0≤λ<b0τk(a0)
−1i
ν0(λ−1B ∩ P (k, i)) (7)
with
C(ν) :=∑
P (k,i)∈P
(− log τ k(a0)i/ log β) ν0(P (k, i)) <∞ (8)
16
if G = βn; n ∈ Z with β > 1.Since ∑
P (k,i)∈P
τ k(a0)i = 1 and∑
P (k,i)∈P
ν0(P (k, i)) = 1,
we have
Hν0(P) := −∑
P (k,i)∈P
log ν0(P (k, i)) · ν0(P (k, i))
≤ −∑
P (k,i)∈P
log τ k(a0)i · ν0(P (k, i)) (9)
by the convexity of − log x. The equality in (??) holds if and only if
ν0(P (k, i)) = τ k(a0)i (∀P (k, i) ∈ P). (10)
By (??)(??)(??), we have
Hν0(P) = −∑
P (k,i)∈P
log ν0(P (k, i)) · ν0(P (k, i)) <∞.
For any ω, ω′ ∈ Ω0 such that Λk(ω) and Λk(ω′) belong to thesame element in P for k = 0, 1, 2, · · · , the horizontal position ofR0(ω), say (x1, x2), coincides with that of R0(ω
′). Therefore, ω andω′ restricted to (x1, x2)× (0, b0) coincide. In the same way, if Λk(ω)and Λk(ω′) belong to the same element in P for any k ∈ Z, thenR0 := R0(ω) = R0(ω
′) holds and all the ancestors of R0 in ω and ω′
coincide as well as their colors. Therefore, ω and ω′ restricted to theregion covered by the ancestors of R0 coincide. Hence, if ω or ω′ doesnot have the separating lines, then ω = ω′ holds.Since ν(Σ) = 0, we have ν0(Σ ∩ Ω0) = 0. Hence, the above argu-
ment implies that P is a generator of the system (Ω0, ν,Λ). Thus,hν0(Λ) = hν0(Λ,P). It follows from (??) that
hν0(Λ) = hν0(Λ,P)
≤ Hν0(P)
≤ −∑
P (k,i)∈P
log τ k(a0)i · ν0(P (k, i)). (11)
17
The equality in the above that
hν0(Λ) = −∑
P (k,i)∈P
log τ k(a0)i · ν0(P (k, i))
holds if and only if (ΛnP)n∈Z is an independent sequence with respectto ν0 satisfying (??).Since
hν(λ1)/ log λ1 =hν0(Λ)∫
Ω0λ0(ω)dν0(ω)
=hν0(Λ)
−∑
P (k,i)∈P log τ k(a0)i · ν0(P (k, i)),
hν(λ1) ≤ log λ1 follows from (??), while the equality holds if and onlyif (ΛnP)n∈Z is an independent sequence with respect to ν0 satisfying(??). Let this probability measure on Ω0 be µ0 and the measure νin (??) for this µ0 be µ. We prove that µ is invariant under theadditive action, then by the uniqueness of such a measure ([?]), µ =µΩ follows.
We may identify ω ∈ Ω0 with the element in P∞−∞ := ∨∞
n=−∞ΛnP .Let ω+ be the elemnet of P∞
0 including ω, and ω− be the elemnetof P−1
−∞ including ω. Therefore, we can write ω = (ω−, ω+). Since(ΛnP)n∈Z is an independent sequence of partitions with respect toµ0, we can write the measure µ0 = µ−
0 × µ+0 corresponding to the
representation ω = (ω−, ω+).Define a function x0 : Ω0 → R by x0(ω) = x1 such that (x1, x2)×
(y1, y2) = R0(ω). Let B ∈ PN0 for some N . Then, a pair (k, i) is
determined such that B is the set of ω ∈ Ω0 satisfying that the (k, i)-descendent of R0(ω) intersects with the y-axis (or touches it from theright). Hence,
B = ω ∈ Ω0; − b0
i∑j=0
τ k(a0)j < x0(ω) ≤ −b0i−1∑j=0
τ k(a0)j.
Moreover, µ0(B) = τ k(a0)i. Take another N′ > N and B ∈ PN ′
0 suchthat B′ ⊂ B. Then, the pair (k′, i′) determined by B′ is such that
18
for any ω ∈ Ω0, the (k′, i′)-descendent of R0(ω) is a descendent of the
(k, i)-descendent of R0(ω). Therefore, we have
−b0i∑
j=0
τ k(a0)j ≤ −b0i′∑
j=0
τ k′(a0)j
< x0(ω) ≤ −b0i′−1∑j=0
τ k′(a0)j ≤ −b0
i−1∑j=0
τ k(a0)j .
This implies that there exists a measurable bijection ψ : P∞0 →
[0, b0) such that ψ(ω+) := −x0(ω) which is actually the limit ofb0∑i−1
j=0 τk(a0)j as above along the element B ∈ PN
0 including ω+.Moreover, the function ψ is uniformly distributed on [0, b0) under µ0
since µ0(B) = τ k(a0)i for any B ∈ PN0 and N .
Let ψ(ω+) + t ∈ [0, b0). Then, it is easy to see that (ω + t)− = ω−and ψ((ω+t)+) = ψ(ω+)+t. This implies that for any measurable setB ⊂ Ω0, if ψ(ω+) + t ∈ [0, b0) holds for any ω ∈ B, then µ0(B + t) =µ0(B).For y ∈ G, let Ωy
0 := ω ∈ Ω; y−1ω ∈ Ω0 and µy0 be the measure
on Ωy0 such that µy
0(B) = µ0(y−1B). Let xy0 be the function on Ωy
0
such that xy0(ω) = yx0(y−1ω). Then just as above, for any y ∈ G,
the function −xy0 restricted to Ωy0 is uniformly distributed on [0, yb0)
with respect to µy0. Moreover, for any measurable set B ⊂ Ωg
0, if−xy0(ω) + t ∈ [0, yb0) holds for any ω ∈ B, then µy
0(B + t) = µy0(B).
Note that (??)(??) implies that for any measurable set B ⊂ Ω1, itholds that
µ(B) = C(µ)−1
∫ ∞
1
µy0(B
1y)dy/y
orµ(B) = C(µ)−1
∑y∈G, y≥1
µy0(B
1y)
where
B1y = ω ∈ B; y is the minimum λ ≥ 1 with λ ∈ G and ω ∈ Ωλ
0.
If we replace 1 by b ∈ G in the above, we have
µ(B) = Cb(µ)−1
∫ ∞
b
µy0(B
by)dy/y
19
orµ(B) = Cb(µ)
−1∑
y∈G, y≥b
µy0(B
by)
where
Bby = ω ∈ B; y is the minimum λ ≥ b with λ ∈ G andω ∈ Ωλ
0.
For any measurable set B ⊂ Ω1, t ∈ R and b ∈ G, let
B′ = ω ∈ B; − xb0(ω) + t ∈ [0, bb0).
Then, µ(B′) ≤ |t|/(bb0) holds. Since µy0(B \ B′) = µy
0((B \ B′) + t)holds for any y ≥ b, we have µ(B \B′) = µ((B \B′) + t). Therefore,we have
µ(B) ≤ µ(B \B′) + |t|/(bb0)= µ((B \B′) + t) + |t|/(bb0) ≤ µ(B + t) + |t|/(bb0).
Taking b→ ∞, we have µ(B) ≤ µ(B+ t). By the symmetry, we haveµ(B + t) ≤ µ(B). Thus, we have µ(B) = µ(B + t) and µ is invariantunder the additive action, which complete the proof of Lemma 2 andTheorem 2.
The following Theorem 3 follows from a known result about thespectrum of unitary operators corresponding to the affine action(Lemma 11.6 of [?], for example). Nevertheless, we give the prooffor to be self-contained.
Theorem 3. Let Ω be a numeration system with G = R+, that is,with the multiplicative R+-action. Then, the additive action on theprobability space Ω with respect to µΩ has a pure Lebesgue spectrum.
Proof. Let Ut (t ∈ R) and Vλ (λ ∈ R+) be the groups of the unitaryoperators on L2(Ω, µΩ) defined by
(Utf)(ω) = f(ω + t) , (Vλf)(ω) = f(λω).
Let
Ut =
∫ ∞
−∞eitudEu (t ∈ R)
20
be the spectral decomposition of Ut ([?]). Since UtVλ = VλUλt, wehave dEuVλ = VλdEλ−1u.Take any f ∈ L2(Ω, µΩ) with
∫fdµΩ = 0 and
∫|f |2dµΩ = 1. Let
m(f) be the measure on R defined by
m(f)(S) =
∫S
∥ dEuf ∥2
for any Borel set S ⊂ R. Then, m(f) is a probability measure withm(f)(0) = 0. Since dEuVλ = VλdEλ−1u, we have
m(Vλf)(S) =
∫S
∥ dEuVλf ∥2=∫S
∥ VλdEλ−1uf ∥2
=
∫S
∥ dEλ−1uf ∥2=∫λ−1S
∥ dEuf ∥2= m(f)(λ−1S).
Moreover, we have
|(f, Vλf)| = |∫
(dEuf, dEuVλf)|
≤∫
∥ dEuf ∥∥ dEuVλf ∥=∫ √
dm(f)√dm(Vλf).
Since limλ→1 |(f, Vλf)| = 1, we have
limλ→1
∫ √dm(f)
√dm(f) λ−1 = lim
λ→1
∫ √dm(f)
√dm(Vλf) = 1.
It follows from this that m(f) is absolutely continuous by the follow-ing well known argument (see [?], for example).Suppose to the contrary that m := m(f) is not absolutely contin-
uous. Take a Borel set S ⊂ R such that S has Lebesgue measure 0while δ := m(S) > 0. Denoting ρ(λ) :=
∫ √dm
√dm λ−1, we have
2(1− ρ(λ)) =
∫ (√dm−
√dm λ−1
)2≥
(√m(S)−
√m λ−1(S)
)2=(√
δ −√m(λ−1S)
)2.
21
Since 2(1− ρ(λ)) → 0 as λ→ 1, there exists ϵ > 0 such that for anyλ with 1− 2ϵ ≤ λ ≤ 1 + 2ϵ, m(λ−1S) > δ/2 holds. Hence,
2δϵ ≤∫ 1+2ϵ
1−2ϵ
m(λ−1S)dλ =
∫ ∫ 1+2ϵ
1−2ϵ
1S(λu)dλdm(u).
This implies that the set of u ∈ R such that∫ 1+2ϵ
1−2ϵ1S(λu)dλ ≥ δϵ
has the measure at least δ/4 with respect to m. Since m(0) = 0,
this implies that there exists u = 0 such that∫ 1+2ϵ
1−2ϵ1S(λu)dλ ≥ δϵ.
Thus, S has Lebesgue measure at least |u|δϵ, which contradicts theassumption that S has Lebesgue measure 0. Thus, m = m(f) isabsolutely continuous.
4 ζ-function
Let Ω := Ω(σ, τ, g) satisfying (3) and (4). For α ∈ C, we define theassociated matrices on the suffix set A× A as follows:
Mα :=
∑i;σ(a)i=a′
τ(a)αi
a,a′∈A
(12)
Mα,+ :=(1σ(a)0=a′ τ(a)
α0
)a,a′∈A
Mα,− :=(1σ(a)|σ(a)|−1=a′ τ(a)
α|σ(a)|−1
)a,a′∈A
Let Θ be the set of closed orbits of Ω with respect to the action ofG. That is, Θ is the family of subsets ξ of Ω such that ξ = Gω forsome ω ∈ Ω with λω = ω for some λ ∈ G with λ > 1. We call λ asabove a multiplicative cycle of ξ. The minimum multiplicative cycleof ξ is denoted by c(ξ). Note that c(ξ) exists since λω = ω for anyω ∈ Ω and λ ∈ G with 1 < λ < minτ(a)−1
i ; a ∈ A, 0 ≤ i < |τ(a)|.We say that ξ ∈ Θ has a separating line if ω ∈ ξ has a separating
line. Note that in this case, the separating line is necessarily the y-axis and is in common among ω ∈ ξ. Denote by Θ0 the set of ξ ∈ Θwith the separating line.
22
Let
L(σ) := (a, k, i); a ∈ A, k = 1, 2, · · · and
0 ≤ i < |σk(a)| with σk(a)i = a.
For (a, k, i) and (a, k′, i′) in L(σ), define the product by
(a, k, i)(a, k′, i′) = (a, k + k′,
i−1∑j=0
|σk′(σk(a)j)|+ i′).
We say that (a, k, i) ∈ L(σ) is irreducible if (a, k, i) = (a, k′, i′)h doesnot hold for any h = 2, 3, · · · and (a, k′, i′) ∈ L(σ).Let ξ ∈ Θ \ Θ0 and ω ∈ ξ. Then, there exists R = (x1, x2) ×
(y1, y2) ∈ dom(ω) with x1 < 0 < x2. Since c(ξ)ω = ω, there existsa descendant R′ of R such that ω(R′) = ω(R) =: a and R = c(ξ)R′.Let R′ be the (k, i)-descendant of R.
Lemma 3. For any ξ ∈ Θ \Θ0 with the above setting, the followingstatements hold.(i) 1 ≤ i < |σk(a)| − 1.(ii) (a, k, i) is in L(σ) and irreducible.(iii) c(ξ) = τ k(a)−1
i .Conversely, any triple (a, k, i) ∈ L(σ) satisfying (i)(ii) determinesξ ∈ Θ \Θ0 and (iii) follows.
Proof. (iii) holds since c(ξ)R′ = R and R′ is the (k, i)-descendantof R.Since R′ is the (k, i)-descendant of R such that τ k(a)−1
i R′ = R, wehave
−x1x2
=
∑j≤i−1 τ
k(a)j∑j≥i+1 τ
k(a)j. (13)
Since 0 < −x1/x2 <∞, (i) follows.If (a, k, i) = (a, k′, i′)ℓ with ℓ ≥ 2, then (k′, i′)-descendant R′′ of R
also satisfies τ k′(a)−1
i′ R′′ = R and that τ k
′(a)−1
i′ = c(ξ)1/ℓ becomes acycle of ξ, contradicting the minimality of c(ξ). Thus, (ii) follows.Let us prove the last statement. Assume that (a, k, i) ∈ L(σ) is
irreducible satisfying (i). Take any ω′ ∈ Ω and R ∈ dom(ω′) with
23
ω′(R) = a. Take t ∈ R such that R + t =: (x1, x2)× (y1, y2) satisfiesthe equation (??). Then x1 < 0 < x2, and τ
k(a)−ni (ω′ + t) converges
as n → ∞ to, say ω ∈ Ω which satisfies that τ k(a)−1i ω = ω. Thus,
we have ξ := Gω ∈ Θ \ Θ0 which is determined by (a, k, i) ∈ L(σ).(iii) follows by the irreducibility of (a, k, i).
Let ξ ∈ Θ \ Θ0 and ω ∈ ξ. Let R and R′ satisfy that R = c(ξ)R′
and that R′ is the (k, i)-descendant of R. Let R0, R1, R2, · · · be thesequence of tiles in ω intersecting with the y-axis such that Ri+1
is the mother of Ri (i = 0, 1, 2, · · · ), R0 = R′ and Rk = R. LetRj be the (k, ij)-descendant of Rj+k for any j = 0, 1, · · · , k − 1.Then, ω(Rj+k) = ω(Rj) holds and the triple (ω(Rj+k), k, ij) (j =0, 1, · · · , k−1) satisfies the condition (i)(ii) in Lemma ?? determiningξ in the sense of Lemma ??. In fact, these k triples are different fromeach other such that they are all that determine ξ, while k is commonamong them which we denote by k(ξ).
Lemma 4. For k = 1, 2, · · · , it holds that
tr(Mkα)− tr(Mk
α,+)− tr(Mkα,−)
=∑
ξ∈Θ\Θ0k(ξ)|k
k(ξ)c(ξ)−k
k(ξ)α. (14)
Proof. Note that
tr(Mkα)− tr(Mk
α,+)− tr(Mkα,−) =
∑a∈A,1≤i<|σk(a)|−1
σk(a)i=a
τ k(a)αi .
For any ξ ∈ Θ \Θ0 with k(ξ)|k, there exist exactly k(ξ) number oftriples (aj, k(ξ), ij) ∈ L(σ) (j = 0, 1, · · · , k(ξ) − 1) determining ξ so
24
that τ k(ξ)(aj)ij = c(ξ)−1 follows. Therefore,∑a∈A,1≤i<|σk(a)|−1
σk(a)i=a
τ k(a)αi =∑
(a,k,i)∈L(σ)
1≤i<|σk(a)|−1
τ k(a)αi
=∑k′|k
∑(a,k′,i)∈L(σ):irreducible
1≤i<|σk′ (a)|−1
τ k′(a)
(k/k′)αi
=∑k′|k
∑ξ∈Θ\Θ0k(ξ)=k′
k′c(ξ)−(k/k′)α
=∑
ξ∈Θ\Θ0k(ξ)|k
k(ξ)c(ξ)−k
k(ξ)α.
The following lemma follows from Lemma 1.
Lemma 5. The set Θ0 is a finite set. In fact, we have
♯Θ0 ≤∑a∈A
(|σ(a)| − 1).
Lemma 6. For α ∈ C with R(α) > 1, where R(α) is the real partof α, we have ∑
ξ∈Θ
|c(ξ)−α| <∞.
Proof. By Lemma ??, it is sufficient to prove that∑ξ∈Θ\Θ0
|c(ξ)−α| <∞.
Sincemaxa∈A
∑0≤i<|τ(a)|
τ(a)R(α)i < 1,
25
there exists δ with 0 < δ < 1 such that the maximal eigen-value ofMR(α) is less than δ. Hence, by (??) we have∑
ξ∈Θ\Θ0
|c(ξ)−α| =∑
ξ∈Θ\Θ0
c(ξ)−R(α)
≤∞∑k=1
∑ξ∈Θ\Θ0k(ξ)|k
k(ξ)c(ξ)−k
k(ξ)R(α)
≤∞∑k=1
tr(MkR(α)) ≤
∞∑k=1
Cδk <∞.
Define the ζ-function of G-action to Ω by
ζΩ(α) :=∏ξ∈Θ
(1− c(ξ)−α)−1, (15)
where the infinite product converges for any α ∈ C with R(α) > 1by Lemma ??. It is extended to the whole complex plane by theanalytic extension.
Theorem 4. We have
ζΩ(α) =det(I −Mα,+) det(I −Mα,−)
det(I −Mα)ζΣ0(α),
whereζΣ0(α) :=
∏ξ∈Θ0
(1− c(ξ)−α)−1
is a finite product with respect to ξ ∈ Θ0.
Proof. By the definition of ζΩ(α) and (??), for any α withR(α) > 1,
26
it holds that
ζΩ(α) = ζΣ0(α) exp
∑ξ∈Θ\Θ0
− log(1− c(ξ)−α)
= ζΣ0(α) exp
∑ξ∈Θ\Θ0
∞∑k=1
(1/k)c(ξ)−kα
= ζΣ0(α) exp
∞∑k=1
∑ξ∈Θ\Θ0k(ξ)|k
(k(ξ)/k)c(ξ)−k
k(ξ)α
= ζΣ0(α) exp
(∞∑k=1
(1/k)(tr(Mkα)− tr(Mk
α,+)− tr(Mkα,−))
)
= ζΣ0(α) exp
(tr(
∞∑k=1
(1/k)(Mkα −Mk
α,+ −Mkα,−))
)= ζΣ0(α) exp (tr(− log(I −Mα) + log(I −Mα,+)
+ log(I −Mα,−)))
=det(I −Mα,+) det(I −Mα,−)
det(I −Mα)ζΣ0(α)
Example 3. For the 2-adic expansion substitution (σ, τ) definedin Remark 1, 1 → (1, 1/2)(1, 1/2), define Ω := Ω(σ, τ) with G =2n;n ∈ Z, g ≡ 1 (see (4)). Then, we have
Mα = 2(1/2)α , Mα,+ =Mα,− = (1/2)α.
Moreover, we have Θ0 = Gω0 with ω0 shown in Figure 6. Sincec(Gω0) = 2 and ζΣ0(α) = (1− 2−α)−1, we have
ζΩ(α) = ζΣ0(α)(1− (1/2)α)2
1− 2(1/2)α
=1− (1/2)α
1− 2(1/2)α
27
x
y
Figure 6: ω0 in the 2-adic expansion
Example 4. Consider the weighted substitution (σ, τ) in Example2, that is 1 → (1, p)(1, 1− p). Then, we have
Mα = pα + (1− p)α , Mα,+ = pα , Mα,− = (1− p)α.
Since Θ0 = ∅, we have
ζΩ(α) =(1− pα)(1− (1− p)α)
1− pα − (1− p)α.
Example 5. For the Thue-Morse substitution (σ, τ) defined in Re-mark 1, 1 → (1, 1/2)(2, 1/2), 2 → (2, 1/2)(1, 1/2), define Ω :=
28
x
y
1 1(2)
2 1(2)
1 1(2)
2 1(2)
Figure 7: ω1(ω2, respectively) in the Thue-Morse substitution
Ω(σ, τ) with G = 2n;n ∈ Z, g ≡ 1. Then, we have
Mα =
((1/2)α (1/2)α
(1/2)α (1/2)α
)Mα,+ =
((1/2)α 0
0 (1/2)α
)Mα,− =
(0 (1/2)α
(1/2)α 0
)Moreover, we have Θ0 = Gω1, Gω2 with ω1 and ω2 shown in Figure7. Since c(Gω1) = c(Gω2) = 4 and ζΣ0(α) = (1− 4−α)−2, we have
ζΩ(α) = ζΣ0(α)(1− (1/2)α)2(1− (1/2)2α)
1− 2(1/2)α
=1− (1/2)α
(1 + (1/2)α)(1− 2(1/2)α)
29
x x
y y
±
±
±±
±
±
±±
6
?vertical size 81/64
6
?vertical size 16/9
∓
∓∓
∓
∓
∓∓
Figure 8: 4 elements in Θ0 in Example 6 (±, respectively)
Example 6. For the weighted substitution (σ, τ) in Example 1
+ → (+, 4/9)(−, 1/9)(+, 4/9)− → (−, 4/9)(+, 1/9)(−, 4/9),
define Ω := Ω(σ, τ), where since B(σ, τ) = R holds, taking g ≡ 1, wehave
Mα =
(2(4/9)α (1/9)α
(1/9)α 2(4/9)α
)Mα,+ = Mα,− =
((4/9)α 0
0 (4/9)α
)There are 4 elements in Θ0 determined as in (iii) of Lemma 1 by thepairs (+, 0), (+, 1), (−, 0), (−, 1). All of them are different as shownin Figure 8. The ± corresponds to the ± in the first coordinate, while0, 1 corresponds to the vertical gap from the left side tiles to the rightside tiles. It is 81/64 for 0 (left side in Figure 8) and 16/9 for 1 (rightside in Figure 8). These 4 elements have the same multiplicative
30
cycle 9/4 . Hence, we have
ζΩ(α) =1
(1− 2(4/9)α − (1/9)α)(1− 2(4/9)α + (1/9)α).
Theorem 5. @(i) ζΩ(α) = 0 if R(α) = 0.(ii) In the region R(α) = 0, α is a pole of ζΩ(α) with multiplicity
k if and only if it is a zero of det(I−Mα) with multiplicity k for anyk = 1, 2, · · · .(iii) 1 is a simple pole of ζΩ(α).
Proof. Since c(ξ) > 1 for any ξ ∈ Θ, 1 − c(ξ)−α = 0 if R(α) = 0.Hence, ζΣ0(α) has neither pole nor zero in the region R(α) = 0.For α with R(α) = 0, suppose that det(I −Mα,+) = 0, so that
Mα,+ξ = ξ holds for some nonzero vector ξ = (ξa)a∈A. Take a0 ∈ Asuch that ξa0 = 0. Then, for some k ≥ 0 and ℓ ≥ 1, σk(a0)0 =σk+ℓ(a0)0 =: a1 holds. Since ξa1 = τ k(a0)
α0 ξa0 = 0 and τ ℓ(a1)
α0 ξa1 =
ξa1 , we have τ ℓ(a1)α0 = 1. This is impossible since 0 < τ ℓ(a1)0 < 1.
Thus, det(I −Mα,+) has no zero in the region R(α) = 0. In thesame way, det(I −Mα,−) has no zero in the region R(α) = 0. Thesefacts with Theorem 4 prove (i)(ii) of Theorem 5.(iii) Since M1
t(1, · · · , 1) = t(1, · · · , 1), 1 is an eigen-value of M1,hence is a zero of det(I−Mα). We prove that it is a simple zero. LetA = a1, · · · , ar and A′ := A \ a1. For a matrix M = (mij)i∈I,j∈Jand I ′ ⊂ I, J ′ ⊂ J , let M [I ′, J ′] := (mi,j)i∈I′,j∈J ′ . Since 1 is themaximum eigen-value ofM1 and σ is primitive, there exists a positiverow vector (ξ1, ξ2, · · · , ξr) with ξ1 = 1 such that (ξ1, ξ2, · · · , ξr)(I −M1) = (0, · · · , 0). Therefore, since
det(I −Mα) = det
1−∑|τ(a1)|−1
i=0 τ(a1)αi
...
1−∑|τ(ar)|−1
i=0 τ(ar)αi
(I −Mα)[A,A′]
,
31
we have
d
dαdet(I −Mα)|α=1
= det
∑
i −τ(a1)i log τ(a1)i...∑
i −τ(ar)i log τ(ar)i(I −M1)[A,A′]
= det
∑
i,j −ξjτ(aj)i log τ(aj)i 0 · · · 0∑i −τ(a2)i log τ(a2)i
...∑i −τ(ar)i log τ(ar)i
(I −M1)[A′,A′]
=
(∑i,j
−ξjτ(aj)i log τ(aj)i
)det((I −M1)[A′,A′]).
We have∑
i,j −ξjτ(aj)i log τ(aj)i > 0 and det((I −M1)[A′,A′]) = 0since the spectral radius of M1[A′,A′] is strictly less than 1. Hence,ddα
det(I −Mα)|α=1 = 0 and 1 is a simple zero of det(I −Mα). By(ii), it is a simple pole of ζΩ.
Theorem 6. For Ω = Ω(σ, η, g), if B(σ, τ) = λn; n ∈ Z withλ > 1, then there exist polynomials p, q ∈ Z[z] such that ζΩ(α) =p(λα)/q(λα). Conversely, if ζΩ(α) = p(λα)/q(λα) holds for somepolynomials p, q ∈ Z[z] and λ > 1, then B(σ, τ) = λkn; n ∈ Z forsome positive integer k.
Proof. Assume that B(σ, η) = λn; n ∈ Z with λ > 1. Let gsatisfies (4). Then for any a ∈ A and i with 0 ≤ i < |σ(a)|, thereexists r(a)i ∈ Z such that τ(a)i =
g(σ(a)i)g(a)
λr(a)i . Hence, we have
Mα = Λ−1α
∑0≤i<|σ(a)|σ(a)i=a′
(λα)r(a)i
a,a′∈A
Λα
Mα,+ = Λ−1α
(1σ(a)0=a′(λ
α)r(a)0)a,a′∈A Λα
Mα,− = Λ−1α
(1σ(a)|σ(a)|−1=a′(λ
α)r(a)|σ(a)|−1
)a,a′∈A
Λα,
32
where Λα = (g(a)α1a′=a)a,a′∈A is a diagonal matrix. Therefore, det(I−Mα), det(I−Mα,+) and det(I−Mα,−) are polynomials in λα dividedpossibly by (λα)n for some positive integer n. Since c(ξ) = λn forsome positive integer n for any ξ ∈ Θ0 and Θ0 is a finite set, ζΣ0(α)
−1
is a polynomial in λ−α. Thus, ζΩ(α) = p(λα)/q(λα) for some polyno-mials p, q ∈ Z[z].Conversely, assume that ζΩ(α) = p(λα)/q(λα) for some polynomi-
als p, q ∈ Z[z]. Then we have
Πξ∈Θ(1− c(ξ)−α) = q(λα)/p(λα)
on R(α) > 1. Comparing their expansions as Dirichlet series in α, itholds that c(ξ) ∈ λn; n ∈ Z for any ξ ∈ Θ ([?]). Since B(σ, τ) isgenerated by c(ξ); ξ ∈ Θ, we have B(σ, τ) ⊂ λn; n ∈ Z. Thus,B(σ, τ) = λkn; n ∈ Z for some positive integer k.
Theorem 7. If B(σ, τ) = λn; n ∈ Z, then λ is an algebraicnumber.
Proof. Let Ω := Ω(σ, τ, g). By Theorem 6, there exist polynomialsp, q ∈ Z[z] such that ζΩ(α) = p(λα)/q(λα). By Theorem 5, 1 is a poleof ζΩ(α). Hence q(λ) = 0, which implies that λ is algebraic.
5 β-expansion system
Let β be an algebraic integer with β > 1 such that 1 has the followingperiodic β-expansion
1 = (b10i1−1b20
i2−1 · · · bk0ik−1)∞
b1, b2, · · · , bk ∈ 1, 2, · · · , ⌊β⌋i1, i2, · · · , ik ∈ 1, 2, · · · ,
where ( )∞ implies the infinite time repetition of ( ). Let n :=i1+ i2+ · · ·+ ik ≥ 1 and assume that n is the minimum period of theabove sequence. Since the above sequence is the expansion of 1, wehave the solution of the following equation in a1, a2, · · · , ak+1 witha1 = ak+1 = 1 and 0 < aj < 1 (j = 2, · · · , k):
aj = bjβ−1 + aj+1β
−ij (j = 1, 2, · · · , k).
33
Let A := 1, 2, · · · , k and define a weighted substitution (σ, τ) by
j → (1, (1/aj)β−1)bj (j + 1, (aj+1/aj)β
−ij)
(j = 1, 2, · · · , k − 1)
k → (1, (1/ak)β−1)bk (1, (ak+1/ak)β
−ik)
where ( , )k implies the k-time repetition of ( , ). Then, σ is primitiveand B(σ, τ) = βn; n ∈ Z. Define g : A → R+ by g(j) := aj. Then,g satisfies (4) and Ω(σ, τ, g) is a numeration system by Theorem 2.We denote Ω(β) := Ω(σ, τ, g) and Ω(β) is called the β-expansionsystem.The β-expansion system is studied by many authors, for example,
S. Ito and Y. Takahashi ([?]) where the ζ-function is obtained. Here,we give the formula again as a corollary of Theorem 4. There is a littledifference between them. In [?], the ζ-function is for the restrictionω+; ω ∈ Ω to the right-quarter space, while ours is for Ω itself, sothat ours is the product of the former and 1−βα
1−β−nα .
Theorem 8. We have
ζΩ(β)(α) =1− β−α
1−∑k
j=1 bjβ−(i1+···+ij−1+1)α − β−nα
.
Proof. We have
det(I −Mα) =∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
1− b1(1
a1β)α −( a2
a1βi1)α 0 0 · · · 0 0
−b2( 1a2β
)α 1. . . 0 · · · 0 0
......
. . . . . ....
......
......
.... . . . . .
......
......
......
. . . . . ....
−bk−1(1
ak−1β)α 0 0 0 · · · 1 −( ak
ak−1βik−1
)α
−bk( 1akβ
)α − ( ak+1
akβik)α 0 0 0 · · · 0 1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 1−
k∑j=1
bjβ−(i1+···+ij−1+1)α − β−nα,
34
det(I −Mα,+) =
∣∣∣∣∣∣∣∣∣∣∣
1− ( 1a1β
)α 0 · · · · · · 0
−( 1a2β
)α 1 0 · · · 0...
.... . .
......
−( 1ak−1β
)α 0 · · · 1 0
−( 1akβ
)α 0 · · · 0 1
∣∣∣∣∣∣∣∣∣∣∣= 1− β−α
and
det(I −Mα,−) =∣∣∣∣∣∣∣∣∣∣∣∣
1 −( a2a1βi1
)α 0 · · · 0
0 1 −( a3a2βi2
)α · · · 0...
.... . . . . .
...0 · · · 0 1 −( ak
ak−1βik−1
)α
−( ak+1
akβik)α 0 · · · 0 1
∣∣∣∣∣∣∣∣∣∣∣∣= 1− β−nα.
Let ξ ∈ Θ0 and ω ∈ ξ. Let R±i (i ∈ Z) be the sequence of tiles
in dom(ω) from down to up intersecting with the line x = ±0 and(±0, 1 + 0) ∈ R±
0 (± respectively). Then, we have
· · ·ω(R+−1)ω(R
+0 )ω(R
+1 )ω(R
+2 ) · · · = · · · · · · 11111 · · · · · ·
· · ·ω(R−−1)ω(R
−0 )ω(R
−1 )ω(R
−2 ) · · · = · · · k · · · 21k · · · 21 · · · .
Hence, the minimum multiplicative cycle of ω on the right quarterplane is β, while on the left quarter plane is βn. Thus, c(ξ) = βn.Moreover, the right half tiling and the left half tiling are synchronizedso that the vertical position of the lower side of any tile R−
i withω(R−
i ) = 1 coincides with that of R+j for some j. Therefore, ξ is the
unique element in Θ0. Hence, ζΣ0(α) = (1− β−nα)−1.Combining these results using Theorem 4, we have
ζΩ(β)(α) =1− β−α
1−∑k
j=1 bjβ−(i1+···+ij−1+1)α − β−nα
.
35
Example 7. Let β = (1 +√5)/2 be the golden number. Then, the
expansion of 1 is (10)∞. Therefore, A = 1 and (σ, τ) is
1 → (1, β−1)(1, β−2).
By Theorem ??, we have
ζΩ(β)(α) =1− β−α
1− β−α − β−2α.
Example 8. Let us consider the β-expansion system with β > 1such that β3 − β2 − β − 1 = 0. Then the expansion of 1 is (110)∞
and the corresponding weighted substitution is
1 → (1, β−1)(2, β−2 + β−3)
2 → (1,β
β + 1)(1,
1
β + 1)
By Theorem ??, we have
ζΩ(β)(α) =1− β−α
1− β−α − β−2α − β−3α.
We will discuss this example in the next section.
6 homogeneous cocycles and fractals
Let Ω := Ω(σ, τ, g) satisfy (3) and (4).A continuous function F : Ω× R → C is called a cocycle on Ω if
F (ω, t+ s) = F (ω, t) + F (ω + t, s) (16)
holds for any ω ∈ Ω and s, t ∈ R. A cocycle F on Ω is called α-homogeneous if
F (λω, λt) = λαF (ω, t)
for any ω ∈ Ω, λ ∈ G and t ∈ R, where α is a given complex number.A cocycle F (ω, t) on Ω is called adapted if there exists a functionΞ : A× R+ → C such that
F (ω, x2)− F (ω, x1) = Ξ(ω(R), x2 − x1) (17)
36
for any ω ∈ Ω and tile R := (x1, x2)× (y1, y2) ∈ dom(ω).In [?], nonzero adapted α-homogeneous cocycles on Ω with 0 <
α < 1 is characterized. In fact, we have
Theorem 9. A nonzero adapted α-homogeneous cocycle on Ω ischaracterized by (??) with α and Ξ satisfying that R(α) > 0 and thereexists a nonzero vector ξ = (ξa)a∈A such that Mαξ = ξ (see (??)) andΞ(ω(R), x2−x1) = (x2−x1)αξω(R) for any tile R := (x1, x2)×(y1, y2) ∈dom(ω). Hence, a nonzero adapted α-homogeneous cocycle exists ifand only if R(α) > 0 and α is a pole of ζΩ(α).
Proof. The last part of the theorem follows from Theorem 5. Thecondition that R(α) > 0 is necessary for the convergence of F (ω, t)with (??) for a general t ∈ R.
It is known [?] that
Theorem 10. Let µΩ be the equilibrium measure on Ω. Let 0 <α < 1. For a nonzero α-homogeneous cocycle F on Ω, we have thefollowing results.(i) There exists a constant C such that
|F (ω, t)− F (ω, s)| ≤ C|t− s|α
for any ω ∈ Ω and s, t ∈ R. That is, the functions F (ω, t) on t forω ∈ Ω are uniformly α-Holder continuous.(ii) For any ω ∈ Ω and t ∈ R,
lim sups↓0
1
sα|F (ω, t+ s)− F (ω, t)| > 0
holds. That is, for any ω ∈ Ω the function F (ω, ·) is nowhere lo-cally α′-Holder continuous for any α′ > α. In particular, F (ω, ·) isnowhere differentiable.(iii) The stochastic process F (ω, t) with time parameter t ∈ R
and random element ω ∈ Ω with respect to µ has a strictly ergodicstationary increment having 0-entropy.(iv) F (ω, λt) has the same law as λαF (ω, t) for any λ ∈ G. Hence,
the process F (ω, t) is α-self similar if G = R+.(v)
∫F (ω, t)dµ(ω) = 0 for any t ∈ R.
37
Example 9. Take Ω in Example 1 and 6. Since
M1/2 =
(4/3 1/31/3 4/3
),
ξ =
(1−1
)is an eigenvector of M1/2 with eigen-value 1. Let F be
the 1/2-homogeneous adapted cocycle on Ω defined by the equation:
F (ω, x2)− F (ω, x1) = ±(x2 − x1)1/2
if there exists a tile (x1, x2)× (y1, y2) in ω with color ±, respectively(see Theorem 9).Then, F (ω, t) is a 1/2-selfsimilar process with respect to the unique
invariant measure µ under the additive action, called N -process,which is discussed in the next section.Consider the family of functions (F (ω+n, 1))n=0,1,2,··· on the prob-
ability space (Ω, µ). Since
nE[F (ω, 1)2] = E[(n1/2F (ω, 1))2] = E[F (nω, n)2] = E[F (ω, n)2]
= E[F (ω, 1)2] + E[(F (ω, n)− F (ω, 1))2]
+2E[F (ω, 1)(F (ω, n)− F (ω, 1))]
= E[F (ω, 1)2] + E[F (ω + 1, n− 1)2]
+2E[F (ω, 1)(F (ω, n)− F (ω, 1))]
= E[F (ω, 1)2] + (n− 1)E[F (ω, 1)2]
+2E[F (ω, 1)(F (ω, n)− F (ω, 1))],
we haveE[F (ω, 1)(F (ω, n)− F (ω, 1))] = 0
for any n = 1, 2, · · · . Therefore,
E[F (ω, 1)(F (ω + n, 1)]
= E[F (ω, 1)(F (ω, n+ 1)− F (ω, n))]
= E[F (ω, 1)(F (ω, n+ 1)− F (ω, 1))]
−E[F (ω, 1)(F (ω, n)− F (ω, 1))]
= 0
38
for any n = 1, 2, · · · . Hence, for any n < m,
E[F (ω + n, 1)(F (ω +m, 1)] = E[F (ω, 1)(F (ω +m− n, 1)] = 0.
This implies that the family of functions (F (ω + n, 1))n=0,1,··· is non-correlated.
Let I(Ω) be the set of ω ∈ Ω such that there exists (x1, x2) ×(y1, y2) ∈ dom(ω) satisfying that x1 = 0 and y1 ≤ 1 < y2. Anelement ω ∈ I(Ω) is called an integer in Ω. Let
II(Ω) := (ω, t) ∈ I(Ω)× R; ω + t ∈ I(Ω).
A continuous function F : II(Ω) → C is called a cocycle on I(Ω)if (??) is satisfied for any ω ∈ I(Ω) and t, s ∈ R such that (ω, t) ∈II(Ω) and (ω, t+ s) ∈ II(Ω).A cocycle F on I(Ω) is called adapted if there exists a function
Ξ : A×R+ → C such that (??) is satisfied for any ω ∈ I(Ω) and tile(x1, x2)× (y1, y2) ∈ dom(ω) with y2 > 1. Let α ∈ C. A cocycle F onI(Ω) is called α-homogeneous if
F (λω, λt) = λαF (ω, t)
for any (ω, t) ∈ II(Ω) and λ ∈ G with (λω, λt) ∈ II(Ω). Note thatif (ω, t) ∈ II(Ω), then for any λ ∈ G with λ > 1, (λω, λt) ∈ II(Ω)holds.A cocycle F on I(Ω) is called a coboundary on I(Ω) if there exists
a continuous function G : I(Ω) → Rk such that
F (ω, t) = G(ω + t)−G(ω)
for any (ω, t) ∈ II(Ω).The following theorem is proved in [?].
Theorem 11. A nonzero adapted α-homogeneous cocycle on I(Ω)with R(α) < 0 is characterized by (??) with Ξ satisfying that thereexists a nonzero vector ξ = (ξa)a∈A such that Mαξ = ξ (see (??)) andΞ(ω(R), x2−x1) = (x2−x1)αξω(R) for any tile R := (x1, x2)×(y1, y2) ∈dom(ω) with y2 > 1. Hence, a nonzero adapted α-homogeneous cocy-cle on I(Ω) with R(α) < 0 exists if and only if α is a pole of ζΩ(α).Moreover, any cocycle as this is a coboundary.
39
Example 10. Let us consider the β-expansion system in Example8. Denote Ω := Ω(β). The associated matrix is
Mα =
(β−α (β−2 + β−3)αβα+1(β+1)α
0
)Let γ be one of the complex solutions of the equation z3−z2−z−1 =0. Then, |γ| < 1. Let α ∈ C be such that γ = βα. Then, R(α) < 0.Since we have
Mα
(1δ
)=
(1δ
)with δ :=
βα + 1
(β + 1)α, there exists an α-homogeneous adapted cocycle
F on I(Ω) satisfying that
F (ω, x2)− F (ω, x1) =
(x2 − x1)
α (ω(R) = 1)δ(x2 − x1)
α (ω(R) = 2)
if there exists R := (x1, x2)× (y1, y2) ∈ dom(ω) with y2 > 1.For ω ∈ I(Ω), let R0(ω) be the tile (x0, x0) × (y0, y0) ∈ dom(ω)
such that x0 = 0 and y0 ≤ 1 < y0. For i = 0, 1, 2, · · · , let Ri be thei-th ancestor of R0(ω). Let Ri =: (xi, xi)× (yi, yi). Let
G(ω) :=∞∑i=0
(xi − xi+1)α. (18)
Since if xi+1 < xi for some i = 0, 1, 2, · · · , then there exists a tile(xi+1, xi)× (y, yi+1) for some y with color 1 in ω, we have
F (ω, xi)− F (ω, xi+1) = (xi − xi+1)α
for any i = 0, 1, · · · .Take any t ∈ R such that (ω, t) ∈ II(Ω). Let (R′
i)i=1,2,··· and(x′i)i=0,1,··· be the sequences of above (Ri)i=1,2,··· and (xi)i=0,1,··· forω + t instead of ω.If there does not exist a seperating line of ω in the region (0, t] ×
(0,∞) (if t > 0) or (t, 0] × (0,∞) (if t < 0), then there exist i0, j0
40
such that R′i0+k = Rj0+k− t for any k = 0, 1, · · · . Then, since x′j0+k =
xi0+k − t for any k = 0, 1, · · · , we have
G(ω + t)−G(ω)
=
j0−1∑i=0
(x′i − x′i+1)α −
i0−1∑i=0
(xi − xi+1)α
= −F (ω + t, x′j0) + F (ω, xi0) = F (ω, t).
Assume that there exists a separating line x = b of ω in the region(0, t] × (0,∞) (if t > 0) or (t, 0] × (0,∞) (if t < 0). Without lossof generality, we assume that t > 0. Let (a, b) × (c, d) be the tile indom(ω) such that c ≤ 1 < d. Then,
F (ω, t) = F (ω, a) + F (ω + a, b− a) + F (ω + b, t− b)
holds and from the above argument,
F (ω, a) = G(ω + a)−G(ω)
F (ω + b, t− b) = G(ω + t)−G(ω + b)
holds. Hence, to prove G(ω + t)−G(ω) = F (ω, t), it is sufficient toprove F (ω + a, b− a) = G(ω + b)−G(ω + a).Denoting ω for ω + b and t for a − b in the above, it is sufficient
to prove G(ω + t) − G(ω) = F (ω, t) in the case that x = 0 is theseperating line of ω and (t, 0)× (c, d) ∈ dom(ω) satisfies that c ≤ 1 <d and t < 0. Note that G(ω) = 0 since x0 = x1 = · · · = 0 in (??).Case 1: (ω + t)(R0(ω + t)) = 1. Since
x′i − x′i+1 =
β3j+1 (i = 2j)β3j+2 (i = 2j + 1)
,
G(ω + t) = βα + β2α + β4α + β5α + β7α + β8α + · · ·= γ + γ2 + γ4 + γ5 + γ7 + γ8 + · · ·= −1 + γ3 + γ4 + γ5 + γ7 + γ8 + · · ·= −1 + γ6 ++γ7 + γ8 + · · · = −1
41
holds by (??), where we used 1 + γ + γ2 = γ3.On the other hand, (t, 0)× (c, d) ∈ dom(ω) satisfies that −t = c ≤
1 < d and has color 1. Since −t ∈ g(1)βn; n ∈ Z = ββn; n ∈ Z,this implies that −t = 1. Hence, F (ω, t) = −(−t)α = −1. Thus,G(ω + t)−G(ω) = F (ω, t) holds.Case 2: (ω + t)(R0(ω + t)) = 2. Since
x′i − x′i+1 =
β3j (i = 2j)β3j+2 (i = 2j + 1)
,
G(ω + t) = 1 + β2α + β3α + β5α + β6α + β8α + β9α + · · ·= 1 + γ2 + γ3 + γ5 + γ6 + γ8 + γ9 · · ·= 1− γ + γ4 + γ5 + γ6 + γ8 ++γ9 · · ·= 1− γ + γ7 + γ8 + γ9 + · · · = 1− γ
holds by (??). On the other hand, (t, 0) × (c, d) ∈ dom(ω) satisfiesthat −t = c ≤ 1 < d and has color 2. Since −t ∈ g(2)βn; n ∈ Z =(β − 1)βn; n ∈ Z, this implies that −t = β − 1. Hence,
F (ω, t) = −δ(−t)α = − βα + 1
(β + 1)α(β − 1)α
= − βα + 1
(β3 − β2)α(β − 1)α = −γ + 1
γ2
= −γ3 − γ2
γ2= 1− γ.
Thus, G(ω + t)−G(ω) = F (ω, t) holds.
Thus, the α-homogeneous cocycle F is a coboundary with cobound-ary function G. The set G(I(Ω)) is known as Rauzy fractal. Forω ∈ I(Ω), let
I(Ω)a = ω ∈ I(Ω); ω(R0(ω)) = a (a = 1, 2)
andGa = G(I(Ω)a) (a = 1, 2).
42
x x x
y y
y
ri ri ri
1
1 2
2
11
2
1
1
Figure 9: βI(Ω)1, βI(Ω)2, β2I(Ω)2 + β
Considering the children of the tile R0(ω), we have a set equationthat
I(Ω)1 = βI(Ω)1 ∪ βI(Ω)2 ∪ (β2I(Ω)2 + β)
I(Ω)2 = βI(Ω)1 + 1.
Hence, the following set equation holds:
G1 = γG1 ∪ γG2 ∪ (γ2G2 + γ)
G2 = γG1 + 1.
G1 ∪G2 is shown in Figure 10.
Remark 2. The above Rauzy fractal is usually introduced by thesubstitution 1 → 12, 2 → 13, 3 → 1 (S. Ito and P. Arnoux [?]). Wemodify it canonically to the weighted substitution in Example 8 (seeRemark 1). The set equation is usually denoted as
G1 = γG1 ∪ γG2 ∪ γG3
G2 = γG1 + 1
G3 = γG2 + 1,
which is equivalent to ours.
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Figure 10: G(I(Ω))
7 N-process
We consider the N-process defined in Example 9. It is defined asa (1/2)-homogeneous cocycle F on the space Ω = Ω(σ, τ) with theweighted substitution (σ, τ) in Example 1. Hence, F is defined by
F (ω, x2)− F (ω, x1) = ±(x2 − x1)1/2 (19)
if there is a tile (x1, x2)× (y1, y2) ∈ dom(ω), where ± corresponds tothe color of the tile.Take ω0 ∈ Ω which has a tile R0 := (0, 1)× (1, 9/4) with ω0(R0) =
+. Then, F (ω0, 1) = F (ω0, 1) − F (ω0, 0) = 1 by (??). Since R0 has3 children R1,0, R1,1, R1,2 with colors +,−,+ and the vertical sizes4/9, 1/9, 4/9, we have F (ω0, 4/9) = 2/3 and F (ω0, 5/9)−F (ω0, 4/9) =−1/3 by (??). Hence, F (ω0, 5/9) = 1/3.Since there is a one-to-one correspondence between the descendants
44
of R0 and those of R1,0 keeping the color given by
(x1, x2)× (y1, y2) → ((4/9)x1, (4/9)x2)× ((4/9)y1, (4/9)y2).
Hence by (??),
F (ω0, (4/9)x2)− F (ω0, (4/9)x1) = (2/3)(F (ω0, x2)− F (ω0, x1))
holds if (x1, x2) × (y1, y2) ∈ dom(ω0) and (x1, x2) ⊂ [0, 1]. By thecontinuity of F (ω0, t) in t, this implies that
F (ω0, (4/9)t) = (2/3)F (ω0, t)
for any t ∈ [0, 1]. By the similar correspondence keeping the colorbetween the descendants of R0 and those of R1,2, we have
F (ω0, (5 + 4t)/9)− F (ω0, 5/9) = (2/3)F (ω0, t)
for any t ∈ [0, 1]. By the similar correspondence changing the colorbetween the descendants of R0 and those of R1,1, we have
F (ω0, (4 + t)/9)− F (ω0, 4/9) = −(1/3)F (ω0, t)
for any t ∈ [0, 1].Hence, the graph Γ of the function F (ω0, t) on t ∈ [0, 1] satisfies
the set equation Γ = ψ(Γ), where for a compact set R,
ψ(R) := ψ0(R) ∪ ψ1(R) ∪ ψ2(R)
with the functions ψ0, ψ1, ψ2 : H → R2 given by
ψ0(x, y) = ((4/9)x, (2/3)y)
ψ1(x, y) = ((x+ 4)/9, (−y + 2)/3)
ψ2(x, y) = ((4x+ 5)/9, (2y + 1)/3).
Then, Γ is obtained as the limit in the sense of Hausdorff metricof ψn(ΓN0) as n → ∞, where ΓN0 denotes the graph of the functionN0(t) := t (t ∈ [0, 1]). For n = 1, 2, · · · , define a function Nn on
45
Figure 11: N1, N2, N3 and N∞
46
[0, 1] by ΓNn = ψn(ΓN0). Then, N1 is a continuous piecewise linearfunction whose graph consists of 3 line segments
ψ0(ΓN0) = (x, y); y = (3/2)x, 0 ≤ x ≤ 4/9ψ1(ΓN0) = (x, y); y = −3x+ 2, 4/9 ≤ x ≤ 5/9ψ2(ΓN0) = (x, y); y = (3/2)x− (1/2), 5/9 ≤ x ≤ 1.
as seen in Figure 11.Since ΓN2 = ψ(ΓN1), N2 is a continuous piecewise linear function
on [0, 1] obtained by replacing 3 line segments in ΓN1 by self-affineimages of ΓN1 or Γ−N1 keeping the 2 end points fixed. Then, thegraph of N2 consists of 9 line segments. In the same way, we obtainNn from Nn−1 for n = 3, 4, · · · . Then, Nn is a continuous piecewiselinear function on [0, 1] whose graph consists of 3n line segments.Let Ξn be the set of closed intervals which are the projection to the
horizontal axis of the 3n line segments consisting of the graph of Nn.Note that Ξn is the set of closed intervals which are the projection tothe vertical axis of the descendents of the n-th generation of the tileR0 in ω0. Let ∆n be the set of the end points of Ξn. Let Ξ = ∪∞
n=0Ξn
and ∆ = ∪∞n=0∆n. Denote N∞(t) := F (ω0, t) (t ∈ [0, 1]). Then, the
function N∞(t) is the pointwise limit of Nn(t) as n→ ∞.
Theorem 12. @(i) For any s, t with 0 ≤ s < t ≤ 1, |N∞(t) − N∞(s)| ≤ |t − s|1/2
holds. Moreover, the equality holds if and only if [s, t] ∈ Ξ.(ii) For any Borel set R ⊂ [0, 1],∫
N∞(t)∈Rdt =
∫R
(2− |4t− 2|
)dt.
(iii) The Hausdorff dimension of the graph ΓN∞ of the functionN∞ satisfies that H-dimΓN∞ = 3/2. In fact, the (3/2)-dimensionalHausdorff measure of it is positive and finite.(iv) N∞(t) is locally minimal or maximal at t = t0 if and only if
t0 ∈ ∆. In this case, there exists δ > 0 such that
(1/√5)|h|1/2 ≤ |N∞(t0 + h)−N∞(t0)| ≤ |h|1/2
47
holds for any h with t0 + h ∈ [0, 1] and |h| < δ.(v) For any t0 ∈ [0, 1] with t0 /∈ ∆ and ϵ > 0, it holds that
H-dims ∈ (t0 − ϵ, t0); N∞(s) = N∞(t0)= H-dims ∈ (t0, t0 + ϵ); N∞(s) = N∞(t0) = 1/2.
In fact, the (1/2)-dimensional Hausdorff measures of the above setsare positive and finite.
Proof. (i) is proved in [?].(ii) Let L be a bounded operator on the Banach space C([0, 1])
defined by
(Lf)(t) := (4/9)f((2/3)t) + (1/9)f((2− t)/3) + (4/9)f((1 + 2t)/3)
for any t ∈ [0, 1] and f ∈ C([0, 1]).Let ν and τ be the probability measures on [0, 1] defined by
ν(S) =
∫N∞(t)∈S
dt∫S
dτ =
∫S
(2− |4t− 2|
)dt
for any Borel set S ⊂ [0, 1]. We prove that ν = τ .
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For any f ∈ C([0, 1]), we have∫Lfdν =
∫(Lf)(N∞(t))dt
=
∫(4/9)f((2/3)N∞(t))dt
+
∫(1/9)f((2−N∞(t))/3)dt
+
∫(4/9)f((1 + 2N∞(t))/3)dt
=
∫(4/9)f(N∞((4/9)t))dt
+
∫(1/9)f(N∞((t+ 4)/9))dt
+
∫(4/9)f(N∞((4t+ 5)/9))dt
=
∫ 4/9
0
f(N∞(t))dt+
∫ 5/9
4/9
f(N∞(t))dt+
∫ 1
5/9
f(N∞(t))dt
=
∫f(N∞(t))dt =
∫fdν.
Thus, ν is invariant under L. We prove that τ is also invariant underL.
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In fact,∫Lfdτ =
∫ (2− |4t− 2|
)(Lf)(t)dt
=
∫ (2− |4t− 2|
)(4/9)f((2/3)t)dt
+
∫ (2− |4t− 2|
)(1/9)f((2− t)/3)dt
+
∫ (2− |4t− 2|
)(4/9)f((1 + 2t)/3)dt
=
∫ 2/3
0
(2− |6t− 2|
)(4/9)f(t)(3/2)dt
+
∫ 2/3
1/3
(2− |12t− 6|
)(1/9)f(t)3dt
+
∫ 1
1/3
(2− |6t− 4|
)(4/9)f(t)(3/2)dt
=
∫ 1/3
0
6t(2/3)f(t)dt
+
∫ 1/2
1/3
((4− 6t)(2/3) + (−4 + 12t)(1/3) + (−2 + 6t)(2/3))f(t)dt
+
∫ 2/3
1/2
((4− 6t)(2/3) + (8− 12t)(1/3) + (−2 + 6t)(2/3)
)f(t)dt
+
∫ 1
2/3
(6− 6t)(2/3)f(t)dt
=
∫ (2− |4t− 2|
)f(t)dt =
∫fdτ.
By the definition of the operator L, we have
(Lf)′(t) = (8/27)f ′((2/3)t)−(1/27)f ′((2−t)/3)+(8/27)f ′((1+2t)/3)
for any f ∈ C1([0, 1]). This implies that ∥ (Lf)′ ∥≤ (17/27) ∥ f ′ ∥.Therefore, we have ∥ (Lnf)′ ∥≤ (17/27)n ∥ f ′ ∥, and hence, (Lnf)′
converges to 0. This implies that there exists a subsequence n′ ⊂
50
n such that Ln′f converges to a constant, say c. Hence, we have∫
fdν = lim
∫Ln′
fdν = c = lim
∫Ln′
fdτ =
∫fdτ
for any f ∈ C1([0, 1]). This implies that ν = τ .
(iii) Let Γ := ΓN∞(t). Since N∞ is uniformly 1/2-Holder continu-ous, the (3/2)-dimensional Hausdorff measure of Γ is finite. We provethat it is positive.Let νΓ be the probability measure supported by Γ such that for
any Borel set S ⊂ [0, 1],
νΓ((S × [0, 1]) ∩ Γ) :=
∫S
dt.
Then by (ii), it holds that
νΓ([0, 1]× S) =
∫S
(2− |4t− 2|
)dt. (20)
Take any cover U = Ui; i = 1, 2, · · · of Γ. Let I3/2(U) :=∑i d(Ui)
3/2, where d(U) denotes the diameter of the set U . Sincefor any set U , we can find a closed rectangle U ′ such that U ⊂U ′ and d(U ′) ≤ 2
√2d(U), we can replace each set U ∈ U by a
closed rectangle as this, so that we have a cover U ′ of Γ consistingof closed rectangles such that I3/2(U ′) ≤ 5I3/2(U). Since for anyinterval [a, b] ⊂ [c, d], we can find at most 2 intervals I1, I2 in Ξ suchthat I1∪I2 ⊃ [a, b] and |I1|+|I2| ≤ 9(b−a). We replace each rectangle[a, b]× [c, d] ∈ U ′ as above by I1 × [c, d] and I2 × [c, d]. Furthermore,we replace them again by Ii × ([c, d] ∩ N∞(Ii)) (i = 1, 2). Let V =Vi; i = 1, 2, · · · be the cover of Γ obtained by this procedure fromU . Then, we have
I3/2(V) ≤ 27I3/2(U ′) ≤ 135I3/2(U).
Take any of Vi =: I × [c, d] ∈ V . By the assumption, I ∈ Ξ and[c, d] ⊂ N∞(I) =: [C,D]. Let N∞(I) =: [C,D]. The graph of Γ
51
restricted to I × [C,D] is the image of Γ by the mapping (x, y) 7→(a+ |I|x, b± |I|1/2y) with some a, b and ±. Hence by (??), we have
νΓ(Vi) = νΓ(I × [c, d])
= |I|νΓ([0, 1]×
[ c− C
D − C,d− C
D − C
])≤ 2|I|x2 − x1
D − C
= 2|I|x2 − x1|I|1/2
= 2|I|1/2(x2 − x1) ≤ 2d(Vi)3/2. (21)
Thus, adding the above inequality, we have
I3/2(V) =∑i
d(Vi)3/2 ≥ (1/2)
∑i
νΓ(Vi) ≥ (1/2)νΓ(Γ) = 1/2,
so that I3/2(U) ≥ 1/270, which completes the proof.
(iv) Let t0 ∈ ∆ \ 0, 1. By (i),
|N∞(t0 + h)−N∞(t0)| ≤ |h|1/2
holds for any h with t0+h ∈ [0, 1]. Therefore, it is sufficient to provethat there exists δ > 0 such that
(1/√5)|h|1/2 ≤ |N∞(t0 + h)−N∞(t0)|
holds for any h with t0 + h ∈ [0, 1] and |h| < δ. There are intervalsI and J in some Γn such that t0 = I ∩ J and I is in the left sideof J . Then, either the piecewise linear function Nn is increasing inI and decreasing in J or Nn is decreasing in I and increasing in J .Without loss of generality, we assume the latter. Then, we have
N∞(t0 − |I|s)−N∞(t0) = |I|1/2(1−N∞(1− s))
N∞(t0 + |J |s)−N∞(t0) = |J |1/2N∞(s).
for any s ∈ [0, 1] by the set equation ψ(Γ) = Γ. Therefore, withh = |J |s
|N∞(t0 + h)−N∞(t0)| ≥1√5|h|1/2
52
follows from the statement that
N∞(s) ≥ s1/2√5
for any s ∈ (0, 1] and with h = −|I|s
|N∞(t0 + h)−N∞(t0)| ≥1√5|h|1/2
follows from the statement that
1−N∞(1− s) ≥ s1/2√5
for any s ∈ (0, 1]. By the symmetry of the graph of N∞ with respectto (1/2, 1/2), the second statement follows from the first statementthat
N∞(s) ≥ s1/2√5
for any s ∈ (0, 1].We prove this inequality. Note that the equality holds for s = 5/9.
Suppose that
N∞(s)/s1/2 <1√5
holds for some s ∈ (0, 1]. Since
c0 := infs∈(0,1]
N∞(s)/s1/2 = mins∈[4/9,1]
N∞(s)/s1/2 = mins∈[5/9,1]
N∞(s)/s1/2,
there exists s0 ∈ [5/9, 1] which attains the minimum. Then,
c0s01/2 = N∞(s0)
= (1/3) +N∞(s0 − (5/9))
> c0(5/9)1/2 + c0(s0 − (5/9))1/2
≥ c0s1/20
which is a contradiction. Thus, N∞(s) ≥ s1/2√5for any s ∈ [0, 1], which
completes the proof of (iv).
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(v) We prove that for any x ∈ (0, 1), the (1/2)-dimensional Haus-dorff measure of (N∞)x is positive and finite, where we denote (N∞)x :=t ∈ [0, 1]; N∞(t) = x. Let
δ := (2− |4x− 2|) ∧ (2− |2(x+ 1)− 2|) > 0.
We prove that for any cover U of (N∞)x, I1/2(U) ≥ (1/10)δ. Wemay assume that U consists of open intervals, so that there exits afinite subcover of U . Therefore, we may assume that U is a finitecover consisting of closed intervals. Moreover, by the same argumentas in the proof of (iii), it is sufficient to prove that for any
V := Ii; i = 1, 2, · · · , K ⊂ Ξ,
I1/2(V) ≥ (1/2)δ. Let ϵ > 0 satisfy that x + 2ϵ < 1 and |Ii| > ϵ forany i = 1, 2, · · · , K. By the same argument as in (??), we have
νΓ(Ii × [x, x+ ϵ]) ≤ 2|Ii|ϵ
|Ii|1/2≤ 2ϵ|Ii|1/2.
Adding the above inequality, we have
νΓ(Γ ∩ ([0, 1]× [x, x+ ϵ])) ≤ 2ϵI1/2(V),
Hence by (??) and the definition of δ, we have
I1/2(V) ≥ (2ϵ)−1νΓ([0, 1]× [x, x+ ϵ]) ≥ (1/2)δ,
which completes the proof that Hausdorff measure of (N∞)x is posi-tive.To prove that it is finite, for any sufficiently small ϵ > 0 and t ∈
(N∞)x, we take V (t) ∈ Ξ such that t ∈ V (t) and 2ϵ ≤ |V (t)| < 18ϵ.Then, there exists a finite subcover V := Vi : i = 1, 2, · · · , K ofV (t); t ∈ (N∞)x such that Vi ∩ Vj is at most one point for anyi = j.For any i = 1, 2, · · · , K, by the same argument as in (??), we have
νΓ(Vi × [x− ϵ1/2, x+ ϵ1/2])
≥ |Vi|∫ (ϵ/|Vi|)1/2
0
(2− |4t− 2|)dt
≥ |Vi|2ϵ/|Vi| ≥ 2ϵ ≥ (2/5)ϵ1/2|Vi|1/2.
54
Adding this inequality together with (??), we have
4ϵ1/2 ≥ νΓ([0, 1]× [x− ϵ1/2, x+ ϵ1/2])
= νΓ(Γ ∩ ([0, 1]× [x− ϵ1/2, x+ ϵ1/2]))
≥ (2/5)ϵ1/2I1/2(V).
Hence, we have I1/2(V) ≤ 10, which completes the proof that (1/2)-dimensional Hausdorff measure of (N∞)x is finite.The statements in (iv) follows from this.
Consider the stochastic process (Nt)t∈R defined byNt(ω) = F (ω, t),where ω comes from the probability space (Ω, µ), µ being the uniqueinvariant probability measure invariant under the additive action.This process was called the N-process and studied in [?]. A pre-diction theory based on the N-process was developed. A processYt = H(Nt, t), where the function H(x, s) is an unknown functionwhich is twice continuously differentiable in x and once continuouslydifferentiable in s and Hx(x, s) > 0 is considered. The aim is topredict the value Yc from the observation YJ := Yt; t ∈ J, whereJ = [a, b] and a < b < c.
Theorem 13. ([?]) There exists an estimator Yc which is a measur-able function of the observation YJ such that
E[(Yc − Yc)2] = O((c− b)2)
as c ↓ b.
Open problems:(1) The systems of the geodesic and horocycle flows on the upper halfplane H/G devided by cocompact discrete groups G ∈ PSL2(R) arenumeration systems which are not isomorphic to those coming fromweighted substitutions (private communications by Prof. BenjaminWeiss). How to characterize the numeration systems coming fromweighted substitutions?(2) Is the condition B(σ, τ) = R+ necessary for the R-action of anumeration system coming from weighted substitutions with respect
55
to the unique invariant probability measure to be weakly mixing?When doess it have a discrete spectrum?(3) What is the multiplicity of the pure Lebesgue spectrum possessedby the R-action of a numeration system coming from a weighted sub-stitution with B(σ, τ) = R+ with respect to the unique invariantprobability measure?(4) When does a numeration system admit an additive group struc-ture consistent with the (R, G)-action?
Acknowledgement: The author would like to thank the anony-mous referee for useful suggestions. The metric in (11) and the proofof Lemma 4 which simplified the original one are suggested by thereferee. The author thanks Prof. Sigeki Akiyama and Prof. PierreArnoux for their useful discussions with the author. The author alsothanks Prof. Benjamin Weiss and Prof. Hillel Furstenberg for teach-ing him the references about Theorem 3.
References
[1] Pierre Arnoux and Shunji Ito, Pisot substitutions and Rauzy frac-tals, Bull. Belg. Math. Soc. 8(2001), pp.181-207.
[2] V. Berthe, S. Ferenczi, C. Mauduit and A. Siegel (ed.), Substitu-tions in dynamics, arithmetics and combinatorics, Lecture Notesin Mathematics 1794, Springer-Verlag, 2002.
[3] N. Dunford and J. T. Schwartz, Linear Operators II, Interscience,1963
[4] T. N. T. Goodman, Relating topological entropy and measureentropy, Bull. London Math. Soc. 3(1971), pp.176-180.
[5] Nertila Gjini and Teturo Kamae, Coboundary on colored tilingspace as Rauzy fractal, Indagationes Mathematicae 10-3 (1999),pp.407-421.
[6] G. H. Hardy and M. Riesz, The general theory of Dirichlet’s series,Cambridge Univ. Press, 1952
56
[7] S. Ito and Y. Takahashi, Markov subshifts and realization of β-expansions, J. Math. Soc. Japan 26(1974), pp.33-55.
[8] Teturo Kamae, Linear expansions, strictly ergodic homogeneouscocycles and fractals, Israel J. Math. 106 (1998), pp.313-337.
[9] Teturo Kamae, Stochastic analysis based on deterministic Brow-nian motion, Israel J. Math. 125 (2001), pp.317-346.
[10] Teturo Kamae, Numeration systems, fractals and stochastic pro-cesses, Israel J. Math. (to appear).
[11] Karl Petersen, Ergodic Theory, Cambridge studies in advancedmathematics 2, Cambridge Univ. Press, 1983.
[12] A. Plessner, Spectral theory of linear operators I, UspekhiMatem. Nauk 9 (1941), pp.3-125.
[13] A. N. Starkov, Dynamical Systems on Homogeneous Spaces,Translations of Mathematical Monographs 190, Amer. Math.Soc., 2000.
[14] Peter Walters, Ergodic Theory− Introductory Lectures, LectureNotes in Mathematics 458. Springer-Verlag, 1975.
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