Linear numeration systems and
homogeneous cocycles
Teturo Kamae
(Osaka City Univ., 558-8585 Japan)
(kamae@apost.plala.or.jp)
Abstract: A numeration system originally implies a digitization
of real numbers, but in this paper it rather implies a compactification
of real numbers as a result of the digitization.
By definition, a linear numeration system with G, where G is a
nontrivial closed multiplicative subgroup of R+ , is a compact metrizable space Ω admitting a continuous (λω + t)-action of (λ, t) ∈ G × R
to ω ∈ Ω, such that the (· + t)-action is strictly ergodic having 0
topological entropy with the unique invariant probability measure
µΩ , which is the unique G-invariant probability measure attaining
the topological entropy | log λ| of the (λ × ·)-action.
We construct a class of numeration systems coming from weighted
substitutions, which contains those coming from substitutions or βexpansions with algebraic β. It also contains those with G = R+ .
We obtained the ζ-function ζΩ of the numeration systems Ω coming from weighted substitutions. For α ∈ C with 0 < Re(α) < 1,
there exists a nontrivial adapted α-homogeneous cocycle on Ω if and
only if ζΩ has a pole at α. These cocycles are generalizations of the
fractal functions of Peano type. Moreover, in the case G = R+ , these
cocycles define self-similar processes of order α under µΩ . We discuss
one 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 set
of integer points I(Ω) in Ω, which is proved to be a coboundary. The
image of I(Ω) under this coboundary function becomes a fractal set
1
of Rauzy type.
This paper is based on [10], changed the way of presentation and
modified on a large scale.
1
Linear numeration systems
By a linear numeration system, we mean a compact metrizable space
Ω 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 such
that λ′ (λω + t) + t′ = λ′ λω + λ′ t + t′ .
(♯2) The (· + t)-action is strictly ergodic having 0 topological entropy with the unique invariant probability measure µΩ called the
equilibrium measure on Ω. Consequently, it is invariant under the
(λω + t)-action of (λ, t) ∈ G × R to ω ∈ Ω as well.
(♯3) The (λ × ·)-action has | log λ| topological entropy which is
attained uniquely by µΩ .
The (ω + t)-action of t ∈ R to ω ∈ Ω is called the additive action
or R-action, while the λω-action of λ ∈ G to ω ∈ Ω is called the
multiplicative action or G-action.
Note that if Ω is a linear numeration system, then Ω is a connected
space with the continuum cardinality. Also, note that the multiplicative group 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 implies
t = 0 for any ω ∈ Ω and t ∈ R.
This is because if there exist ω1 ∈ Ω and t1 6= 0 such that ω1 + t1 =
ω1 , then take a sequence λn in G such that λn → 0 and λn ω1 converges
as n → ∞. Let ω∞ := limn→∞ λn ω1 . For any t ∈ R, let an be a
sequence of integers such that an λn t1 → t as n → ∞. Then we have
ω∞ + t = lim (λn ω1 + λn an t1 )
n→∞
=
lim λn (ω1 + an t1 ) = lim λn ω1 = ω∞ .
n→∞
n→∞
Thus, ω∞ becomes a fixed point of the (ω+t)-action of t ∈ R to ω ∈
Ω. Since this action is minimal, we have Ω = {ω∞ }, contradicting
2
with that Ω has at least 2 elements.
An example of a linear numeration system is the set {0, 1}Z with
the product topology divided by the closed equivalence relation ∼
such that
(· · · 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 the
same statement with α and β exchanged. Let Ω(2) := {0, 1}Z/ ∼ and
the equivalencePclass containing (· · · α−2 , α−1 ; α0 , α1 α2 · · · ) ∈ {0, 1}Z
n
is denoted by ∞
n=−∞ αn 2 ∈ Ω(2). Then, Ω(2) is an additive topological group with the addition as follows:
∞
X
αn 2 n +
n=−∞
∞
X
βn 2n =
n=−∞
∞
X
γ n 2n
n=−∞
if and only if there exists (· · · η−2 , η−1 ; η0 , η1 , η2 · · · ) ∈ {0, 1}Z satisfying that
2ηn+1 + γn = αn + βn + ηn (∀n ∈ Z).
This is isomorphic to the 2-adic solenoidal group which is by definition the projective limit of the projective system θ : R/Z → R/Z
with θ(α) = 2α (α ∈ R/Z).
Moreover, R is imbedded in Ω(2) continuously as a dense additive
subgroup
that a nonnegative
real number α is identified
PN
P∞ in the way
n
with n=−∞ αn 2 such that α = n=−∞ αn 2n and αn = 0 (∀n > N)
for some N ∈ Z,
a negative real number −α with α as above is
Pwhile
∞
identified with n=−∞ (1−αn )2n . Then, R acts additively to Ω(2) by
this addition. Furthermore, G := {2k ; k ∈ Z} acts multiplicatively
to Ω(2) by
∞
∞
X
X
n
k
αn 2 =
αn−k 2n .
2
n=−∞
n=−∞
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.
3
Theorem 1. Ω(2) is a linear numeration system with G = {2n ; n ∈
Z}.
We can express Ω(2) in the following different way. By a partition
of the upper half plane H := {z = x + iy; y > 0}, we mean a disjoint
family of open sets such that the union of their closures coincides
with H. Let us consider the space Ω(2)′ of partitions ω of H by open
squares of the form (x1 , x2 ) × (y1 , y2 ) with x2 − x1 = y2 − y1 = y1 and
y1 ∈ G such that (x1 , x2 ) × (y1 , y2 ) ∈ ω implies
and
(x1 , (x1 + x2 )/2) × (y1 /2, y1) ∈ ω (type 0)
(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 the
squares 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 the
squares in ω intersecting with
the line segment from +0 + i to +0.
P∞
n
Then, ω is identified
P∞ with n n=−∞ αn 2 . Note that replacing +0
by −0, we get
n=−∞ βn 2 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 such
that limn→∞ ρ(R, Rn ) = 0, where ρ is the Hausdorff metric between
sets R, R′ ⊂ H
ρ(R, R′ ) := max{sup inf
|z − z ′ |, sup inf |z − z ′ |}.
′
′
z ′ ∈R′ z∈R
z∈R z ∈R
(2)
For ω ∈ Ω(2)′ , t ∈ R and λ ∈ {2n ; n ∈ R}, ω + t ∈ Ω(2)′ and
λω ∈ 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}.
4
y
iu
x
Figure 1: the tiling corresponding to · · · 01.101 · · ·
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+
t) = κ(ω) + t and κ(λω) = λκ(ω) for any ω ∈ Ω(2)′ , t ∈ R and
λ ∈ {2n ; n ∈ Z}. Thus, Ω(2)′ is isomorphic to Ω(2) as a linear
numeration system 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)
5
x
Figure 2: admissible tiles
in H is called an admissible tile if
x2 − x1 = y1
is satisfied (see Figure 2). In another word, an admissible tile is a
rectangle (x1 , x2 ) × (y1 , y2 ) in H such that the lower side has the
hyperbolic 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) 6= (R′ , a′ ),
and
(2) ∪a∈A ∪(R,a)∈ω R = 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 tile
R 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 {ωn }n∈I ⊂ Ω(A) converges to
6
ω ∈ Ω(A) if for every (R, a) ∈ ω, there exists (Rn , an ) ∈ ωn such that
an = a for any sufficiently large n ∈ I and lim ρ(R, Rn ) = 0,
n→∞
where ρ is the Hausdorff metric defined in (2).
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) ∈ ω}
λω = {(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 of Ω(A)
corresponding to weighted substitutions which are linear numeration
systems. Though ♯A ≥ 2 is assumed in [8], we consider the case
♯A = 1 as well.
2
Remarks on the notations and terminology
In this paper, the notations and terminology are changed in a large
scale from the previous papers [8], [9] and [10] of the author. The
main changes are as follows:
(1) In [10], the author called the linear numeration system just “numeration system”. Since the word “numeration system” is widely
used in more general sense, we change the terminology here.
(2) Here and in [10], the colored tilings are defined on the upper half
plane H and the scaling action is defined to be multiplication, different from in the [8], [9]. (3) Here and in [10], the set of colors is
7
denoted by A instead of Σ. Colors are denoted by a, a′ , ai (etc.)
instead of σ, σ ′ , σi (etc.).
(4) Here and in [10], the weighted substitution is denoted by (σ, τ )
instead of (ϕ, η).
(5) Here and in [10], admissible tiles are denoted by R, R′ , Ri , Ri
(etc.) instead of S, S ′ , Si , S i (etc.).
(6) Here and in [10], the terminology “primitive” for substitutions is
used instead of “mixing” in [8] and [9].
3
Weighted substitutions
A
σ on a set A is a mapping A → A+ , where A+ =
S∞substitution
ℓ
+
ℓ
ℓ=1 A . For ξ ∈ A , we denote |ξ| := ℓ if ξ ∈ A , and ξ with |ξ| = ℓ
is usually denoted by ξ0 ξ1 · · · ξℓ−1 with ξi ∈ A. We can extend σ to
be 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 σ 2 , σ 3 , · · · as the compositions of σ : A+ → A+ .
A weighted substitution (σ, τP
) 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. Note
that σ is a substitution on A. We define τ n : A → (0, 1)+ (n = 2, 3, ...)
(depending on σ) inductively by
τ n (a)k = τ (a)i τ n−1 (σ(a)i )j
for any a ∈ A and i, j, k with
0 ≤ i < |σ(a)|, 0 ≤ j < |σ n−1 (σ(a)i )|, k =
X
h<i
|σ n−1(σ(a)h )| + j.
Then, (σ n , τ n ) 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, σ n (a)i = a′ holds for some i
with 0 ≤ i < |σ n (a)|.
8
For a weighted substitution (σ, τ ) on A, we always assume that
the substitution σ is primitive.
(3)
We define the base set B(σ, τ ) as the closed, multiplicative subgroup
of R+ generated by the set
n
τ (a)i ; a ∈ A, n = 0, 1, · · · and 0 ≤ i < |σ n (a)|
.
such that σ n (a)i = a
+
:
4/9
:
1/9
4/9
+
+
−
+
+
+
+
−
+
-
+
−
+
-
+
+
−+−
-
+
+
-
+
Figure 3: the weighted substitution in Example 1
9
x
Example 1. Let A = {+, −} and (σ, τ ) be a weighted substitution
such that
+ → (+, 4/9)(−, 1/9)(+, 4/9)
− → (−, 4/9)(+, 1/9)(−, 4/9),
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. Moreover, 1/81 ∈ B(σ, τ ) since σ 2 (+)4 = + and τ 2 (+)4 = 1/81. Since
4/9 and 1/81 do not have a common multiplicative base, we have
B(σ, τ ) = R+ . This weighted substitution is discussed in the following sections. The repetition of this weighted substitution starting at
+ is shown in Figure 3 by colored tiles. The substituted word of a
color is represented as the sequence of colors of the connected tiles in
below in order from left. The horizontal (additive) sizes of tiles are
proportional to the weights and the vertical (multiplicative) sizes are
the inverse of the weights.
Let G := B(σ, τ ). Then, there exists a function g : A → R+ such
that
g(σ(a)i )G = g(a)τ (a)i G
(4)
for any a ∈ A and 0 ≤ i < |σ(a)|. Note that if G = R+ , then we
can take g ≡ 1. In the other case, we can define g by g(a0 ) = 1 and
g(a) := τ n (a0 )i for some n and i such that σ n (a0 )i = a, where a0 is
any 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 ω
in 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
R
i
:= (x1 + (x2 − x1 )
×(τ (a)i y1 , y1 ).
i−1
X
j=0
τ (a)j , x1 + (x2 − x1 )
10
i
X
j=0
τ (a)j )
A vertical line γ := {x} × (−∞, ∞) is called a separating line of
ω ∈ Ω(σ, τ, g)′ if for any (R, a) ∈ ω, R ∩ γ = ∅. Let Ω(σ, τ, g)′′ be
the set of all ω ∈ Ω(σ, τ, g)′ which do not have a separating line and
Ω(σ, τ, g) be the closure of Ω(σ, τ, g)′′ . Then, the action of G × R
on Ω(σ, τ, g) satisfies (♯1). We usually denote Ω(σ, τ, 1) simply by
Ω(σ, τ ).
+
Remark 1. [8] A nontrivial
P 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 M
and ξ := (ξa )a∈A be a positive column vector such that Mξ = λξ.
Define weight τ by
ξσ(a)i
τ (a)i =
,
λξa
which is called the natural weight of σ. Thus, we get a weighted substitution (σ, τ ) which admits weight 1. We modify (σ, τ ) if necessary
in the following way. If there exists a ∈ A with |σ(a)| = 1, so that
a → (a′ , 1) is a part of (σ, τ ), then we replace all the occurrences of
a in the right hand side of “→” by a′ and remove a from A together
with the rule a → (a′ , 1) from (σ, τ ). We continue this process until
no a ∈ A satisfies |σ(a)| = 1. After that if there exist a, a′ ∈ A such
that (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). The
Thue-Morse substitution 1 → 12, 2 → 21 corresponds to the weighted
substitution 1 → (1, 1/2)(2, 1/2), 2 → (2, 1/2)(1, 1/2). The Fibonacci substitution 1 → 12, 2 → 1 corresponds√ to the weighted
substitution 1 → (1, λ−1)(1, λ−2 ), where λ = (1 + 5)/2.
The weighted substitution (σ, τ ) obtained in this way satisfies that
B(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 discussed by many authors (see [2], for example). Our weighted substitutions are a generalization of them.
11
Let (σ, τ ) be a weighted substitution on A satisfying (3). Let g
satisfy (4). Consider Ω(σ, τ, g). We call the tile Ri in (II) the i-th
child of the tile (x1 , x2 ) × (y1 , y2 ), and (x1 , x2 ) × (y1 , y2) the mother of
Ri . Note that the vertical size S(Ri ) of Ri coincides with the inverse
of 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
the i-th tile among the set of the k-th descendants of Rk counting as
0, 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 linear numeration system with
G = B(σ, τ ).
Proof. We have already proved (♯1) and (♯2) in Theorem 3 of [8].
Here we prove (♯3). Let Ω := Ω(σ, τ, g) and µΩ be the equilibrium
measure. Since µΩ is the unique invariant probability measure under
the additive action, it is also invariant under the multiplicative action.
By Goodman [4], it is sufficient to prove that for any λ ∈ G with
λ 6= 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 line}
Σ0 := {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 individually with respect to the G-action. Hence, Gω with respect to the
G-action is either cyclic or conjugate to a 2-dimensional irrational
rotation 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 of
pairs a ∈ A and i with 0 ≤ i < i + 1 < |σ(a)| onto the set of minimal
sets in Σ0 .
12
Proof. (i) If the line x = u is the separating line of ω ∈ Ω, then
x = λu is the separating line of λω. Hence, Σ \ Σ0 is dissipative.
(ii) Let ω ∈ Σ0 . Denote by ω + the restriction of ω to the right
quarter plane (0, ∞) × (0, ∞), while by ω − the restriction of ω to the
left quarter plane (−∞, 0) × (0, ∞). Let (Ri± )i∈Z be the sequence of
tiles in dom(ω) such that Ri± intersects with the upper half lines of
±
x = ±0, and Ri± is a child of Ri+1
for any i ∈ Z (± respectively). Let
±
±
±
ai := ω(Ri ) be the colors of Ri (± respectively). Define mappings
σ± from A to A by σ+ (a) = σ(a)0 and σ− (a) = σ(a)|σ(a)|−1 . Since
±
±
σ± (a±
i ) = ai−1 (i ∈ Z) (± respectively), the sequence (ai )i∈Z is
±
±
periodic, which also implies that the vertical sizes S(Ri ) of Ri , which
coincide with the inverses of the weights τ (ai+1 )± , are also periodic
in i ∈ Z with the period, say r ± which is the minimum period of
+
−1
+
(a±
:= τ r (a+
is the minimum
0 )0
i )i∈Z (± respectively). Then, λ
−
−1
(multiplicative) cycle of ω + , while λ− := τ r (a−
)
0 |σr − (a− )|−1 is the
0
minimum (multiplicative) cycle of ω − , that is, λω ± = ω ± holds for
λ = 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 cycle
of ω is the minimum positive number x such that x = (λ+ )n =
(λ− )m holds for some positive integers n, m. Otherwise, the G-action
to Gω is conjugate to an 2-dimensional irrational rotation with a
multiplicative 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 ′ ∈ Ω
having a tile R ∈ dom(ω ′ ) with ω ′(R) = a such that the y-axis
passes 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
n
n
(σ−
(σ(a)i ), σ+
(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 assume
13
that there exists a pair (a, i) such that for any n = 1, 2, · · · , there
exists R ∈ dom(ωn ) with a = ωn (R) such that the y-axis separetes
the i-th child of R and the i + 1-th child of R. Then, ω ∈ ψ(a, i),
which proves that ψ is a mapping from the set of pairs (a, i) with
a ∈ A and 0 ≤ i < i + 1 < |σ(a)| onto the set of minimal sets in Σ0
with respect to the G-action.
y
vertical sizes
1/(1 − p)
vertical sizes
1/p
x
Figure 4: an element in Σ0 in Example 2
Example 2. Let p with 0 < p < 1 satisfy that log p/ log(1 − p) is
irrational. Let (σ, τ ) be a weighted substitution on A = {1} such that
1 → (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 as
shown in Figure 4. Then, the dynamical system (Σ0 , λ (λ ∈ R+ )) is
isomorphic 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 6=
14
1 and ν be any other G-invariant probability measure on Ω, then
hν (λ) < | log λ|.
Proof. To prove the lemma, it is sufficient to prove the statements
for λ > 1. Take any G-invariant probability measure ν on Ω which
attains the topological entropy of the multiplication by λ1 ∈ G with
λ1 > 1, that is, hν (λ1 ) = log λ1 . We assume also that the G-action to
Ω is ergodic with respect to ν. Then by Lemma 1, either ν(Σ0 ) = 1 or
ν(Ω \ Σ) = 1. In the former case, hν (λ) = 0 holds for any λ ∈ G since
the G-action on Σ0 is equicontinuous by Lemma 1, which contradicts
with 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 := min{b ≤ 1; b ∈ g(a0 )G} (see (4)). Let
Ω1 := { ω ∈ Ω; the set {λ ∈ G; λω(R0 (λω)) = a0 }
is 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 that
λω ∈ Ω0 . Define a mapping Λ : Ω0 → Ω0 by Λ(ω) := λ0 (ω)ω.
For k = 0, 1, 2, · · · and i = 0, 1, · · · , |σ k (a0 )| − 1, let
P (k, i) := { ω ∈ Ω0 ; λ0 (ω)−1 R0 (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 < |σ k (a0 )| }
be a measurable partition of Ω0 . Note that λ0 (ω) = τ k (a0 )−1
if
i
ω ∈ P (k, i).
15
y
a0
b0 s
a0
s
(3/13)b0
x
Figure 5: ω ∈ P (3, 4) with λ0 (ω) = 13/3
Since ν(Ω1 ) = 1 by the ergodicity and
[
[
λP (k, i),
Ω1 =
P (k,i)∈P
1≤λ<τ k (a0 )−1
i
λ∈G
there exists a unique Λ-invariant probability measure ν0 on Ω0 such
that for any Borel set B ⊂ Ω, we have
X Z b0 τ k (a0 )−1
i
−1
ν0 (λ−1 B ∩ P (k, i))dλ/λ
(5)
ν(B) = C(ν)
P (k,i)∈P
with
C(ν) :=
X
P (k,i)∈P
if G = R+ and
ν(B) = C(ν)−1
X
b0
− log τ k (a0 )i ν0 (P (k, i)) < ∞
P (k,i)∈P
X
λ∈G
b0 ≤λ<b0 τ k (a0 )−1
i
16
ν0 (λ−1 B ∩ P (k, i))
(6)
(7)
with
C(ν) :=
X
P (k,i)∈P
(− log τ k (a0 )i / log β) ν0 (P (k, i)) < ∞
if G = {β n ; n ∈ Z} with β > 1.
Since
X
τ k (a0 )i = 1 and
P (k,i)∈P
X
(8)
ν0 (P (k, i)) = 1,
P (k,i)∈P
we have
Hν0 (P) := −
≤ −
X
P (k,i)∈P
X
P (k,i)∈P
log ν0 (P (k, i)) · ν0 (P (k, i))
log τ k (a0 )i · ν0 (P (k, i))
(9)
by the convexity of − log x. The equality in (9) holds if and only if
ν0 (P (k, i)) = τ k (a0 )i (∀P (k, i) ∈ P).
(10)
By (6)(8)(9), we have
Hν0 (P) = −
X
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 the
same element in P for k = 0, 1, 2, · · · , the horizontal position of
R0 (ω), say (x1 , x2 ), coincides with that of R0 (ω ′ ). Therefore, ω and
ω ′ 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, then
R0 := 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 the
region covered by the ancestors of R0 coincide. Hence, if ω or ω ′ does
not have the separating lines, then ω = ω ′ holds.
Since ν(Σ) = 0, we have ν0 (Σ ∩ Ω0 ) = 0. Hence, the above argument implies that P is a generator of the system (Ω0 , ν, Λ). Thus,
17
hν0 (Λ) = hν0 (Λ, P). It follows from (9) that
hν0 (Λ) = hν0 (Λ, P)
≤ Hν0 (P)
X
≤ −
log τ k (a0 )i · ν0 (P (k, i)).
(11)
P (k,i)∈P
The equality in the above that
X
hν0 (Λ) = −
log τ k (a0 )i · ν0 (P (k, i))
P (k,i)∈P
holds if and only if (Λn P)n∈Z is an independent sequence with respect
to ν0 satisfying (10).
Since
hν0 (Λ)
λ (ω)dν0 (ω)
Ω0 0
hν (λ1 )/ log λ1 = R
=
−
hν0 (Λ)
,
k
P (k,i)∈P log τ (a0 )i · ν0 (P (k, i))
P
hν (λ1 ) ≤ log λ1 follows from (11), while the equality holds if and only
if (Λn P)n∈Z is an independent sequence with respect to ν0 satisfying
(10). Let this probability measure on Ω0 be µ0 and the measure ν in
(5) for this µ0 be µ. We prove that µ is invariant under the additive
action, then by the uniqueness of such a measure ([8]), µ = µΩ follows.
∞
n
We may identify ω ∈ Ω0 with the element in P−∞
:= ∨∞
n=−∞ Λ P.
Let ω+ be the elemnet of P0∞ including ω, and ω− be the elemnet
−1
of P−∞
including ω. Therefore, we can write ω = (ω− , ω+ ). Since
n
(Λ P)n∈Z is an independent sequence of partitions with respect to
+
µ0 , we can write the measure µ0 = µ−
0 × µ0 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 ∈ P0N 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 the
18
right). Hence,
B = {ω ∈ Ω0 ; − b0
i
X
j=0
k
τ (a0 )j < x0 (ω) ≤ −b0
i−1
X
j=0
τ k (a0 )j }.
′
Moreover, µ0 (B) = τ k (a0 )i . Take another N ′ > N and B ∈ P0N such
that B ′ ⊂ B. Then, the pair (k ′ , i′ ) determined by B ′ is such that
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
−b0
i
X
j=0
′
i
X
k
τ (a0 )j ≤ −b0
< x0 (ω) ≤ −b0
i′ −1
X
j=0
′
τ k (a0 )j
j=0
′
τ k (a0 )j ≤ −b0
i−1
X
τ k (a0 )j .
j=0
This implies that there exists a measurable bijection ψ : P0∞ →
[0,P
b0 ) such that ψ(ω+ ) := −x0 (ω) which is actually the limit of
k
N
b0 i−1
j=0 τ (a0 )j as above along the element B ∈ P0 including ω+ .
Moreover, the function ψ is uniformly distributed on [0, b0 ) under µ0
since µ0 (B) = τ k (a0 )i for any B ∈ P0N 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 set
B ⊂ Ω0 , if ψ(ω+ ) + t ∈ [0, b0 ) holds for any ω ∈ B, then µ0 (B + t) =
µ0 (B).
For y ∈ G, let Ωy0 := {ω ∈ Ω; y −1ω ∈ Ω0 } and µy0 be the measure
on Ωy0 such that µy0 (B) = µ0 (y −1B). Let xy0 be the function on Ωy0
such that xy0 (ω) = yx0 (y −1 ω). 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 ⊂ Ωg0 , if
−xy0 (ω) + t ∈ [0, yb0 ) holds for any ω ∈ B, then µy0 (B + t) = µy0 (B).
Note that (5)(7) implies that for any measurable set B ⊂ Ω1 , it
holds that
Z ∞
−1
µ(B) = C(µ)
µy0 (By1 )dy/y
1
or
µ(B) = C(µ)−1
X
y∈G, y≥1
19
µy0 (By1 )
where
By1 = {ω ∈ B; y is the minimum λ ≥ 1 with λ ∈ G and ω ∈ Ωλ0 }.
If we replace 1 by b ∈ G in the above, we have
Z ∞
−1
µ(B) = Cb (µ)
µy0 (Byb )dy/y
b
or
µ(B) = Cb (µ)−1
X
µy0 (Byb )
y∈G, y≥b
where
Byb = {ω ∈ B; y is the minimum λ ≥ b with λ ∈ G andω ∈ Ωλ0 }.
For any measurable set B ⊂ Ω1 , t ∈ R and b ∈ G, let
B ′ = {ω ∈ B; − xb0 (ω) + t 6∈ [0, bb0 )}.
Then, µ(B ′ ) ≤ |t|/(bb0 ) holds. Since µy0 (B \ B ′ ) = µy0 ((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 have
µ(B + t) ≤ µ(B). Thus, we have µ(B) = µ(B + t) and µ is invariant
under the additive action, which complete the proof of Lemma 2 and
Theorem 2.
The following Theorem 3 follows from a known result about the
spectrum of unitary operators corresponding to the affine action
(Lemma 11.6 of [13], for example). Nevertheless, we give the proof
for to be self-contained.
Theorem 3. Let Ω be a linear numeration system with G = R+ ,
that is, with the multiplicative R+ -action. Then, the additive action
on the probability space Ω with respect to µΩ has a pure Lebesgue
spectrum.
20
Proof. Let Ut (t ∈ R) and Vλ (λ ∈ R+ ) be the groups of the unitary
operators on L2 (Ω, µΩ ) defined by
(Ut f )(ω) = f (ω + t) , (Vλ f )(ω) = f (λω).
Let
Ut =
Z
∞
−∞
eitu dEu (t ∈ R)
be the spectral decomposition of Ut ([3]). Since Ut Vλ = Vλ Uλt , we
have dEu Vλ = Vλ dEλ−1 u .
R
R
Take any f ∈ L2 (Ω, µΩ ) with f dµΩ = 0 and |f |2dµΩ = 1. Let
m(f ) be the measure on R defined by
Z
k dEu f k2
m(f )(S) =
S
for any Borel set S ⊂ R. Then, m(f ) is a probability measure with
m(f )({0}) = 0. Since dEu Vλ = Vλ dEλ−1 u , we have
Z
Z
2
m(Vλ f )(S) =
k dEu Vλ f k =
k Vλ dEλ−1 u f k2
S
S
Z
Z
2
=
k dEλ−1 u f k =
k dEu f k2 = m(f )(λ−1 S).
λ−1 S
S
Moreover, we have
Z
|(f, Vλ f )| = | (dEu f, dEu Vλ f )|
Z
Z p
p
≤
k dEu f kk dEu Vλ f k=
dm(f ) dm(Vλ f ).
Since limλ→1 |(f, Vλ f )| = 1, we have
Z p
Z p
p
p
−1
dm(f ) dm(f ) ◦ λ = lim
dm(f ) dm(Vλ f ) = 1.
lim
λ→1
λ→1
It follows from this that m(f ) is absolutely continuous by the following well known argument (see [12], for example).
21
Suppose to the contrary that m := m(f ) is not absolutely continuous. Take a Borel set S ⊂ R such thatR S√ has√Lebesgue measure 0
while δ := m(S) > 0. Denoting ρ(λ) :=
dm dm ◦ λ−1 , we have
Z √
2
√
2(1 − ρ(λ)) =
dm − dm ◦ λ−1
2 √
2
p
p
p
−1
−1
m(S) − m ◦ λ (S) =
δ − m(λ S) .
≥
Since 2(1 − ρ(λ)) → 0 as λ → 1, there exists ǫ > 0 such that for any
λ with 1 − 2ǫ ≤ λ ≤ 1 + 2ǫ, m(λ−1 S) > δ/2 holds. Hence,
Z 1+2ǫ
Z Z 1+2ǫ
−1
2δǫ ≤
m(λ S)dλ =
1S (λu)dλdm(u).
1−2ǫ
1−2ǫ
R 1+2ǫ
This implies that the set of u ∈ R such that 1−2ǫ 1S (λu)dλ ≥ δǫ
has the measure at least δ/4 with respect to m. Since m({0}) = 0,
R 1+2ǫ
this implies that there exists u 6= 0 such that 1−2ǫ 1S (λu)dλ ≥ δǫ.
Thus, S has Lebesgue measure at least |u|δǫ, which contradicts the
assumption that S has Lebesgue measure 0. Thus, m = m(f ) is
absolutely continuous.
4
ζ-function
Let Ω := Ω(σ, τ, g) satisfying (3) and (4). For α ∈ C, we define the
associated matrices on the suffix set A × A as follows:


X
τ (a)αi 
(12)
Mα := 
i;σ(a)i =a′
Mα,+ :=
Mα,− :=
a,a′ ∈A
α
1σ(a)0 =a′ τ (a)0 a,a′ ∈A
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 of
G. That is, Θ is the family of subsets ξ of Ω such that ξ = Gω for
some ω ∈ Ω with λω = ω for some λ ∈ G with λ > 1. We call λ as
22
above a multiplicative cycle of ξ. The minimum multiplicative cycle
of ξ is denoted by c(ξ). Note that c(ξ) exists since λω 6= ω for any
ω ∈ Ω and λ ∈ G with 1 < λ < min{τ (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 yaxis and is in common among ω ∈ ξ. Denote by Θ0 the set of ξ ∈ Θ
with the separating line.
Let
L(σ) := {(a, k, i); a ∈ A, k = 1, 2, · · · and
0 ≤ i < |σ k (a)| with σ k (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
X
j=0
′
|σ k (σ k (a)j )| + i′ ).
We say that (a, k, i) ∈ L(σ) is irreducible if (a, k, i) = (a, k ′ , i′ )h does
not 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 exists
a 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 following
statements hold.
(i) 1 ≤ i < |σ k (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) determines
ξ ∈ Θ \ Θ0 and (iii) follows.
Proof. (iii) holds since c(ξ)R′ = R and R′ is the (k, i)-descendant
of R.
′
Since R′ is the (k, i)-descendant of R such that τ k (a)−1
i R = R, we
have
P
k
−x1
j≤i−1 τ (a)j
=P
.
(13)
k
x2
j≥i+1 τ (a)j
23
Since 0 < −x1 /x2 < ∞, (i) follows.
If (a, k, i) = (a, k ′ , i′ )ℓ with ℓ ≥ 2, then (k ′ , i′ )-descendant R′′ of R
′
−1
′′
k′
1/ℓ
also satisfies τ k (a)−1
becomes a
i′ R = R and that τ (a)i′ = c(ξ)
cycle 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
ω ′(R) = a. Take t ∈ R such that R + t =: (x1 , x2 ) × (y1 , y2) satisfies
′
the equation (13). Then x1 < 0 < x2 , and τ k (a)−n
i (ω + t) converges
as n → ∞ to, say ω ∈ Ω which satisfies that τ k (a)−1
i 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 the
sequence 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. Let
Rj 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 3 determining
ξ in the sense of Lemma 3. In fact, these k triples are different from
each other such that they are all that determine ξ, while k is common
among them which we denote by k(ξ).
Lemma 4. For k = 1, 2, · · · , it holds that
k
k
tr(Mαk ) − tr(Mα,+
) − tr(Mα,−
)
X
k
=
k(ξ)c(ξ)− k(ξ) α .
(14)
ξ∈Θ\Θ0
k(ξ)|k
Proof. Note that
k
k
tr(Mαk ) − tr(Mα,+
) − tr(Mα,−
)=
X
τ k (a)αi .
a∈A,1≤i<|σ k (a)|−1
σ k (a)i =a
For any ξ ∈ Θ \ Θ0 with k(ξ)|k, there exist exactly k(ξ) number of
triples (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,
X
X
τ k (a)αi =
a∈A,1≤i<|σ k (a)|−1
σ k (a)i =a
τ k (a)αi
(a,k,i)∈L(σ)
1≤i<|σ k (a)|−1
X
=
k ′ |k
k ′ |k
′
k ′ c(ξ)−(k/k )α
ξ∈Θ\Θ0
k(ξ)=k′
X
=
(k/k ′ )α
τ k (a)i
(a,k′ ,i)∈L(σ):irreducible
′
1≤i<|σ k (a)|−1
X X
=
′
X
k
k(ξ)c(ξ)− k(ξ) α .
ξ∈Θ\Θ0
k(ξ)|k
The following lemma follows from Lemma 1.
Lemma 5. The set Θ0 is a finite set. In fact, we have
X
♯Θ0 ≤
(|σ(a)| − 1).
a∈A
Lemma 6. For α ∈ C with R(α) > 1, where R(α) is the real part
of α, we have
X
|c(ξ)−α | < ∞.
ξ∈Θ
Proof. By Lemma 5, it is sufficient to prove that
X
|c(ξ)−α| < ∞.
ξ∈Θ\Θ0
Since
max
a∈A
X
R(α)
τ (a)i
0≤i<|τ (a)|
25
< 1,
there exists δ with 0 < δ < 1 such that the maximal eigen-value of
MR(α) is less than δ. Hence, by (14) we have
X
ξ∈Θ\Θ0
|c(ξ)−α | =
X
c(ξ)−R(α)
ξ∈Θ\Θ0
≤
∞
X
X
≤
∞
X
k=1
k=1
k
k(ξ)c(ξ)− k(ξ) R(α)
ξ∈Θ\Θ0
k(ξ)|k
k
tr(MR(α)
)
≤
∞
X
k=1
Cδ k < ∞.
Define the ζ-function of G-action to Ω by
Y
ζΩ (α) :=
(1 − c(ξ)−α )−1 ,
(15)
ξ∈Θ
where the infinite product converges for any α ∈ C with R(α) > 1 by
Lemma 6. It is extended to the whole complex plane by the analytic
extension.
Theorem 4. We have
ζΩ (α) =
det(I − Mα,+ ) det(I − Mα,− )
ζΣ0 (α),
det(I − Mα )
where
ζΣ0 (α) :=
Y
ξ∈Θ0
(1 − c(ξ)−α )−1
is a finite product with respect to ξ ∈ Θ0 .
Proof. By the definition of ζΩ (α) and (14), for any α with R(α) > 1,
26
it holds that

ζΩ (α) = ζΣ0 (α) exp 
X
ξ∈Θ\Θ0

= ζΣ0 (α) exp 

− log(1 − c(ξ)−α)
∞
X X
ξ∈Θ\Θ0 k=1

(1/k)c(ξ)−kα


∞
k
X X

= ζΣ0 (α) exp 
(k(ξ)/k)c(ξ)− k(ξ) α 
k=1
= ζΣ0 (α) exp
∞
X
k=1
= ζΣ0 (α) exp tr(
ξ∈Θ\Θ0
k(ξ)|k
k
k
(1/k)(tr(Mαk ) − tr(Mα,+
) − tr(Mα,−
))
∞
X
k=1
k
k
(1/k)(Mαk − Mα,+
− Mα,−
))
!
!
= ζΣ0 (α) exp (tr(− log(I − Mα ) + log(I − Mα,+ )
+ log(I − Mα,− )))
det(I − Mα,+ ) det(I − Mα,− )
ζΣ0 (α)
=
det(I − Mα )
Example 3. For the 2-adic expansion substitution (σ, τ ) defined
in 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. Since
c(Gω0 ) = 2 and ζΣ0 (α) = (1 − 2−α )−1 , we have
(1 − (1/2)α)2
1 − 2(1/2)α
1 − (1/2)α
=
1 − 2(1/2)α
ζΩ (α) = ζΣ0 (α)
27
y
x
Figure 6: ω0 in the 2-adic expansion
Example 4. Consider the weighted substitution (σ, τ ) in Example
2, 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 Remark 1, 1 → (1, 1/2)(2, 1/2), 2 → (2, 1/2)(1, 1/2), define Ω :=
28
y
1
1(2)
2
1(2)
1
1(2)
2 1(2)
x
Figure 7: ω1 (ω2 , respectively) in the Thue-Morse substitution
Ω(σ, τ ) with G = {2n ; n ∈ Z}, g ≡ 1. Then, we have
(1/2)α (1/2)α
Mα =
(1/2)α (1/2)α
(1/2)α
0
Mα,+ =
0
(1/2)α
0
(1/2)α
Mα,− =
(1/2)α
0
Moreover, we have Θ0 = {Gω1 , Gω2 } with ω1 and ω2 shown in Figure
7. Since c(Gω1 ) = c(Gω2 ) = 4 and ζΣ0 (α) = (1 − 4−α )−2 , we have
(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)
ζΩ (α) = ζΣ0 (α)
29
y
±
vertical size 81/64
y
±
6
?
∓
∓
±
±
vertical size 16/9
±
∓
6
?
±
± ∓
±
∓
∓
∓
x
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, we
have
2(4/9)α (1/9)α
Mα =
(1/9)α 2(4/9)α
(4/9)α
0
Mα,+ = Mα,− =
0
(4/9)α
There are 4 elements in Θ0 determined as in (iii) of Lemma 1 by the
pairs (+, 0), (+, 1), (−, 0), (−, 1). All of them are different as shown
in Figure 8. The ± corresponds to the ± in the first coordinate, while
0, 1 corresponds to the vertical gap from the left side tiles to the right
side tiles. It is 81/64 for 0 (left side in Figure 8) and 16/9 for 1 (right
side in Figure 8). These 4 elements have the same multiplicative
30
x
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. r+:qec~?ecea
(i) ζΩ (α) 6= 0 if R(α) 6= 0.
(ii) In the region R(α) 6= 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 any
k = 1, 2, · · · .
(iii) 1 is a simple pole of ζΩ (α).
Proof. Since c(ξ) > 1 for any ξ ∈ Θ, 1 − c(ξ)−α 6= 0 if R(α) 6= 0.
Hence, ζΣ0 (α) has neither pole nor zero in the region R(α) 6= 0.
For α with R(α) 6= 0, suppose that det(I − Mα,+ ) = 0, so that
Mα,+ η = η holds for some nonzero vector η = (ηa )a∈A . Take a0 ∈ A
such that ηa0 6= 0. Then, for some k ≥ 0 and ℓ ≥ 1, σ k (a0 )0 =
σ k+ℓ (a0 )0 =: a1 holds. Since ηa1 = τ k (a0 )α0 ηa0 6= 0 and τ ℓ (a1 )α0 ηa1 =
ηa1 , 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) 6= 0. In the
same way, det(I − Mα,− ) has no zero in the region R(α) 6= 0. These
facts 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. Let
A = {a1 , · · · , ar } and A′ := A \ {a1 }. For a matrix M = (mij )i∈I,j∈J
and I ′ ⊂ I, J ′ ⊂ J, let M[I ′ , J ′ ] := (mi,j )i∈I ′ ,j∈J ′ . Since 1 is the
maximum eigen-value of M1 and σ is primitive, there exists a positive
row vector (ξ1 , ξ2 , · · · , ξr ) with ξ1 = 1 such that (ξ1 , ξ2 , · · · , ξr )(I −
M1 ) = (0, · · · , 0). Therefore, since


P|τ (a1 )|−1
τ (a1 )αi
1 − i=0


..
det(I − Mα ) = det 
(I − Mα )[A, A′ ] ,
.
P|τ (ar )|−1
1 − i=0
τ (ar )αi
31
we have
d
det(I − Mα )|α=1
dα 

P
i −τ (a1 )i log τ (a1 )i


..
= det 
(I − M1 )[A, A′ ]
.
P
i −τ (ar )i log τ (ar )i
 P
0 ··· 0
i,j −ξj τ (aj )i log τ (aj )i
 P −τ (a ) log τ (a )
2 i
2 i

i
= det 
..
(I − M1 )[A′ , A′ ]

.
P
i −τ (ar )i log τ (ar )i
!
X
=
−ξj τ (aj )i log τ (aj )i det((I − M1 )[A′ , A′ ]).





i,j
P
We have i,j −ξj τ (aj )i log τ (aj )i > 0 and det((I − M1 )[A′ , A′ ]) 6= 0
since the spectral radius of M1 [A′ , A′ ] is strictly less than 1. Hence,
d
det(I − Mα )|α=1 6= 0 and 1 is a simple zero of det(I − Mα ). By
dα
(ii), it is a simple pole of ζΩ .
Theorem 6. For Ω = Ω(σ, η, g), if B(σ, τ ) = {λn ; n ∈ Z} with
λ > 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 some
polynomials p, q ∈ Z[z] and λ > 1, then B(σ, τ ) = {λkn ; n ∈ Z} for
some positive integer k.
Proof. Assume that B(σ, η) = {λn ; n ∈ Z} with λ > 1. Let g
satisfies (4). Then for any a ∈ A and i with 0 ≤ i < |σ(a)|, there
i ) r(a)i
exists r(a)i ∈ Z such that τ (a)i = g(σ(a)
λ
. Hence, we have
g(a)



 X
Λα
(λα )r(a)i 
Mα = Λ−1
α 
0≤i<|σ(a)|
σ(a)i =a′
r(a)0
a,a′ ∈A
Λ
Mα,+ = Λ−1
1σ(a)0 =a′ (λα )
α
a,a′ ∈A α
α r(a)|σ(a)|−1
′
(λ
)
Mα,− = Λ−1
1
σ(a)|σ(a)|−1 =a
α
32
a,a′ ∈A
Λα ,
where Λα = (g(a)α 1a′ =a )a,a′ ∈A is a diagonal matrix. Therefore, det(I−
Mα ), det(I − Mα,+ ) and det(I − Mα,− ) are polynomials in λα divided
possibly by (λα )n for some positive integer n. Since c(ξ) = λn for
some 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 polynomials p, q ∈ Z[z].
Conversely, assume that ζΩ (α) = p(λα )/q(λα ) for some polynomials 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, it
holds that c(ξ) ∈ {λn ; n ∈ Z} for any ξ ∈ Θ ([6]). Since B(σ, τ ) is
generated 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 algebraic
number.
Proof. Let Ω := Ω(σ, τ, g). By Theorem 6, there exist polynomials
p, q ∈ Z[z] such that ζΩ (α) = p(λα )/q(λα). By Theorem 5, 1 is a pole
of 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 following
periodic β-expansion
1 = (b1 0i1 −1 b2 0i2 −1 · · · bk 0ik −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 the
above sequence. Since the above sequence is the expansion of 1, we
have the solution of the following equation in a1 , a2 , · · · , ak+1 with
a1 = ak+1 = 1 and 0 < aj < 1 (j = 2, · · · , k):
aj = bj β −1 + aj+1 β −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 primitive
and B(σ, τ ) = {β n ; n ∈ Z}. Define g : A → R+ by g(j) := aj . Then,
g satisfies (4) and Ω(σ, τ, g) is a linear numeration system by Theorem
2. We denote Ω(β) := Ω(σ, τ, g) and Ω(β) is called the β-expansion
system.
The β-expansion system is studied by many authors, for example,
S. Ito and Y. Takahashi ([7]) where the ζ-function is obtained. Here,
we give the formula again as a corollary of Theorem 4. There is a little
difference between them. In [7], the ζ-function is for the restriction
{ω + ; ω ∈ Ω} to the right-quarter space, while ours is for Ω itself, so
1−β α
that ours is the product of the former and 1−β
−nα .
Theorem 8. We have
ζΩ(β) (α) =
Proof. We have
1−
Pk
j=1 bj β
1 − β −α
−(i1 +···+ij−1 +1)α
det(I − Mα ) =
1 − b1 ( 1 )α
−( a1aβ2i1 )α 0
a1 β
..
.
−b2 ( a21β )α
1
..
..
..
.
.
.
..
..
..
.
.
.
..
..
..
.
.
.
−b ( 1 )α
0
0
k−1 ak−1 β
ak+1 α
1
α
−bk ( a β ) − ( a β ik ) 0
0
k
k
=1−
k
X
j=1
0
0 ···
.
..
. ..
.. ..
.
.
.. . .
.
.
0 ···
0 ···
bj β −(i1 +···+ij−1 +1)α − β −nα ,
34
···
− β −nα
.
0
0
..
..
.
.
..
..
.
.
..
..
.
.
1 −( a aβkik−1 )α k−1
0
1
0
0
and
det(I − Mα,+ ) = 1 − ( a11β )α
−( a21β )α
..
.
1
)α
−( ak−1
β
−( ak1β )α
0 ··· ···
1 0 ···
.. . .
..
.
.
.
0 ··· 1
0 ··· 0
0
0
..
.
= 1 − β −α
0 1 det(I − Mα,− ) =
1
−( a1aβ2i1 )α
0
···
0
a3
α
0
0
1
−( a2 β i2 ) · · ·
..
.
.
.
.
..
..
..
..
.
ak
0
···
0
1 −( a β ik−1 )α
k−1
α
−( aak+1
0
·
·
·
0
1
ik )
β
k
= 1 − β −nα .
Let ξ ∈ Θ0 and ω ∈ ξ. Let Ri± (i ∈ Z) be the sequence of tiles
in dom(ω) from down to up intersecting with the line x = ±0 and
(±0, 1 + 0) ∈ R0± (± respectively). Then, we have
+
· · · ω(R−1
)ω(R0+ )ω(R1+ )ω(R2+ ) · · · = · · · · · · 11111 · · · · · ·
−
· · · ω(R−1
)ω(R0− )ω(R1− )ω(R2− ) · · · = · · · k · · · 21k · · · 21 · · · .
Hence, the minimum multiplicative cycle of ω on the right quarter
plane is β, while on the left quarter plane is β n . Thus, c(ξ) = β n .
Moreover, the right half tiling and the left half tiling are synchronized
so that the vertical position of the lower side of any tile Ri− with
ω(Ri− ) = 1 coincides with that of Rj+ 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−
1 − β −α
Pk
−(i1 +···+ij−1 +1)α − β −nα
j=1 bj β
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 8, we have
ζΩ(β) (α) =
1 − β −α
.
1 − β −α − β −2α
Example 8. Let us consider the β-expansion system with β > 1
such that β 3 − β 2 − 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 )
β
1
2 → (1,
)(1,
)
β+1
β+1
By Theorem 8, 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 αhomogeneous if
F (λω, λt) = λα F (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 function
Ξ : A × R+ → C such that
F (ω, x2) − F (ω, x1 ) = Ξ(ω(R), x2 − x1 )
36
(17)
for any ω ∈ Ω and tile R := (x1 , x2 ) × (y1 , y2 ) ∈ dom(ω).
In [8], 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 is
characterized by (17) with α and Ξ satisfying that R(α) > 0 and there
exists a nonzero vector ξ = (ξa )a∈A such that Mα ξ = ξ (see (12)) and
Ξ(ω(R), x2 −x1 ) = (x2 −x1 )α ξω(R) for any tile R := (x1 , x2 )×(y1 , y2) ∈
dom(ω). Hence, a nonzero adapted α-homogeneous cocycle exists if
and only if R(α) > 0 and α is a pole of ζΩ (α).
Proof. The last part of the theorem follows from Theorem 5. The
condition that R(α) > 0 is necessary for the convergence of F (ω, t)
with (17) for a general t ∈ R.
It is known [8] 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 the
following 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 for
ω ∈ Ω are uniformly α-Hölder continuous.
(ii) For any ω ∈ Ω and t ∈ R,
lim sup
s↓0
1
|F (ω, t + s) − F (ω, t)| > 0
sα
holds. That is, for any ω ∈ Ω the function F (ω, ·) is nowhere locally α′ -Hölder continuous for any α′ > α. In particular, F (ω, ·) is
nowhere 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 ergodic
stationary increment having 0-entropy.
(iv) F (ω, λt) has the same law as λα F (ω, t) for any λ ∈ G. Hence,
the process
F (ω, t) is α-self similar if G = R+ .
R
(v) F (ω, t)dµ(ω) = 0 for any t ∈ R.
37
Example 9. Take Ω in Example 1 and 6. Since
4/3 1/3
,
M1/2 =
1/3 4/3
1
is an eigenvector of M1/2 with eigen-value 1. Let F be
ξ=
−1
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 probability space (Ω, µ). Since
nE[F (ω, 1)2] = E[(n1/2 F (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))]
2
= E[F (ω, 1) ] + E[F (ω + 1, n − 1)2 ]
+2E[F (ω, 1)(F (ω, n) − F (ω, 1))]
2
= E[F (ω, 1) ] + (n − 1)E[F (ω, 1)2]
+2E[F (ω, 1)(F (ω, n) − F (ω, 1))],
we have
E[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 noncorrelated.
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 . An
element 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 (16) 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 (17) is satisfied for any ω ∈ I(Ω) and tile
(x1 , x2 ) × (y1 , y2 ) ∈ dom(ω) with y2 > 1. Let α ∈ C. A cocycle F on
I(Ω) is called α-homogeneous if
F (λω, λt) = λα F (ω, t)
for any (ω, t) ∈ II(Ω) and λ ∈ G with (λω, λt) ∈ II(Ω). Note that
if (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 [10].
Theorem 11. A nonzero adapted α-homogeneous cocycle on I(Ω)
with R(α) < 0 is characterized by (17) with Ξ satisfying that there
exists a nonzero vector ξ = (ξa )a∈A such that Mα ξ = ξ (see (12)) and
Ξ(ω(R), x2 −x1 ) = (x2 −x1 )α ξω(R) for any tile R := (x1 , x2 )×(y1 , y2) ∈
dom(ω) with y2 > 1. Hence, a nonzero adapted α-homogeneous cocycle 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 Example
8. Denote Ω := Ω(β). The associated matrix is
−α
β
(β −2 + β −3 )α
Mα =
β α +1
0
(β+1)α
Let γ be one of the complex solutions of the equation z 3 −z 2 −z −1 =
0. Then, |γ| < 1. Let α ∈ C be such that γ = β α . Then, R(α) < 0.
Since we have
1
1
Mα
=
δ
δ
βα + 1
, there exists an α-homogeneous adapted cocycle
(β + 1)α
F on I(Ω) satisfying that
(x2 − x1 )α
(ω(R) = 1)
F (ω, x2) − F (ω, x1) =
α
δ(x2 − x1 )
(ω(R) = 2)
with δ :=
if there exists R := (x1 , x2 ) × (y1 , y2 ) ∈ dom(ω) with y2 > 1.
For ω ∈ I(Ω), let R0 (ω) be the tile (x0 , x̃0 ) × (y0 , ỹ0) ∈ dom(ω)
such that x0 = 0 and y0 ≤ 1 < ỹ0 . For i = 0, 1, 2, · · · , let Ri be the
i-th ancestor of R0 (ω). Let Ri =: (xi , x̃i ) × (yi , ỹi ). Let
G(ω) :=
∞
X
(xi − xi+1 )α .
(18)
i=0
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 (Ri′ )i=1,2,··· and
′
(xi )i=0,1,··· be the sequences of above (Ri )i=1,2,··· and (xi )i=0,1,··· for
ω + 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 Ri′ 0 +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
iX
0 −1
X
(x′i − x′i+1 )α −
(xi − xi+1 )α
=
i=0
i=0
= −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 loss
of generality, we assume that t > 0. Let (a, b) × (c, d) be the tile in
dom(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 to
prove 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 the
seperating 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 (18).
Case 1: (ω + t)(R0 (ω + t)) = 1. Since
x′i
−
G(ω + t) =
=
=
=
x′i+1
=
β 3j+1 (i = 2j)
,
β 3j+2 (i = 2j + 1)
β α + β 2α + β 4α + β 5α + β 7α + β 8α + · · ·
γ + γ2 + γ4 + γ5 + γ7 + γ8 + · · ·
−1 + γ 3 + γ 4 + γ 5 + γ 7 + γ 8 + · · ·
−1 + γ 6 + +γ 7 + γ 8 + · · · = −1
41
holds by (18), where we used 1 + γ + γ 2 = γ 3 .
On the other hand, (t, 0) × (c, d) ∈ dom(ω) satisfies that −t = c ≤
1 < d and has color 1. Since −t ∈ g(1){β n; n ∈ Z} = β{β n ; 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
G(ω + t) =
=
=
=
−
x′i+1
=
β 3j
(i = 2j)
,
3j+2
β
(i = 2j + 1)
1 + β 2α + β 3α + β 5α + β 6α + β 8α + β 9α + · · ·
1 + γ2 + γ3 + γ5 + γ6 + γ8 + γ9 · · ·
1 − γ + γ 4 + γ 5 + γ 6 + γ 8 + +γ 9 · · ·
1 − γ + γ7 + γ8 + γ9 + · · · = 1 − γ
holds by (18). On the other hand, (t, 0) × (c, d) ∈ dom(ω) satisfies
that −t = c ≤ 1 < d and has color 2. Since −t ∈ g(2){β n; n ∈ Z} =
(β − 1){β n ; n ∈ Z}, this implies that −t = β − 1. Hence,
βα + 1
(β − 1)α
(β + 1)α
γ+1
βα + 1
(β − 1)α = − 2
= − 3
2
α
(β − β )
γ
3
2
γ −γ
= −
= 1 − γ.
γ2
F (ω, t) = −δ(−t)α = −
Thus, G(ω + t) − G(ω) = F (ω, t) holds.
Thus, the α-homogeneous cocycle F is a coboundary with coboundary function G. The set G(I(Ω)) is known as Rauzy fractal. For
ω ∈ I(Ω), let
I(Ω)a = {ω ∈ I(Ω); ω(R0(ω)) = a} (a = 1, 2)
and
Ga = G(I(Ω)a ) (a = 1, 2).
42
y
y
2
y
1
1
1
i
2
2
i r
r
1
1
1
x
i r
x
Figure 9: βI(Ω)1 , βI(Ω)2 , β 2 I(Ω)2 + β
Considering the children of the tile R0 (ω), we have a set equation
that
I(Ω)1 = βI(Ω)1 ∪ βI(Ω)2 ∪ (β 2 I(Ω)2 + β)
I(Ω)2 = βI(Ω)1 + 1.
Hence, the following set equation holds:
G1 = γG1 ∪ γG2 ∪ (γ 2 G2 + γ)
G2 = γG1 + 1.
G1 ∪ G2 is shown in Figure 10.
Remark 2. The above Rauzy fractal is usually introduced by the
substitution 1 → 12, 2 → 13, 3 → 1 (S. Ito and P. Arnoux [1]). We
modify it canonically to the weighted substitution in Example 8 (see
Remark 1). The set equation is usually denoted as
G1 = γG1 ∪ γG2 ∪ γG3
G2 = γG1 + 1
G3 = γG2 + 1,
which is equivalent to ours.
43
x
Figure 10: G(I(Ω))
7
N -process
We consider the N-process defined in Example 9. It is defined as
a (1/2)-homogeneous cocycle F on the space Ω = Ω(σ, τ ) with the
weighted 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 to
the 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 (19). Since R0 has
3 children R1,0 , R1,1 , R1,2 with colors +, −, + and the vertical sizes
4/9, 1/9, 4/9, we have F (ω0 , 4/9) = 2/3 and F (ω0 , 5/9)−F (ω0, 4/9) =
−1/3 by (19). 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 (19),
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 the
continuity 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 color
between 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 color
between 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 metric
of ψ n (ΓN0 ) as n → ∞, where ΓN0 denotes the graph of the function
N0 (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 = ψ n (ΓN0 ). Then, N1 is a continuous piecewise linear
function whose graph consists of 3 line segments
ψ0 (ΓN0 ) = {(x, y); y = (3/2)x, 0 ≤ x ≤ 4/9}
ψ1 (ΓN0 ) = {(x, y); y = −3x + 2, 4/9 ≤ x ≤ 5/9}
ψ2 (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-affine
images of ΓN1 or Γ−N1 keeping the 2 end points fixed. Then, the
graph of N2 consists of 9 line segments. In the same way, we obtain
Nn from Nn−1 for n = 3, 4, · · · . Then, Nn is a continuous piecewise
linear 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 to
the vertical axis of the descendents of the n-th generation of the tile
R0 in ω0 . Let ∆n be the set of the end points of Ξn . Let Ξ = ∪∞
n=0 Ξn
∞
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. r+:qec~?ecea
(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],
Z
Z
dt =
2 − |4t − 2| dt.
N∞ (t)∈R
R
(iii) The Hausdorff dimension of the graph ΓN∞ of the function
N∞ satisfies that H-dimΓN∞ = 3/2. In fact, the (3/2)-dimensional
Hausdorff 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 ǫ > 0, it holds that
H-dim{s ∈ (t0 − ǫ, t0 ); N∞ (s) = N∞ (t0 )}
= H-dim{s ∈ (t0 , t0 + ǫ); N∞ (s) = N∞ (t0 )} = 1/2.
In fact, the (1/2)-dimensional Hausdorff measures of the above sets
are positive and finite.
Proof. (i) is proved in [9].
(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
Z
ν(S) =
dt
N∞ (t)∈S
Z
Z
dτ =
2 − |4t − 2| dt
S
S
for any Borel set S ⊂ [0, 1]. We prove that ν = τ .
48
For any f ∈ C([0, 1]), we have
Z
Z
Lf dν =
(Lf )(N∞ (t))dt
Z
=
(4/9)f ((2/3)N∞(t))dt
Z
+ (1/9)f ((2 − N∞ (t))/3)dt
Z
+ (4/9)f ((1 + 2N∞ (t))/3)dt
Z
=
(4/9)f (N∞ ((4/9)t))dt
Z
+ (1/9)f (N∞ ((t + 4)/9))dt
Z
+ (4/9)f (N∞ ((4t + 5)/9))dt
Z 4/9
Z 5/9
Z 1
=
f (N∞ (t))dt +
f (N∞ (t))dt +
f (N∞ (t))dt
0
4/9
5/9
Z
Z
=
f (N∞ (t))dt = f dν.
Thus, ν is invariant under L. We prove that τ is also invariant under
L.
49
In fact,
Z
Z
Lf dτ =
2 − |4t − 2| (Lf )(t)dt
Z
=
2 − |4t − 2| (4/9)f ((2/3)t)dt
Z
+
2 − |4t − 2| (1/9)f ((2 − t)/3)dt
Z
+
2 − |4t − 2| (4/9)f ((1 + 2t)/3)dt
Z 2/3
=
2 − |6t − 2| (4/9)f (t)(3/2)dt
0
Z 2/3
+
2 − |12t − 6| (1/9)f (t)3dt
1/3
+
=
Z
1/3
Z
1
1/3
2 − |6t − 4| (4/9)f (t)(3/2)dt
6t(2/3)f (t)dt
0
+
1/2
Z
(4 − 6t)(2/3) + (−4 + 12t)(1/3) + (−2 + 6t)(2/3))f (t)dt
1/3
+
Z
2/3
1/2
1
(6 − 6t)(2/3)f (t)dt
Z
2 − |4t − 2| f (t)dt = f dτ.
+
=
Z
Z
(4 − 6t)(2/3) + (8 − 12t)(1/3) + (−2 + 6t)(2/3) f (t)dt
2/3
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 ∈ C 1 ([0, 1]). This implies that k (Lf )′ k≤ (17/27) k f ′ k.
Therefore, we have k (Ln f )′ k≤ (17/27)n k f ′ k, and hence, (Ln f )′
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
Z
Z
Z
Z
n′
n′
f dν = lim L f dν = c = lim L f dτ = f dτ
for any f ∈ C 1 ([0, 1]). This implies that ν = τ .
(iii) Let Γ := ΓN∞ (t) . Since N∞ is uniformly 1/2-Hölder continuous, the (3/2)-dimensional Hausdorff measure of Γ is finite. We prove
that it is positive.
Let νΓ be the probability measure supported by Γ such that for
any Borel set S ⊂ [0, 1],
Z
νΓ ((S × [0, 1]) ∩ Γ) :=
dt.
S
Then by (ii), it holds that
νΓ ([0, 1] × S) =
Z
S
2 − |4t − 2| dt.
(20)
cover U = {Ui ; i = 1, 2, · · · } of Γ. Let I3/2 (U) :=
PTake any
3/2
, where d(U) denotes the diameter of the set U. Since
i d(Ui )
for 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 consisting
of closed rectangles such that I3/2 (U ′ ) ≤ 5I3/2 (U). Since for any
interval [a, b] ⊂ [c, d], we can find at most 2 intervals I1 , I2 in Ξ such
that 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 from
U. 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|1/2 y) with some a, b and ±. Hence by (20), we have
νΓ (Vi ) = νΓ (I × [c, d])
= |I|νΓ [0, 1] ×
c − C d − C ,
D−C D−C
x2 − x1
D−C
x2 − x1
= 2|I|
|I|1/2
= 2|I|1/2 (x2 − x1 ) ≤ 2d(Vi )3/2 .
≤ 2|I|
(21)
Thus, adding the above inequality, we have
X
X
I3/2 (V) =
d(Vi )3/2 ≥ (1/2)
νΓ (Vi ) ≥ (1/2)νΓ (Γ) = 1/2,
i
i
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 prove
that 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 intervals
I and J in some Γn such that {t0 } = I ∩ J and I is in the left side
of J. Then, either the piecewise linear function Nn is increasing in
I 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|1/2 N∞ (s).
for any s ∈ [0, 1] by the set equation ψ(Γ) = Γ. Therefore, with
h = |J|s
1
|N∞ (t0 + h) − N∞ (t0 )| ≥ √ |h|1/2
5
52
follows from the statement that
s1/2
N∞ (s) ≥ √
5
for any s ∈ (0, 1] and with h = −|I|s
1
|N∞ (t0 + h) − N∞ (t0 )| ≥ √ |h|1/2
5
follows from the statement that
s1/2
1 − N∞ (1 − s) ≥ √
5
for any s ∈ (0, 1]. By the symmetry of the graph of N∞ with respect
to (1/2, 1/2), the second statement follows from the first statement
that
s1/2
N∞ (s) ≥ √
5
for any s ∈ (0, 1].
We prove this inequality. Note that the equality holds for s = 5/9.
Suppose that
1
N∞ (s)/s1/2 < √
5
holds for some s ∈ (0, 1]. Since
c0 := inf N∞ (s)/s1/2 = min N∞ (s)/s1/2 = min N∞ (s)/s1/2 ,
s∈(0,1]
s∈[4/9,1]
s∈[5/9,1]
there exists s0 ∈ [5/9, 1] which attains the minimum. Then,
c0 s0 1/2 = N∞ (s0 )
= (1/3) + N∞ (s0 − (5/9))
> c0 (5/9)1/2 + c0 (s0 − (5/9))1/2
1/2
≥ c0 s0
which is a contradiction. Thus, N∞ (s) ≥
completes the proof of (iv).
53
1/2
s√
5
for any s ∈ [0, 1], which
(v) We prove that for any x ∈ (0, 1), the (1/2)-dimensional Hausdorff 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. We
may assume that U consists of open intervals, so that there exits a
finite subcover of U. Therefore, we may assume that U is a finite
cover consisting of closed intervals. Moreover, by the same argument
as 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 ǫ > 0 satisfy that x + 2ǫ < 1 and |Ii | > ǫ for
any i = 1, 2, · · · , K. By the same argument as in (21), we have
ǫ
νΓ (Ii × [x, x + ǫ]) ≤ 2|Ii | 1/2 ≤ 2ǫ|Ii |1/2 .
|Ii |
Adding the above inequality, we have
νΓ (Γ ∩ ([0, 1] × [x, x + ǫ])) ≤ 2ǫI1/2 (V),
Hence by (20) and the definition of δ, we have
I1/2 (V) ≥ (2ǫ)−1 νΓ ([0, 1] × [x, x + ǫ]) ≥ (1/2)δ,
which completes the proof that Hausdorff measure of (N∞ )x is positive.
To prove that it is finite, for any sufficiently small ǫ > 0 and t ∈
(N∞ )x , we take V (t) ∈ Ξ such that t ∈ V (t) and 2ǫ ≤ |V (t)| < 18ǫ.
Then, there exists a finite subcover V := {Vi : i = 1, 2, · · · , K} of
{V (t); t ∈ (N∞ )x } such that Vi ∩ Vj is at most one point for any
i 6= j.
For any i = 1, 2, · · · , K, by the same argument as in (21), we have
νΓ (Vi × [x − ǫ1/2 , x + ǫ1/2 ])
Z (ǫ/|Vi |)1/2
(2 − |4t − 2|)dt
≥ |Vi |
0
≥ |Vi |2ǫ/|Vi | ≥ 2ǫ ≥ (2/5)ǫ1/2 |Vi |1/2 .
54
Adding this inequality together with (20), we have
4ǫ1/2 ≥ νΓ ([0, 1] × [x − ǫ1/2 , x + ǫ1/2 ])
= νΓ (Γ ∩ ([0, 1] × [x − ǫ1/2 , x + ǫ1/2 ]))
≥ (2/5)ǫ1/2 I1/2 (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 by Nt (ω) = F (ω, t),
where ω comes from the probability space (Ω, µ), µ being the unique
invariant probability measure invariant under the additive action.
This process was called the N-process and studied in [9]. A prediction theory based on the N-process was developed. A process
Yt = H(Nt, t), where the function H(x, s) is an unknown function
which is twice continuously differentiable in x and once continuously
differentiable in s and Hx (x, s) > 0 is considered. The aim is to
predict the value Yc from the observation YJ := {Yt ; t ∈ J}, where
J = [a, b] and a < b < c.
Theorem 13. ([9]) There exists an estimator Ŷc which is a measurable function of the observation YJ such that
E[(Ŷc − Yc )2 ] = O((c − b)2 )
as c ↓ b.
Open problems:
(1) The systems of the geodesic and horocycle flows on the upper half
plane H/G devided by cocompact discrete groups G ∈ P SL2 (R) are
linear numeration systems which are not isomorphic to those coming
from weighted substitutions (private communications by Prof. Benjamin Weiss). How to characterize the linear numeration systems
coming from weighted substitutions?
(2) Is the condition B(σ, τ ) = R+ necessary for the R-action of a
55
linear numeration system coming from weighted substitutions with
respect 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 possessed by the R-action of a linear numeration system coming from a
weighted substitution with B(σ, τ ) = R+ with respect to the unique
invariant probability measure?
(4) When does a linear numeration system admit an additive group
structure consistent with the (R, G)-action?
References
[1] Pierre Arnoux and Shunji Ito, Pisot substitutions and Rauzy fractals, Bull. Belg. Math. Soc. 8(2001), pp.181-207.
[2] V. Berthe, S. Ferenczi, C. Mauduit and A. Siegel (ed.), Substitutions in dynamics, arithmetics and combinatorics, Lecture Notes
in 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 measure
entropy, Bull. London Math. Soc. 3(1971), pp.176-180.
[5] Nertila Gjini and Teturo Kamae, Coboundary on colored tiling
space as Rauzy fractal, Indagationes Mathematicae 10-3 (1999),
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[6] G. H. Hardy and M. Riesz, The general theory of Dirichlet’s series,
Cambridge Univ. Press, 1952
[7] S. Ito and Y. Takahashi, Markov subshifts and realization of βexpansions, J. Math. Soc. Japan 26(1974), pp.33-55.
[8] Teturo Kamae, Linear expansions, strictly ergodic homogeneous
cocycles and fractals, Israel J. Math. 106 (1998), pp.313-337.
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[9] Teturo Kamae, Stochastic analysis based on deterministic Brownian motion, Israel J. Math. 125 (2001), pp.317-346.
[10] Teturo Kamae, Numeration systems, fractals and stochastic processes, Israel J. Math. 149 (2005), pp.87-136.
[11] Karl Petersen, Ergodic Theory, Cambridge studies in advanced
mathematics 2, Cambridge Univ. Press, 1983.
[12] A. Plessner, Spectral theory of linear operators I, Uspekhi
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[13] A. N. Starkov, Dynamical Systems on Homogeneous Spaces,
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[14] Peter Walters, Ergodic Theory − Introductory Lectures, Lecture
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