Non-dense orbits of the times q-map and
Furstenberg’s conjecture
by Jonas Lindstrøm Jensen, 2003 3834
Supervisor: Simon Kristensen
Department of Mathematical Sciences – University of Aarhus
Contents
Contents
1
1 Foreword
3
2 Non-dense orbits of the times
2.1 Furstenberg’s conjecture . .
2.2 Proof of Rudolph’s theorem
2.3 Hausdorff dimension . . . .
2.4 Lower bounded orbits . . . .
3 On
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
lower bounded orbits
Introduction . . . . . .
Basic definitions . . . .
Proof outline . . . . .
Part and residue . . .
Minimality . . . . . . .
Induction mapping . .
Constant dimension . .
Numerical plots . . . .
Asymptotics . . . . . .
q-map
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4 Cantor-like constructions of Fc
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4.1 Geometric construction . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.2 Specific examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5 Generalizations to higher dimension
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5.1 Endomorphims of the torus . . . . . . . . . . . . . . . . . . . . . . . 41
5.2 Cantor construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6 Ternary expansions of powers of 2
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6.1 A conjecture of Erdős . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
6.2 Hausdorff dimension of certain Cantor-sets . . . . . . . . . . . . . . . 45
Bibliography
51
1
Chapter 1
Foreword
I began my PhD studies at the Department of Mathematical sciences at the University of Aarhus in August 2009 under supervision of Simon Kristensen. The title of
the project is ”Furstenbergs 2x, 3x (mod 1) problem” which states that any ergodic
measure for the joint action of the times-2 and times-3 map on the unit interval
must be the Lebesgue measure. In chapter 2 i briefly introduce this problem and
some basic definitions of ergodicity and entropy and a proof due to B. Host [6] of
Rudolph’s theorem, which is a partial answer to Furstenberg’s conjecture, which also
assumes that the measure has positive entropy.
Chapter 3 is a ’copy-paste’ of a paper I have written, ’On lower bounded orbits
of the times-q map’. In this paper I consider the lower bounded orbits of the times-q
map on the unit interval as a subshift of finite type. This subshift is characterized by
a transition matrix, and the main theorem in the paper is a complete characterization
of the characteristic polynomials of these matrices. One application of this is that
it gives a way of finding the Hausdorff dimension of the lower bounded orbits. The
paper has been submitted to Journal of Uniform Distribution Theory on the 30th
of September.
In chapter 4, another way of calculating the dimension of the lower bounded
orbits is introduced. Constructing these sets iteratively in a geometric way, we see
that they are actually Cantor sets, and using the self-similarity of the construction we
calculate the dimension in some special cases. These constructions can be described
by recursively defined sequences, and the Fibonacci sequence and the Golden Ratio
pops up in one of the examples.
In chapter 5 I present an idea for a generalization, namely by considering the two
dimesional case, namely by letting a 2 × 2 matrix act on the torus. It turns out that
the idea presented in chapter 4 is more fruitful here than the one from the paper in
chapter 3, because the natural way to consider this problem in higher dimension is
not as lower bounded orbits, but instead as orbits avoiding certain parallelograms
from the natural cover given by the lattice generated by eigenvectors of the matrix.
Both chapter 4 and 5 are quite heuristic, but I hope to formalize them further in
the near future.
Chapter 6 is a short introduction to a paper by prof. Jeff Lagarias, who considers
a conjecture of Erdős which in some sense is related to the Furstenberpag conjecture.
He asks himself when the ternary expansion of a power of two omits the digit 2. As
3
4
CHAPTER 1. FOREWORD
a way of approaching this, Lagarias studies for which λ > 0, there is infinitely many
n such that the ternary expansion of the integer bλ2n c omits the digit 2. He proved,
and I present the proof here, that this set has Hausdorff dimension log3 2.
I would like to thank my supervisor Simon Kristensen for great help and discussions during the first half of my PhD-studies.
Chapter 2
Non-dense orbits of the times
q-map
2.1
Furstenberg’s conjecture
Furstenberg’s conjecture [4] has to do with the joint action of T2 , T3 : T → T where
T = R/Z and
Tq (x) = qx
for any integer q ≥ 2. With abuse of notation we will identify T with the unit
interval [0, 1).
We recall the following definition.
Definition 2.1.1. Let X be a measurable space and let S : X → X be a transformation on X. A probability measure µ on x is S-invariant if
µ(B) = µ(S −1 B)
for all Borel sets B. If furthermore
S −1 B = B ⇒ µ(B) ∈ {0, 1}
then we say that µ is ergodic with respect to S.
A measure can also be ergodic with respect to the joint action of two or more
maps. We will only need it in a specific case, namely with respect to the joint action
of T2 and T3 , but we still give the general definition.
Definition 2.1.2. A probability measure µ on X is ergodic with respect to the joint
action of T and S if it is T - and S-invariant and
T −n S −m B = B for all n, m ∈ N ⇒ µ(B) ∈ {0, 1}.
Now Furstenberg’s conjecture states the following.
Conjecture 2.1.3. The only ergodic measures with respect to the joint action of T2
and T3 are the Lebesgue measure and measures supported on a finite set.
5
6
CHAPTER 2. NON-DENSE ORBITS OF THE TIMES Q-MAP
In 1990 Dan Rudolph [12] proved that the conjecture is true if we also assume
that the measure has positive entropy with respect to either of the maps. Recall
that we define entropy of a measure as follows.
Definition 2.1.4. Let T : X → X be a measure preserving map on a probability
space (X, B, µ). Then the entropy of µ with respect to T is
h(T ) = sup h(T, α)
where the supremum is over all partitions α of X such that,
X
H(α) = −
µ(A) log µ(A) < ∞,
A∈α
and
Hn (α)
n→∞
n
h(T, α) = lim
where
−i
Hn (α) = H(∨n−1
i=0 T α).
and
−i
−1
∨n−1
Ar1 ∩ · · · ∩ T −(n−1) Arn−1 | Ari ∈ α for all i}.
i=0 T α = {Ar0 ∩ T
We will also need the following definition, which sometimes makes it easier to
calculate the entropy.
Definition 2.1.5. We say that a partition α = {A1 , . . . , An } is generating for a
system of T acting on (X, B, µ) if the smallest σ-algebra containing all the sets
T n Cj , n ∈ Z, 1 ≤ j ≤ n is B.
One can prove that if α is generating then h(T ) = h(T, α), see for instance [15]
or [11].
2.2
Proof of Rudolph’s theorem
We follow a paper by B. Host [6] who, among other things, proves Rudolph’s theorem.
Host’s assumptions are a bit different than Rudolph’s since he uses the notion of
a meausere being conservative. Later on we will prove how this relates to entropy
and ergodicity.
Definition 2.2.1. Let D ⊂ [0, 1) be a countable additive subgroup. A probability
measure µ on [0, 1) is conservative for D if for all Borel set A ⊆ [0, 1) with µ(A) > 0
there is α ∈ D \ {0} such that
µ(A ∩ (A + α)) 6= 0.
We will now prove the following theorem.
2.2. PROOF OF RUDOLPH’S THEOREM
7
Theorem 2.2.2. Let p, q > 1 be relatively prime integers and µ a probability measure
on T that is conservative for the p-adic rationals
i
, n ∈ N, 0 ≤ i < n
D=
pn
and invariant for Tq . Then µ is the Lebesgue measure.
First we introduce some notation. For n ∈ N we let
j
n
Dn =
|0≤j<p
pn
so
D = ∪n∈N Dn .
Now let Bn be the smallest σ-algebra complete for µ that contains all Borel sets
invariant under the translation x 7→ x + p1n . In other words, B ∈ Bn if
x∈B ⇒x+
1
∈ B.
pn
Since unions and complements of these sets also satisfies the above condition, the
σ-algebra contains of exactly these sets.
Furthermore let
X
µ ∗ δα
ωn =
α∈Dn
be the sum of the convolutions where δα is the Dirac measure, hence
X
µ(A + α).
ωn (A) =
α∈Dn
Notice that if ωn (A) = 0 then µ(A + α) = 0 for all α ∈ Dn , in particular µ(A) = 0,
so the measures are absolutely continuous and we let φn : T → R be the RadonNikodym derivative
dµ
.
φn =
dωn
Now let p, q > 1 be coprime integers and let a 6= 0 be an integer. For each n the
sequence (aq k (mod pn )), k ≥ 0 is periodic – let Tn be the length of the period. Since
p and q are coprime, we have that Tn ≈ pn . Actually Tn = Cpn for sufficiently large
n.
For each N > 0 define gN : R → R
N −1
1 X
e(aq k x)
gN (x) =
N k=0
where e(x) = e2πix . We now prove the following lemmas.
Lemma 2.2.3. For 1 ≤ N ≤ Tn we have
Z
|gN (x)|2
pn
dµ(x) ≤
φn (x)
N
where the integral is over T.
8
CHAPTER 2. NON-DENSE ORBITS OF THE TIMES Q-MAP
Proof. Let n ∈ N be given and let 1 ≤ N ≤ Tn . We have
Z
Z
|gN (x)|2
dµ(x) = |gN (x)|2 dωn (x)
φn (x)
n−1 2
Z pX
j
gN x +
dµ(x)
=
n
p
j=0
Z
= h(x)dµ(x).
where
pn −1 2
X j h(x) =
gN x + pn j=0
pn −1 N −1 N −1 1 XXX
j
j
k
l
= 2
e aq x + n
e −aq x + n
.
N j=0 k=0 l=0
p
p
pn −1 N −1 N −1
X
1 XX
k
l
k
l j
e(a(q − q )x)
e a(q − q ) n .
= 2
N k=0 l=0
p
j=0
Since 0 ≤ k, l < N ≤ Tn we must have aq k 6= aq l (mod pn ) whenever k 6= l so for
these we have
pn −1 X
k
l j
e a(q − q ) n = 0.
p
j=0
For the remaining N terms we have q l = q k and each contributes with pn so
Z
pn
h(x)dµ(x) =
N
as desired.
Lemma 2.2.4. For µ-almost every x ∈ T we have
φn (x) → 0 for n → ∞
if and only if µ is conservative for D.
Proof. For each n ∈ N, the derivative φn (x) ≥ 0, and it can only be zero on a
µ-nullset. So let
K = {x ∈ T | φn (x) > 0}.
By countable stability we have µ(K) = 1 and
φn (x) = 1K (x)φn (x)
for all n.
Now suppose for contradiction that φn (x) → 0 for µ-almost every x ∈ T does
not hold. Then there is a set C ⊆ K with µ(C) > 0 and ε > 0 such that
φn (x) > ε for all x ∈ C.
2.2. PROOF OF RUDOLPH’S THEOREM
9
This implies that
µ(C)
=
ε
Z
C
1
dµ(x) >
ε
Z
C
1
dµ(x) = ωn (C).
φn (x)
Now recalling the definition of ωn we have
Z X
1C (x + α)dµ(x).
ωn (C) =
α∈Dn
This implies
Z X
α∈Dn
1C (x + α)dµ(x) ≤
µ(C)
ε
∞
so the sequence
1
(x
+
α)
must be bounded for µ-almost every x ∈ T,
C
α∈Dn
n=1
and so there must exist n ∈ N and B ⊆ C with µ(B) > 0 such that
P
1C (x + α) = 0
for all x ∈ B and α ∈ D \ Dn , or in other words x + α ∈
/ C and hence x + α ∈
/ B, so
B ∩ (B + α) = ∅
for all α ∈ D \ Dn . Since µ(B) > 0 there must exist some interval such that
1
>0
µ B ∩ β, β + n
p
for β ∈ Dn . So let A be this intersection. Then
A ∩ (A + α) ⊆ B ∩ (B + α)
which is empty for α ∈ D \Dn and for any non-zero α ∈ Dn we have A∩(A+α) = ∅,
so A is a contradiction against µ being conservative for D.
On the other hand, suppose for contradiction that there is a Borel set A such
that µ(A) > 0 and
µ(A ∩ (A + α)) = 0
for all non-zero α ∈ D. We may suppose that A ⊆ K and that
A ∩ (A + α) = ∅
for α ∈ D. Now for all n ∈ N, the sets A + α, α ∈ Dn are disjoint since if
x ∈ (A + α) ∩ (A + β) we must have x ∈ A ∩ (A + α − β) which is a contradiction.
So for n ∈ N we have
!
Z
X
[
1
dµ(x),
µ
(A + α) =
µ(A + α) = ωn (A) =
A φn (x)
α∈D
α∈D
n
n
This expression is bounded from above by 1, since µ is a probability measure, and
this is a contradiction against φn (x) → 0 for µ-almost every x.
10
CHAPTER 2. NON-DENSE ORBITS OF THE TIMES Q-MAP
We are now ready to prove theorem 2.2.2.
Proof of theorem 2.2.2. Suppose that µ is a probability measure conservative for D
and invariant under Tq . Using Cauchy-Schwarz we get
Z
2 Z
Z
|gTn (x)|2
gTn (x)dµ(x) ≤
dµ(x) φn (x)dµ(x),
φn (x)
and using the above lemma, we get that the first factor is less than
to C for sufficiently large n, so
pn
Tn
which is equal
Z
2
Z
gTn (x)dµ(x) ≤ C φn (x)dµ(x),
which tends to 0 by lemma 2.2.4.
Since µ is Tq -invariant we have
Z
1
gTn (x)dµ(x) =
Tn
Z TX
n −1
e(aq k x)dµ(x)
k=0
Z TX
n −1
1
=
e(ax)dµ(x)
Tn
k=0
Z
= e(ax)dµ(x)
for all sufficiently large n. So we must have
Z
e(ax)dµ(x) = 0,
(2.2.1)
and since this is true for all a 6= 0, we claim that µ must be the Lebesgue measure,
since the functions (x → e(ax), a ∈ Z \ {0}) are measure separating. More precisely,
(2.2.1) is certainly true for the Lebesgue measure λ. Now let U ⊆ T be a Borel set,
and let
X
1U (x) =
ca e(kx)
a∈
Z
be the Fourier series of the indicator function of U . Now
Z
Z
µ(U ) = 1U (x)dµ(x) = c0 = 1U (x)dλ(x) = λ(U ),
so µ is the Lebesgue measure.
The following lemma shows why theorem 2.2.2 implies Rudolph’s theorem.
Lemma 2.2.5. Let p > 1 be an integer and let µ be a probability measure on T that
invariant for Tp . Then µ is conservative for the p-adic rationals, D, if and only if
each ergodic component of µ with respect to Tp has positive entropy.
2.2. PROOF OF RUDOLPH’S THEOREM
11
Proof. Suppose that µ is invariant for T = Tp , that is
µ(A) = µ(T −1 A)
for all Borel sets A ∈ B. For all n ∈ N we have
Bn = T −n B0
and we also claim that
φn (x) = φ1 (x)φ1 (T x)φ1 (T 2 x) · · · φ1 (T n−1 x).
(2.2.2)
Recalling the chain rule we see that to prove (2.2.2) it is enough to prove that
dωi
(x) = φ1 (T i x).
dωi+1
(2.2.3)
We prove this for p = 2 and i = 1, but the general case is similar. Note that to
prove (2.2.3) we need to prove that
Z
φ(T x)dω2 = ω1 (A)
A
for all Borel sets A, so let such an A be given. Then
Z
1A (x)φ1 (T x)dω2
Z
=
Z
+
1A (x)φ1 (T x)dµ +
Z
1
4
1
2
1A (x + )φ1 (T x + )dµ
Z
1
3
1
1A (x + )φ1 (T x)dµ + 1A (x + )φ1 (T x + )dµ
2
4
2
Z
Z
1
=
φ1 (T x)dµ +
φ1 (T x + )dµ. (2.2.4)
2
A∪A+ 12
A+ 41 ∪A+ 43
We see that
x∈A∪A+
1
⇐⇒ T x ∈ T A
2
and likewise that
x∈A+
3
1
1
∪ A + ⇐⇒ T x + ∈ T A,
4
4
2
so (2.2.4) can be rewritten as
Z
Z
1
1
1T A (T x)φ1 (T x)dµ + 1T A (T x + )φ1 (T x + )dµ,
2
2
and since µ is invariant for T this is
Z
Z
1
1
1T A (x)φ1 (x)dµ + 1T A (x + )φ1 (x + )dµ,
2
2
(2.2.5)
12
CHAPTER 2. NON-DENSE ORBITS OF THE TIMES Q-MAP
which by the definition of ω1 is
Z
φ1 (x)dω1 (x).
TA
Recall that φ1 =
as
dµ
dω1
so this is equal to µ(T A), and using (2.2.5) this can be written
1
µ(T A) = µ(A) + µ(A + ) = ω1 (A)
2
as desired.
We now wish to prove that
Z
h=
φ1 (x)dµ(x).
(2.2.6)
For all f ∈ L1 (µ) we have that the conditional expectation is
X
E(f | Bn )(x) =
f (x + α)φn (x + α)
α∈Dn
since for B ∈ Bn we have
Z X
Z
Z
f (x + α)φn (x + α)dµ(x) =
f (x)φn (x)dωn (x) =
f (x)dµ(x).
B α∈D
n
B
B
So for α ∈ Dn and J = [α, α + 1/pn ) we have
µ(J | Bn )(x) = E(1J (x) | Bn ) = φn (x)
(2.2.7)
for µ-almost every x ∈ J. We now see that the partition ξ1 where
j j+1
i
ξi =
,
|0≤j<p
pi pi
is a generating partition for T acting on the system (T, µ, B), and hence the entropy
can be found as
1
h = lim H(ξn )
n→∞ n
since
ξn = ∨ni=1 T −n ξ1 .
We can define the conditional entropy as we defined the entropy, but instead using
the conditional measure, so
XZ
H(ξn | Bn ) = −
log µ(x | Bn )(x)dµ(x).
J∈ξn
J
Using (2.2.7) we see that
H(ξn | Bn ) = −
Z
φn (x)dµ(x)
2.3. HAUSDORFF DIMENSION
13
and using (2.2.2) that
Z
Z
Z
H(ξn | Bn ) = − φ1 (x)dµ(x) − φ1 (T x)dµ(x) − · · · − φ1 (T n−1 x)dµ(x)
Z
= −n φ1 (x)dµ(x)
since µ is T -invariant. So in order to prove (2.2.6) we need to prove that
1
1
H(ξn ) = lim H(ξn | Bn ).
n→∞ n
n→∞ n
lim
This follows from the Martingale convergence theorem since
\
Bn = {∅, T}.
n≥0
From (2.2.6) We see that the entropy is zero if and only if φ1 (x) = 1 for µ-almost
every x ∈ T, and due to (2.2.2) this is equivalent with φn (x) = 1 for µ-almost every
x ∈ T for n ≥ 1, which by lemma 2.2.4 is equivalent with µ not being conservative
for D. This proves the first implication.
If T is ergodic with respect to µ we can use (2.2.2) and apply the ergodic theorem
on x 7→ log φ1 (x) to get
Z
log φn (x)
→ − log φ1 (x)dµ(x) = h
n
as n → ∞ and hence
φn (x)1/n 7→ e−h
for n → ∞. So if h > 0 we must have that φn (x) converges as e−nh and hence
φn (x) → 0 which by lemma 2.2.4 proves that µ is conservative for D.
2.3
Hausdorff dimension
The notion of Hausdorff dimension will be important in the following chapters, so
here is a short introduction. See [3] for more..
Let A ⊆ Rn . The following generalizes straight away to general metric spaces,
but we will only need it on the real numbers. We define the diameter of a bounded
set as
diam(U ) = sup |x − y| ,
x,y∈U
and we call a collection of sets {Ui } a δ-cover of A if the union of the sets cover A
and diam(Ui ) ≤ δ. Now let for s ≥ 0
X
Hδs (A) = inf{
diam(Ui )s | {Ui } is a δ-cover of A}.
i
As δ decreases, this value increases, and we let
H s (A) = lim Hδs (A)
δ→0
14
CHAPTER 2. NON-DENSE ORBITS OF THE TIMES Q-MAP
H s (A)
∞
s
dim H A
Figure 2.1: A plot of H s (A) as a function of s.
which may be +∞. We call this the s-dimensional Hausdorff measure of A. This is
in fact a measure, and they generalize the Lebesgue measure λn on Rn , since
H n (A) =
1
λn (A),
λn (B(0, 12 ))
where B(0, 21 ) ∈ Rn is the n-dimensional ball with center in the origin and diameter
1. Now the Hausdorff dimension of A is in some sense the appropriate s to pick,
when trying to measure A with the Hausdorff measure. We see that for t > s we
have
Hδs (A) ≤ δ t−s Hδt (A),
so letting δ → 0 we see that if H t (A) < ∞ we must have H s (A) = 0. This proves
that the graph of s 7→ H s looks something like figure 2.1.
The critical value s, on where the graph jumps from ∞ to 0 is called the Hausdorff
dimension. In this jump, the Hausdorff measure might be 0, ∞ or any positive real
number.
2.4
Lower bounded orbits
I have been studying certain non dense orbits of Tq , namely the lower bounded ones.
This is still far from Furstenberg’s conjecture since this conjecture has to do with
joint actions and my paper only concerns one action at a time.
2.4. LOWER BOUNDED ORBITS
15
This work is inspired by an article [9] of Johan Nilsson, who as me considered
the set1
Fc = {x ∈ [0, 1) | q n x ≥ c for all n ≥ 0}.
Considering the base q expansion of the numbers, he proves that the Hausdorff
dimension of this set as a map of c is continuous but constant almost everywhere,
and characterizes the intervals on where the map is constant.
My paper gives an easy way of calculating dimH Fc whenever c is a q-adic rational.
It can be found as the spectral radius of a transition matrix of a subshift of finite
type, and hence as the Perron root of its characteristic polynomial. The paper gives
an easy way of finding the coefficients of this polynomial based only on elementary
methods, namely integer division with residue.
1
In his paper he assumes q = 2, but in his ph.d-thesis he considers the general case q ≥ 2 as
well as for non-integer bases.
Chapter 3
On lower bounded orbits of the
times q-map
3.1
Introduction
In this paper we study the set
Fcq = {x ∈ [0, 1) | q n x ≥ c for all n ≥ 0}
where q ≥ 2 is an integer. This set is related to badly approximable numbers in
diophantine approximation, and has been studied by Nilsson [9], who studied the
Hausdorff dimension of the set as a map of c, and in more generality by Urbanski
[14] who considered the orbit of an expanding map on the circle.
As Nilsson did we will consider Fcq as a subshift of finite type which enables us
to see it as a problem in dynamical systems. When studied as a subshift of finite
type we can find the dimension of Fcq using the spectral radius of the corresponding
transition matrix, and this motivates the theorem of this paper which characterizes
the characteristic polynomial of this matrix.
The author would like to his PhD supervisor Simon Kristensen and he would
also like to thank Johan Nilsson for reading and commenting on this paper.
3.2
Basic definitions
We fix an integer q ≥ 2 and begin with the definition of part and residue which
comes from elementary integer division with residue.
Proposition 3.2.1. For integers n ∈ N and m ≥ 0 there are unique integers
hn, mi ∈ N (part) and 0 ≤ [n, m] < q m (residue) such that
n = q m hn, mi + [n, m].
We note that if we write n = nk · · · n1 in base q it is easy to find the part and
the residue, since [n, m] = nm · · · n1 and hn, mi = nk · · · nm+1 .
The matrix defined below will be of great importance in this paper, since it and
its submatrices turns out to be the transition matrices of the dynamical systems we
consider.
17
18
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Definition 3.2.2. For m ≥ 1 we define a 0-1 matrix Am of size q m × q m by
(Am )ij = 1 ⇐⇒ [i − 1, m − 1] = hj − 1, 1i.
We let Am (P ) with P ⊆ {1, 2, . . . , q m } be the #P × #P matrix made from picking
only the rows and columns from Am corresponding to the elements in P and for
0 ≤ k ≤ m we let Am (k) be the m − k × m − k matrix where we have removed the
first k rows and columns from Am .
We will often omit the dependency on m when it is not confusing. Considering
i − 1 and j − 1 in base q we see that (Am )ij = 1 if and only if the first m − 1 digits
i
we see that
of j − 1 are equal to the last m − 1 digits of i − 1. So when c = qm
m
q
the base q expansions of the numbers in Fc can be seen as a subshift of finite type
with transition matrix Am (i)m . The metric of the subshift and the unit interval are
equivalent so the dimensional properties are the same. In particular, finding the
Hausdorff dimension of Fcq now boils down to finding the spectral radius ρ(Am (k)),
since
ρ(Am (i)m )
ρ(Am (i))
dimH F (c) =
.
(3.2.1)
=
m
log q
log q
The second equality is simple arithmetic, and for a proof of the first see [10]. This
is the main reason we were interested in finding the characteristic polynomials of
Am (i). The main theorem of this paper is a complete characterization of these
polynomials, and to formulate this theorem we need the following definition.
Definition 3.2.3. For n, m ≥ 1 with 0 ≤ n < q m we define
lm (n) = min{1 ≤ j ≤ m | hn, ji ≥ [n, m − j]}.
Using this definition we let
nm = n − [n, m − lm (n)] = q m−lm (n) hn, m − lm (n)i
be the minimal prefix of n.
This is well defined since [n, 0] = hn, mi = 0 for any n with 0 ≤ n < q m . The
notion of minimal prefix is taken from Nilsson [9], but is here defined somewhat
different since we only consider finite sequences.
Let us consider some examples.
Example 3.2.4. Let q = 3, m = 3. Then
h11, 1i = 3 ≥ 2 = [11, 2]
so l3 (11) = 1 and
If we let n = 7 we have
and
but
so l3 (7) = 3 and 73 = 7.
113 = 11 − [11, 2] = 9.
h7, 1i = 2 < 7 = [7, 2]
h7, 2i = 0 < 1 = [7, 1]
h7, 3i = 0 = [7, 0]
3.3. PROOF OUTLINE
19
We are now ready to state the main theorem.
Theorem 3.2.5. Let 0 < i < q m and let fim (x) be the characteristic polynomial of
Am (i). Then
m
fim (x) = gim (x)xq −m−i
where
gim (x) = xm − a1 xm−1 − · · · − am
and a1 a2 . . . am is the base q expansion of q m − im .
Notice that this implies the nice equality
gim (q) = im .
3.3
Proof outline
First recall that we can find the characteristic polynomial f (x) = xq
− · · · − aqm −i of Am (i) as
X
ak = (−1)k
det Am (P ),
m −i
− a1 x q
m −i−1
(3.3.1)
#P =k, min P >i
or as
ak =
1
trace Am (i)k + a1 trace Am (i)k−1 + · · · + ak−1 trace Am (i) .
k
(3.3.2)
The first formula is sometimes used as the definition of the characteristic polynomial,
and for a proof of the latter see [2]. We now try to outline the proof that essentially is
the construction of an algorithm that calculates both the characteristic polynomial
of Am (i) and im .
• We prove that all the submatrices A(P ) that gives non-zero principal minors
are permutations, so when removing rows and columns from the first to the
last, we only change the characteristic polynomial when removing rows and
columns corresponding to the smallest element of a cycle.
• If lm (i) = m then i is the smallest element of an m-cycle and this is the only
permutation of size ≤ m that has i as an element. So removing i decreases
the m’th coefficient of the characteristic polynomial by 1 and leaves all the
preceding coefficients unchanged. On the other hand, if lm (i) = n < m, then
the nontrivial part of the characteristic polynomial, gim (x), can be found as
n
xm−n ghi,m−ni
(x) since we have (3.3.2) and can prove that
trace Am (i)k = trace An (hi, m − ni)k
for all k ≤ m.
• If lm (i) = m, then im = i + 1m − 1, and if lm (i) = n < m then im =
q m−n hi, m − nin , so we see that i and the characteristic polynomials follow
the same pattern.
20
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
• Since the theorem is true for m = 1, we can now use induction if lm (i) < m.
If not, we increase i until we have lm (i) < m, which happens at some point
since lm (q m − 1) = 1.
• The m + 1’st, m + 2’nd, . . . , q m ’th coefficient of fim (x) are all zero, because
we have found the first M coefficients of the characteristic polynomial for any
M , so we pick K such that lM (K) = m and hK, M − mi = k, then we see that
M
gK
(x) has its m + 1’th, m + 2’th, . . . , M ’th coefficients equal to zero, which
will then also be true for gkm (x). This finishes the proof of the theorem.
3.4
Part and residue
The results in this sections explain some properties of the part and residue functions
and gives a characterization of the powers of A. We will use these results throughout
the paper, often without specifically stating so. The proofs in this section are rather
straightforward and may be skipped on a first read.
Proposition 3.4.1.
1. For j, k, n ≥ 0 we have [[n, j], k] = [n, min{j, k}] and
hn, ki, j = hn, k + ji.
2. For j > k we have
h[n, j], ki = [hn, ki, j − k].
Proof. Let us first prove the two equalities in 1. Since [n, k] is the same as
n (mod q)k
we have the first equality. Now assume that j + k ≤ m. Now hn, ki = q j hn, ki, j +
[hn, ki, j], so
n = q k hn, ki + [n, k] = q k+j hn, ki, j + q k [hn, ki, j] + [n, k],
but since [hn, ki, j] < q j and [n, k] < q k we have
q k [hn, ki, j] + [n, k] ≤ q k (q j − 1) + q k − 1 = q k+j − 1 < q k+j ,
and by the uniqueness of the residue and parts we see that hhn, ki, ji = hn, k + ji.
Now consider 2., so let j > k. From 1. we have
hn, ki = q j−k hn, ki, j − k + [hn, ki, j − k] = q j−k hn, ji + [hn, ki, j − k]
and
So
[n, j] = q k h[n, j], ki + [n, j], k = q k h[n, j], ki + [n, k].
n = q k hn, ki + [n, k]
= q j hn, ji + q k [hn, ki, j − k] − q k h[n, j], ki + [n, j]
= q j hn, ji + [n, j] + q k ([hn, ki, j − k] − h[n, j], ki)
and since n = q j hn, ji + [n, j] this implies that
[hn, ki, j − k] = h[n, j], ki.
0
0
3
0
0
1
1
1
i
i
a1
a2
a3
l(i)
Ai+1,i+1
A2i+1,i+1
0
0
3
2
2
2
1
1
0
0
3
2
2
1
2
2
0
1
2
2
2
0
3
3
0
0
3
2
1
2
4
4
0
0
3
2
1
1
5
5
0
1
2
2
1
0
6
6
0
0
3
2
0
2
7
7
0
0
3
2
0
1
8
8
0
0
1
2
0
0
9
9
0
1
1
2
0
0
9
10
0
0
1
2
0
0
9
11
0
0
1
2
0
0
9
12
1
1
1
2
0
0
9
13
0
0
3
1
1
1
14
14
0
0
2
1
1
0
15
15
0
1
2
1
1
0
15
16
0
0
3
1
0
1
17
17
0
0
1
1
0
0
18
18
0
0
1
1
0
0
18
19
0
1
1
1
0
0
18
20
0
0
1
1
0
0
18
21
0
0
1
1
0
0
18
22
0
1
1
1
0
0
18
23
0
0
1
1
0
0
18
24
0
0
1
1
0
0
18
25
1
1
1
0
0
0
18
26
Table 3.1: Calculation of the characteristic polynomials of A3 (i) when q = 3. We let gi (x) = x3 − a1 x2 − a2 x − a3 . We also give
the minimal prefix and the length of the minimal prefix. The numbers in bold indicates that we consider a minimal number (see
Definition 3.5.2) with non-maximal prefix length l(i) < 3.
3.4. PART AND RESIDUE
21
22
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Lemma 3.4.2. Let 1 ≤ k ≤ m. Then Akij = 1 if and only if
[i − 1, m − k] = hj − 1, ki.
Proof. We will prove this by induction. For k = 1 it is the definition of A, so assume
that 1 < k ≤ m. We assume that the lemma is true for all smaller k. If Akij = 1
there must exist some n with 0 ≤ n < q m and Anj = 1 and Ak−1
= 1. Using the
in
induction hypothesis we get
[i − 1, m − k + 1] = hn − 1, k − 1i and [n − 1, m − 1] = hj − 1, 1i
(3.4.1)
for this n. Now by part 2. of the above proposition we have
[hn − 1, k − 1i, m − k] = h[n − 1, m − 1], k − 1i,
and using (3.4.1) we get
[i − 1, m − k + 1], m − k = hj − 1, 1i, k − 1 ,
and using part 1. of the proposition we get
[i − 1, m − k] = hj − 1, ki
as desired.
Now assume that [i − 1, m − k] = hj − 1, ki. Let
n − 1 = q k−1 [i − 1, m − k + 1] + [hj − 1, 1i, k − 1].
This is a positive integer smaller than q m . By the uniqueness of the residue and
parts we see that
[i − 1, m − k + 1] = hn − 1, k − 1i
(3.4.2)
and
[hj − 1, 1i, k − 1] = hn − 1, k − 1i.
Ak−1
in
(3.4.3)
From (3.4.2) and the induction hypothesis we see that
= 1. We now want to
prove that Anj = 1. Recall that we assume [i − 1, m − k] = hj − 1, ki, so
h[n − 1, m − 1], k − 1i = [hn − 1, k − 1i, m − k]
= [i − 1, m − k + 1], m − k
= [i − 1, m − k]
= hj − 1, ki.
Using this and (3.4.3) we see that
[n − 1, m − 1] = q k−1 h[n − 1, m − 1], k − 1i + [n − 1, m − 1], k − 1
= q k−1 hj − 1, ki + [n − 1, k − 1]
= q k−1 hj − 1, ki + [hj − 1, 1i, k − 1]
= q k−1 hj − 1, 1i, k − 1 + [hj − 1, 1i, k − 1]
= hj − 1, 1i.
3.5. MINIMALITY
23
This proves that Ak−1
= 1 and Anj = 1 which implies that Akij > 0. Now assume
in
that there is another n0 such that Ak−1
in0 = 1 and An0 j = 1. Then
[i − 1, m − k + 1] = hn0 − 1, k − 1i
and
[hj − 1, 1i, k − 1] = hn0 − 1, k − 1i
so
n0 − 1 = q k−1 hn0 − 1, k − 1i + [n0 − 1, k − 1]
= q k−1 [i − 1, m − k + 1] + [n0 − 1, m − 1], k − 1
= q k−1 [i − 1, m − k + 1] + [hj − 1, 1i, k − 1]
= n − 1,
which proves that there can be only one such n, so Akij = 1.
Lemma 3.4.3. If a, b, k is such that [a, k] < [b, k] and ha, ki = hb, ki, then
[a, k + j] < [b, k + j]
for all 0 ≤ j ≤ m − k.
Proof. If ha, ki = hb, ki then
ha, ki, j = hb, ki, j ,
and hence
ha, k + ji = hb, k + ji.
Since a < b we thus have
[a, k + j] < [b, k + j]
as desired.
3.5
Minimality
We now prove the following rather simple lemma which states that the only non-zero
principal minors can be found as submatrices of A who are permutations.
Lemma 3.5.1. If det A(P ) 6= 0 then the corresponding matrix is a permutation
matrix.
Proof. Assume that we choose P such that one of the rows of A(P ) has two ones.
In other words there are i, j1 , j2 ∈ P such that
Aij1 = Aij2 = 1.
Using the definition of A this implies that
hj1 − 1, 1i = [i − 1, m − 1] = hj2 − 1, 1i.
24
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Now let k ∈ P be arbitrary. Then Akj1 = 1 if and only if [k − 1, m − 1] = hj1 − 1, 1i,
which is true if and only if
[k − 1, m − 1] = hj2 − 1, 1i,
so Akj1 = Akj2 for all k ∈ P , so the j1 ’th and j2 ’nd column are equal and so
det A(P ) = 0. The proof is similar when we assume that there are two ones in one
column.
Recall that if A(P ) is a permutation, then P = P1 ∪ · · · ∪ Pn where ∩i Pi = ∅
and A(Pi )’s are all cycles. This motivates the following two theorems, where we
characterize the subsets P where A(P ) is a cycle. We are interested in the smallest
elements of cycles, since the whole cycle is removed when we remove this element,
which we will prove is exactly the numbers that are minimal.
Definition 3.5.2. We say that 0 ≤ n ≤ q m is m-minimal if
l(n)
An+1,n+1 = 1,
or equivalently using lemma 3.4.2 if
[n, m − l(n)] = hn, l(n)i.
Theorem 3.5.3. Let P ⊂ {1, 2, . . . , q m } be such that A(P ) is a k-cycle for some
1 ≤ k ≤ m. Then min P − 1 is minimal with lm (min P − 1) = k.
Proof. Let P = {i1 , i2 , . . . , ik } be a k-cycle with Aji1 ij+1 = 1 for 1 ≤ j < k and
Aki1 i1 = 1. Without loss of generality we can assume that min P = i1 . Using
lemma 3.4.2 we get that
[i1 − 1, m − j] = hij+1 − 1, ji,
for 1 ≤ j < k and
[i1 − 1, m − k] = hi1 − 1, ki
so we need to prove that hij+1 − 1, ji > hi1 − 1, ji for j = 1, 2, k − 1. We have the
non-strict inequality since i1 < ij . So assume for contradiction that
hi1 − 1, ji = hij+1 − 1, ji.
Now since i1 < ij+1 we have
[i1 − 1, j] < [ij+1 − 1, j],
and due to lemma 3.4.3 we have
[i1 − 1, m − k + j] < [ij+1 − 1, m − k + j]
(3.5.1)
since k ≤ m. Since Ak−j
ij+1 i1 = 1 we have [ij+1 − 1, m − k + j] = hi1 − 1, k − ji. Using
(3.5.1) we get
[i1 − 1, m − k + j] < hi1 − 1, k − ji.
3.5. MINIMALITY
25
Now consider ik−j+1 . Since j < k we have Ak−j
i1 ik−j+1 = 1 so
[i1 − 1, m − k + j] = hik−j+1 − 1, k − ji,
and hence
hik−j+1 − 1, k − ji < hi1 − 1, k − ji.
This implies that ik−j+1 < i1 which is a contradiction against i1 being the least
element in P .
Theorem 3.5.4. Assume that i − 1 is minimal. Then there is a unique P ⊆
{1, 2, . . . , q m } such that min P = i and A(P ) is a l(i − 1)-cycle.
Proof. We let P = {i, i2 , i3 , . . . , ik } where
i2 − 1 = q[i − 1, m − 1] + hi − 1, m − 1i
i3 − 1 = q 2 [i − 1, m − 2] + hi − 1, m − 2i
..
.
ik − 1 = q k−1 [i − 1, m − k + 1] + hi − 1, m − k + 1i.
We now need to prove that An−1
iin = 1 and that i < in for all n = 2, 3, . . . , k. Using
the uniqueness of the part and residue we see that
hin − 1, n − 1i = [i − 1, m − n + 1]
and
[in − 1, n − 1] = hi − 1, m − n + 1i
for n = 2, 3, . . . , k. The first of these equations implies that An−1
iin = 1.
Since lm (i − 1) = k we know that
hi − 1, ni < [i − 1, m − n]
for n = 1, 2, . . . , k − 1. This implies that
in+1 − 1 = q n [i − 1, m − n] + hi − 1, m − ni > q n hi − 1, ni + [i − 1, n] = i − 1
since both hi − 1, m − ni and [i − 1, n] are smaller than q n .
We now need to prove that this P is unique. Assume that we have P 0 =
{i, i02 , . . . , i0k }, where we order the elements such that An−1
ii0n = 1. This implies that
[i − 1, m − n + 1] = hi0n − 1, n − 1i
for all n = 2, 3, . . . , k. Since A(P ) is a k-cycle, we furthermore know that Ak−n+1
= 1,
i0n i
so
[i0n − 1, m − k + n − 1] = hi − 1, k − n + 1i.
Now we want to prove that i0n = in , so let 2 ≤ n ≤ k be given. We have
i0n − 1 = q n−1 hi0n − 1, n − 1i + [i0n − 1, n − 1]
26
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
and hi0n − 1, n − 1i = [i − 1, m − n + 1], so we just need to prove that
[i0n − 1, n − 1] = hi − 1, m − n + 1i.
We have
[i0n − 1, n − 1] = [i0n − 1, m − k + n − 1], n − 1
= [hi − 1, k − n + 1i, n − 1]
= [in − 1, m − k + n − 1], n − 1
= [in − 1, n − 1]
= hi − 1, m − n + 1i
so in = i0n for all n, and so P = P 0 .
Corollary 3.5.5. If lm (i − 1) = m then there is exactly one P ⊆ {1, 2, . . . , q m } such
that min P = i and A(P ) is a m-cycle.
Proof. This follows from the fact that Am
ij = 1 for all i, j. In particular we have
=
1
for
all
i.
Am
ii
Now compare this corollary with the following lemma.
Lemma 3.5.6. If lm (i − 1) = m, then im = i − 1m + 1.
Proof. It is enough to prove that i = i, since we certainly have i − 1 = i − 1. Using
the definition we see that this is equivalent with [i, m − l(i)] = 0. If l(i) = m we are
done, so assume that l(i) < m. Now either [i, m − l(i)] = 0, in which case we are
done, or [i, m − l(i)] = [i − 1, m − l(i)] + 1. Now since l(i − 1) = m we have
[i − 1, m − l(i)] < hi − 1, l(i)i,
since l(i) < m = l(i − 1), but
[i − 1, m − l(i)] = [i, m − l(i)] − 1 ≤ hi, l(i)i − 1 ≤ hi − 1, m − l(i)i,
which is a contradiction.
Recalling the idea of the proof we here see that if lm (i − 1) = m and we remove
the i’th row and column of Am , then we remove exactly one permutation of size
≤ m, namely a m-cycle, which increases the m’th coefficient of the characteristic
polynomial by one, and we also see that it increases the m’th digit of the base q
expansion of i by one.
3.6
Induction mapping
In the following chapter we will no longer suppress the dependency on m, since we
are interested in mapping permutations between matrices of different sizes while
3.6. INDUCTION MAPPING
27
preserving cycles. We will illustrate the idea with an example. If q = 3, and we
write all numbers in base 3 we see that
012, 120, 201
(3.6.1)
is a 3-cycle in A3 (012). We now map this up to
0120, 1201, 2012
which is a 3-cycle in A4 (0120). On the other hand we could also map (3.6.1) down
to
01, 12, 20
which is a 3-permutation in A2 (01). In this section we will formally define these
maps, and also prove that they map cycles to cycles. We begin with the ’down’ map
which is defined in the following way.
Definition 3.6.1. If 0 ≤ i < q m+1 then we define
Dm (i) = hi, 1i.
For M > m and 0 ≤ i ≤ q M we let
Dm,M (i) = Dm ◦ · · · ◦ DM −1 (i) = hi, M − mi.
We now prove the following lemma.
Lemma 3.6.2. If lM (i) = m < M we have
lm (Dm,M (i)) = m.
Proof. We have [i, M − m] ≥ hi, mi and [i, M − j] < hi, ji for all 1 ≤ j < m, and
we need to prove that [i, m − j] < hi, ji for all 1 ≤ j ≤ m. But this is clearly the
case since m < M , so
[i, m − j] < [i, M − j] < hi, ji
for all 1 ≤ j < m.
Corollary 3.6.3. Let 0 ≤ i < q M . If lM (i) = m < M , then
iM = q M −m Dm,M (i)m .
Proof. This follows from the definition of the minimal prefix.
We saw earlier that the characteristic polynomial of a matrix can be found by
considering the trace of the powers of the matrix. So if we can map permutations
bijectively between two transition matrices we must have the same characteristic
polynomials. As before we only need to consider cycles as all permutations are
products of cycles.
First we formally define what we mean by a cycle in a matrix.
28
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Definition 3.6.4. An ordered k-tuple of distinct elements, (i1 , . . . , ik ) with 0 ≤
ij ≤ q m for all j = 1, 2, . . . , k is a k-cycle in Am (c) if Am (c)ij ,ij+1 = 1 for all
j = 1, 2, . . . , k − 1, and Am (c)ik ,i1 = 1. In other words, if we have
[ij , m − 1] = hij , 1i
for j = 1, 2, . . . , k − 1 and [ik , m − 1] = hi1 , 1i and ij ≥ c for all j = 1, 2, . . . , k.
We have a ‘down’ map, mapping from large matrices to smaller and we now
define an ’up’ map, mapping from smaller to larger.
Definition 3.6.5. Let P = (i1 , . . . , ik ) be a k-cycle in Am (c). Then we let
Um (P ) = (qi1 + [i2 , 1], · · · , qik + [i1 , 1]),
and for M > m we let Um,M = UM −1 ◦ UM −2 ◦ · · · ◦ Um .
Lemma 3.6.6. Let m = lM (c) and let P = (i1 , i2 , . . . , ik ) be a k-cycle in AM (c).
Then
Dm,M (P ) = (Dm,M (i1 ), · · · , Dm,M (ik ))
is a k-cycle in Am (Dm,M (c)). Furthermore, if Q = (j1 , . . . , jk ) is a k-cycle in
Am (Dm,M (c)), then Um,M (Q) is a k-cycle in AM (c).
Proof. To prove that Dm,M (P ) is a k-cycle in Am (Dm,M (c)) can be done by straightforward calculations. We also get that Um,M (Q) is a k-cycle in AM (q M −m hc, M −mi)
rather straightforward. The problem is to prove that it actually is a k-cycle in AM (c),
or in other words that there are no k-cycles with its smallest element in the interval between q M −m hc, M − mi and c. Recalling the definition of cM and that the
least element of a cycle always is minimal we thus need to prove that if we have
cM ≤ n < c, then n cannot be minimal.
We get that nM = cM and lM (n) = lM (c) so
[c, M − m] − [n, M − m] = c − n
so if we assume that n is minimal we get
hc, mi ≥ [c, M − m] = [n, M − m] + c − n = hn, mi + c − n
which is a contradiction. This finishes the proof of the theorem.
These two lemmas now lead to the following theorem regarding the invariance of
the traces.
Theorem 3.6.7. Let m, k ≤ M . Then
trace Am (c)k = trace AM (q M −m c)k .
More generally we have
trace Am (hc, M − mi)k = trace AM (c)k
whenever lM (c) ≥ m.
3.6. INDUCTION MAPPING
29
Proof. Each k-cycle contributes to the trace, and since the maps used in the lemmas
maps all k-cycles injectively, we get the theorem.
Newton’s formula for the characteristic polynomial gives us, that if
fim (x) = xn − a1 xn−1 − · · · − an = det(xI − Am (k))
is the characteristic polynomial of Am (i) where n = q m − k, then
ak =
1
trace Am (i)k − a1 trace Am (i)k−1 − · · · − at−1 trace Am (i)
k
so the above theorem gives us that
fiM (x) = xM −m fqmM −m i (x).
Combining this with the simple lemma below gives us the proof of the main theorem.
Lemma 3.6.8. Let 0 ≤ n < q m . Then
qnm = qnm+1 .
Proof. We see that
qnm = q(n − [n, m − lm (n)]) = qn − [qn, m + 1 − lm (n)],
so we just need to prove that lm+1 (qn) = lm (n). Assume that j = lm (n). Then
hqn, ji ≥ q hqn, ji, 1 = qhqn, j + 1i = qhn, ji ≥ q[n, m − j] = [qn, m].
Now assume that hqn, ji ≥ [qn, m + 1 − j] for some j > lm (n). Then
q[n, j] = [qn, j] ≤ hqn, m + 1 − ji
so
[n, j] ≤ hhqn, m + 1 − ji, 1i = hn, m − ji
which is a contradiction.
We are now ready to prove the main theorem, so let us restate it.
Theorem 3.6.9. Let 1 ≤ i ≤ q m and let fim (x) be the characteristic polynomial of
Am (i). Then
m
fim (x) = gim (x)xq −m−i
where
gim (x) = xm − a1 xm−1 − · · · − am
and a1 a2 . . . am is the base q expansion of q m − im .
30
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Proof. We prove this theorem using induction. If m = 1 it is certainly true since
i1 = i for all 0 ≤ i < q and A1 is the all one matrix of size q × q.
We see that when choosing m and i > 0 we have two possibilities: Either we
have l(i − 1) = m or l(i − 1) < m. In the first case removing the i’th column
and row only removes one non-zero minor, namely the unique m-cycle with i as its
minimal element given in theorem 3.5.4. In this case we also have that the last digit
of i − 1m is [i − 1, 1] which must be non-zero, so here we just decrease am with 1,
so the first m coefficients of the characteristic polynomial changes in the right way
due to lemma 3.5.6.
If we have l(i − 1) = n < m we see that we can find the characteristic polynomial
of the smaller matrix of size q n instead and multiply it by xm−n . As we see in
Corrolary 3.6.3 this is also the case for i. So by induction we are done.
Now we need to prove that the remaining coefficients are all zero. To prove this
we once again use lemma 3.6.7 to see that the M ’th coefficient of fim must be equal
to the M ’th coefficient of fqMM −m i for any M > m. And here we see that the m + 1’th,
m + 2’th, . . . , and M ’th coefficient all are zero, since the M ’th digit of the base q
expansion of
q M − q M −m iM = q M −m (q m − im )
is zero. This finishes the proof of the theorem.
3.7
Constant dimension
Now define φ : c 7→ dimH F (c). Recall from (3.2.1) that when c has finite base q
expansion we can calculate φ(c). Nilsson [9] proved that this function is continuous
and constant almost everywhere. Using the theorem we see that if we have 0 ≤ i <
j < q m such that im = j m then
i
j
φ m =φ m
q
q
and since φ is a decreasing function it must be constant on the interval
i j
,
.
qm qm
Now let 0 ≤ i < q be given and let
j(m) =
m
X
iq n−1 .
n=1
We now claim that
q m−1 im = j(m)m .
To prove this we see that lm (q m−1 i) = 1 and so q m−1 im = q m−1 i. Now lm (j(m)) = 1
and
j(m)m = iq m−1
3.8. NUMERICAL PLOTS
31
which proves the claim. This gives us
j(m)
i
=φ
φ
q
qm
for all m and letting m → ∞ we get that φ is constant on the interval
i
i
,
.
q q−1
Now letting m = 1 we find
gi1 (x) = x − i1 = x − i
which has one root, x = i, so we get
i
log i
φ
=
q
log q
on this interval.
A bit more work allows us to calculate φ(x) for x =
need to solve polynomial equations of degree n.
3.8
i
qn
for larger n since we here
Numerical plots
Calculating the spectral radii of Am (i), we can make numerical plots of the function φ. The plots in figure 1–3 was made using GNU Octave.
1
q=2
q=3
q=5
q=7
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
Figure 3.1: Numerical plots of φ for q ∈ {2, 3, 5, 7}.
1
32
3.9
CHAPTER 3. ON LOWER BOUNDED ORBITS OF THE TIMES Q-MAP
Asymptotics
We now want to consider φ as q → ∞. We consider the function ψ : [0, 1) → [0, 1)
where
0 ≤ c < q−1
1 + log(1−c)
log q
q
ψ(c) =
0
otherwise.
and wish to prove that φ and ψ are somewhat asymptotically similar. This can also
be expressed by saying that ρ(Ac ) behaves somewhat like q − qc, which is true in the
starting point of the intervals where φ is constant, so we get the following theorem.
1
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
Figure 3.2: Plots of φ and ψ when q = 7.
Theorem 3.9.1. For all c ∈ [0, 1) we have
φ(c)
→1
ψ(c)
as q → ∞.
Proof. Let c ∈ [0, 1) be given. Then if we let i = bqcc we have
i
i+1
≤c≤
.
q
q
Now
i+1
i
φ
≥ φ(c) ≥ φ
q
q
and likewise for ψ since both functions are decreasing. Due to the result we got
earlier on constant intervals we have
log(q − i)
log(q + 1 − i)
≥ φ(c), ψ(c) ≥
log q
log q
3.9. ASYMPTOTICS
33
so recalling the definition of i we have
log(q − i)
φ(c)
log(q − i + 1)
≥
≥
log(q − i + 1)
ψ(c)
log(q − i)
and since i → ∞ as q → ∞, both the lower and upper bound converges to 1. This
finishes the proof.
Since we also see that ψ(c) → 1 as q → ∞, we also have the following corollary.
Corollary 3.9.2. For all c ∈ [0, 1) we have
φ(c) → 1 as q → ∞.
The convergence is very slow though – since φ and ψ are equal on q points we
can just look at the convergence of
log(1 − c)
log q
to zero which is easy to calculate.
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0
0.2
0.4
0.6
Figure 3.3: Plot of φ when q = 50000.
0.8
1
Chapter 4
Cantor-like constructions of Fc
4.1
Geometric construction
One idea I recently came up with, is to construct the sets Fc in a Cantor-like fashion.
It gives a representation of these sets as morphic words. The discussion in this
chapter is rather informal but will be further formalized, if the idea turns out to
give some interesting results.
Let
Fc (m) = {x ∈ [0, 1) | q n x (mod 1) ≤ c for all 0 ≤ n ≤ m}.
Defining Tq : [0, 1) → [0, 1) by Tq (x) = qx (mod q) we have
Fp/qk (m) =
m
\
Tq−j
j=0
\
m
p
,
1
=
Jp/qk (j)
qk
j=0
where
p
Jp/qk (j) = Tq−j [ k , 1)
q
p 1
p + qk 2
p + (q j − 1)q k
= k+j , j ∪
,
∪ ··· ∪
,1
q
q
q k+j q j
q k+j
qj [
p + (i − 1)q k i
=
, j .
k+j
q
q
i=1
Figure 4.1 shows the first couple of iterations for c ∈ { 91 , 29 , 93 , 49 , 59 } and q = 3.
We can calculate the Hausdorff dimension in each case using the self similarity1 .
4.2
Specific examples
We fix q = 3 and consider the examples from figure 4.1.
Example 4.2.1. In the case c = 95 we see that in each iteration each interval of
length 4/3m is split into two intervals of length 4/3m+1 and 3/3m+1 , and each interval
1
See [3].
35
36
CHAPTER 4. CANTOR-LIKE CONSTRUCTIONS OF FC
Figure 4.1: Fc (4) for c ∈ { 19 , 29 , 39 , 94 , 59 }.
of length 3/3m gives an interval of length 4/3m+1 . Seen as iterations on words on
{3, 4} we can write it as 4 → 43 and 3 → 4, and then the first couple of iterations
are given as
4, 43, 434, 43443, 43443434, 4344343443443, 434434344344343443434, · · ·
and in each step we can find λ(Fc (m)) as the sum of the digits divided by 3m+1 . We
see that this sequence is actually a Fibonacci sequence,
am = am−2 + am−1 , a0 = 3, a1 = 4
and am = 3m+1 λ(Fc (m)). Using the scaling property of the Hausdorff measure we
get
s
2s
1
1
s
s
H (Fc (m)) =
H (Fc (m − 1)) +
H s (Fc (m − 2))
3
3
and letting m → ∞ and assuming that 0 < H s (Fc ) < ∞ for some s we get
1=
Solving this as an equation in
1
3s
1
1
+ 2s .
s
3
3
gives
1
=
3s
√
5−1
,
2
and hence
s = log3 φ
4.2. SPECIFIC EXAMPLES
where
37
√
5+1
φ=
2
since
1
= φ + 1.
φ
So under the assumption that Fc is actually an s-set (has positive and finite sdimensional Hausdorff measure), we have proven that
dimH (F 5 ) = log3 (φ).
9
We see that we also have
φ = lim
n→∞
an
,
an−1
and this is no coincidence, because letting
fn =
an
an−1
we get
fn =
an−1 + an−2
1
an
=
=1+
,
an−1
an−1
fn−1
and hence
1=
1
1
+ 2,
f
f
where
an
,
n→∞ an−1
so f has to satisfy the same equation as 3s and hence
f = lim
s = log3 f.
Comparing with the the result from chapter 3 we see that 52 = 5, and since the
ternary expansion of 9 − 5 is 11, the characteristic polynomial of the corresponding
matrix is x2 − x − 1, which has Perron root φ and hence also gives that the Hausdorff
dimension is log3 φ.
The proof of the following lemma is exactly as the above example and will be
left to the reader.
Lemma 4.2.2. Assume that
F =
\
Fn
n≥0
where
F0 ⊃ F1 ⊃ F2 ⊃ · · ·
and Fn is a union of an intervals (maybe not disjoint at the endpoint of the intervals)
of length λ1n where λ is a positive integer. Suppose furhtermore that the sequence
(an ) satisfies a recurrence relation, that is there is c1 , . . . , cm such that
an = c1 an−1 + c2 an−2 + · · · + cm an−m
38
CHAPTER 4. CANTOR-LIKE CONSTRUCTIONS OF FC
for all n. Suppose furthermore that
0 < H sF < ∞
(4.2.1)
for some s. Then
s = logλ
an
lim
n→∞ an−1
(4.2.2)
assuming that the limit exists. In particular this gives the formula
an
dimH F = logλ lim
.
n→∞ an−1
Furthermore s can be found by solving the equation
1=
m
X
cj λjs .
j=1
The assumption (4.2.1) can probably be justified to be true always in this case
and the assumption that the limit exists in (4.2.2) is well known.
Example 4.2.3. Using the Cantor construction from above we see, why F3/9 and
F4/9 has the same dimension. The iterations of F3/9 can be described by 6 → 66 and
the iterations of F4/9 is described by 5 → 55, so both their Hausdorff dimension is
log3 (2). This is the same result we get from finding the characteristic polynomials
of the two, which both are x2 − 2x since 42 = 32 = 3, and 9 − 3 is 20 in base 3.
Example 4.2.4. Now let c = 29 . Using the same notations as in the example with
c = 5/9 we get
7 → 377, 3 → 7,
starting with 7. The first couple of iterations are
7, 377, 7377377, 37773773777377377,
73773773777377377737737737773773777377377, . . . .
The sum of these digits are
7, 17, 41, 99, 239, . . .
and can be calculated using the recursive formula as2
an = 2an−1 + an .
Letting fn = an /an−1 we have
2+
1
fn−1
=2+
an−2
2an−1 + an−2
an
=
=
= fn
an−1
an−1
an−1
and assuming that the limit f = limn→∞ fn exists we get
f =2+
2
1
f
This is also the convergents of the continued fraction of
√
2.
4.2. SPECIFIC EXAMPLES
39
and hence
which has the Perron root
√
2+ 8
2
f 2 − 2f − 1 = 0,
which yields
dimH F2/9 = log3
√ !
2+ 8
.
2
Words defined iteratively in this way are Morphic words and using the theory
that already exists about these, maybe more interesting things could be said about
this.
Choosing c = qin with minimal prefix equal to it self, seems to give a morphic
definition on n symbols and hence a recursive formula depending on the n preceding
digits and hence a characteristic polynomial of degree n. Letting q = 2 and c = 1/4
we get
3 → 23, 2 → 3
which once again yeilds the Fibonacci sequence. Letting c = 3/8 we get the sequence
5 → 25, 2 → 4, 4 → 5
starting with a 5, which yields the recursion formula
an = an−1 + an−3 .
Letting c = 1/16 we get the sequence (written in base 16)
F → EF, E → CF, C → 8F, 8 → EF.
Note that we can characterize many more Cantor set constructed in this way,
and not only the ones made from bounding the orbit from below. Let U ⊆ [0, 1) be
a union of disjoint q-adic intervals that we want the orbit of the times-q map to stay
in. Then
\
FU =
Tq−n U
(4.2.3)
n≥0
where Tq (x) = qx (mod 1). The Cantor middle third set can then be characterized
by 3 → 33 where q = 3, so an /an−1 = 2 for all n so we get the well the known result
that the Hausdorff dimension of the Cantor middle third set is log3 2.
Chapter 5
Generalizations to higher
dimension
5.1
Endomorphims of the torus
Since I finished the paper, I have been working how to generalize the results. One
idea is to instead consider lower bounded orbits of the dynamical system of a 2 × 2
matrix A acting on the torus. That is
Fc = {x ∈ [0, 1)2 | |An x (mod 1)| ≥ c for all n ≥ 0}
for some norm |·|. This turns out not to be so easy, because in the one dimensional
case I use the fact that a certain partition of the unit interval, namely in smaller
intervalse, behaves nicely with this lower bound. It seems that the right way to
generalize the results to higher dimensions is to remove sets from a certain nice
partition, namely one generated by the eigenvectors of the transformation, since
parallel to these vectors, the matrix acts like a scaling like in the one dimensional
case. It turns out that the approach in the preceding chapter, where we construct
the sets as Cantor sets, works out in a nice way here.
This idea has been added shortly before I finished the report and is very heuristic
and mainly based on graphic considerations, but the ideas should formalize easily
and it shows in some sense that the ’right’ way to consider the sets Fc from chapter
3 is not as lower bounded orbits, but orbits avoiding some parallelograms of a lattice
generated by the eigenvectors of the transformation.
5.2
Cantor construction
We let
A=
3 0
2 2
.
Then the eigenvalues of A are 2 and 3 and we choose corresponding eigenvectors are
1
1
, v=
,1 .
u = 0,
2
2
41
42
CHAPTER 5. GENERALIZATIONS TO HIGHER DIMENSION
Figure 5.1: We choose some parallelograms (everything but the blue one) from the
cover U0 to form a set U that the orbit should stay in.
We define for each n ≥ 0 a cover Un of the torus, by considering the lattice generated
by 21n u and 31n v, and letting each set in the cover be the parallelograms from the
lattice. Now we choose a union of sets from this cover U , that we wish the orbit to
stay in, say everything but the blue area in figure 5.1. We now know that
\
FU = {x ∈ [0, 1)2 | T n x ∈ U for all n ≥ 1} =
T −n U,
n≥1
where T x = Ax (mod 1). Since the sides of these parallelograms are made from
eigenvectors of A we know how this preimage looks like, and we get a picture like
in figure 5.2. We now consider the intersection of these covers, where we every time
remove the blue ones. We then see that the green area in figure 5.1 gives four new
allowable areas in figure 5.2, and the magenta and pink both gives five new areas.
This pattern continues, so we represent the pink and magenta by 5 and the green
by 4 and get a substitution scheme as
4 → 4455,
5 → 45555,
and as before, calculating the sum of the digits in each step gives the numbers
14, 66, 312, 1476, 6984, . . .
and we can represent them with the recursive formula
an = 6an−1 − 6an−2 .
5.2. CANTOR CONSTRUCTION
43
Figure 5.2: Each color of the sets in U1 represents the preimage of the parallelogram
in the preceding figure.
This gives us
√
an
→ 3 + 3 as n → ∞
an−1
assuming that the limit, and since we scale the parallelograms by a factor
time, we get the dimension
√
log(3 + 3)
√
≈ 1.735008197
dimH FU =
log 6
√
6 every
where we have used lemma 4.2.21
It seems that this method should generalize to any dimension. We should assume
though that the eigenvalues of the matrix are positive integers in order to be able
to refine the lattice in a nice way. We also need to be able to pick an eigenvector
with integer coordinates, but that is implied if we require the matrix to have integer
entries as well.
To obtain a general result on Hausdorff dimension in higher dimension, one need
a general combinatorial argument on how the inersections behave, depending on the
collection from the cover that we wish the orbits to stay in.
1
This lemma only talks about the one dimensional case, but should generalize straight away.
Chapter 6
Ternary expansions of powers of 2
6.1
A conjecture of Erdős
Recently I stumbled upon a paper of professor Jeffrey Lagarias [7] who considers an
old conjecture of Erdős
When does the ternary expansion of 2n omit the digit 2?
Erdős [1] conjectured that it does not happen for n ≥ 9 (20 , 22 and 28 does), but
the conjecture is still unsolved. Lagarias then introduces a parameter λ > 0 and
considers the set
Nλ (X) = #{n ≤ X | (bλ2n c)3 omits the digit 2}
where (m)3 for an integer m denotes the ternary expansion of m. Erdős conjecture
then states that N1 (X) = 3 for X > 9, but Lagarias instead considers general λ,
and asks for how many λ this set is infinite. It has been proven [8] that
N1 (X) ≤ 1.62xlog3 2
and that there are no solutions for n < 4374 [5] than other then the three already
mentioned. Lagarias proves in his paper that
Nλ (X) ≤ 25X 0.9725
for sufficiently large n, depending on λ. He also proves that the set is not always
bounded – there is a sequence (nk ), which we will study more in the proof of theorem 6.2.1, such that for uncountably many λ, all the integers bλ2nk c have ternary
expansions omitting the digit 2.
6.2
Hausdorff dimension of certain Cantor-sets
Lagarias also proves the following theorem.
Theorem 6.2.1. Let
E = {λ > 0 | (bλ2n c)3 omits the digit 2 for infinitely many n},
45
46
CHAPTER 6. TERNARY EXPANSIONS OF POWERS OF 2
then
dimH E = log3 2.
This is a related to the Furstenbergs conjecture since it links the ternary expansion of a number with some multiplicative properties by 2. It is also related to the
paper I have written since it has to do with omitting digits from an expansion in
some way and then considering the size and dimension of the resulting Cantor-like
set. One huge difference is though, that he only requires infinitely many expansions
to omit the digit, so the connection between his paper and mine is especially clear
when Lagarias considers the sets
C(M1 , . . . , Mk ) = {λ | (bλMi c)3 omits the digit 2 for all i}.
These sets can be constructed as
k \
1
Σ3,2
C(M1 , . . . , Mk ) =
M
j
j=1
where Σ3,2 is the Cantor set of all real numbers that omits the digit 2 in its√ base
3 expansion. Lagarias proves that dimH C(1, 7) = log3 (φ) where φ = 1+2 5 by
considering the transition matrix of the corresponding subshift of finite type – just
as in my paper for F5/9 when q = 3.
I am in contact with prof. Lagarias and I hope to study the connection between
his paper and mine and try to understand the connection between Lagarias work
and Furstenbergs conjecture.
Let us consider the proof of theorem 6.2.1 from Lagarias paper [7]. The lower
bound relies heavily on the following rather technical lemma that describes how the
set Σ̃ can be constructed as a Cantor set. This construction can be seen in the proof
of theorem 1.2 in [7].
Lemma 6.2.2. There is a sequence (mk ) such that
Σ̃ = {λ > 0 | (bλ2mk c)3 omits the digit 1} ⊆ [1, 2]
is an uncountable set. Letting sk = bmk log3 2c for all k ≥ 1 we get a decreasing
sequence of sets
X1 ⊃ X2 ⊃ · · ·
such that
σ̃ =
∞
\
Xs ,
s=0
and for sk ≤ s < sk+1 , Xs is a union of disjoint intervals of length 3sk −s 2mk , except
for possibly two intervals adjacent to the ends that has length between 12 3s and 3s .
The distance between two intervals in Xs is at least 12 3−s , and for s0 ≥ s, each
0
interval in Xs contains at least 2s −s intervals from Xs0 . In particular, Xs consists
of at least 2s−1 intervals.
6.2. HAUSDORFF DIMENSION OF CERTAIN CANTOR-SETS
47
Proof of theorem 6.2.1. We first prove that dimH E ≤ log3 2. Since E ⊆ R+ we
have
∞
[
1
E=
E∩
,M ,
M
M =2
so it suffices to prove that
dimH
1
E∩
,M
M
≤ log3 2.
To show such an upper bound, we need to find a suitable covering of the set, so let
Sn (M ) =
∞
[
Σj
j=n
1
,M
M
,
where
Σj
1
,M
M
=
1
j
≤ λ ≤ M | (bλ2 c)3 omits the digit 2 .
M
We then see that for each n ≥ 1 we have
1
E∩
, M ⊆ Sn (M ).
M
Now pick n = log3 M + 2. This yields in particular that
λ2j ≥ 1
for j ≥ n. Fix such a j. Note that the map χ(λ) = bλ2j c is constant on intervals of
length 21j , and letting letting λ vary over the interval [1/M, M ] we see that χ(λ) varies
over a subset of {1, 2, . . . , M 2j }. Every time one of these integers gives an admissible
ternary expansion (one that omits the digit 2), we get an admissible interval of length
j
1/2j . In {1, 2, . . . , M 2j } there is at most 2dlog3 (2 M )e such admissible integers – two
choices for each digit. This is certainly less than
j M )+2
2log3 (2
≤ 2j log3 2+log3 M +2 ≤ 4M 2j log3 2 .
Now each of these integers k yields an interval on which χ(λ) = k is constant. Let
Ij (M ) be the collection of these at most 4M 2j log3 2 intervals of length 1/2j . This
collection covers the set Σj ([1/M, M ]). Now let
I(n, M ) =
∞
[
Ij (M ),
j=n
then E ∩ [1/M, M ] is covered by I(n, M ), and still every interval in this cover has
length 1/2j . Now let ε > 0. Letting |I| denote the length of an interval I we have
48
CHAPTER 6. TERNARY EXPANSIONS OF POWERS OF 2
that the Hausdorff measure of dimension log3 2 + ε is
∞
X
X
X 1 log3 2+ε
log3 2+ε
|I|
=
2j
j=n I∈Ij (M )
I∈I(n,M )
log3 2+ε
∞
X
1
j log3 2
≤
4M 2
2j
j=n
= 4M
∞
X
1
2jε
j=n
= 4M
2−nε
.
1 − 2−ε
Now let n → ∞. Then the diameter of each interval in the covering I(n, M ) goes
to zero, so
1
, M ) ≤ log3 2 + ε,
dimH (E ∩
M
and letting ε → 0 gives us the desired upper bound.
Now we wish to prove the lower bound, so we let Σ̃, mk , and sk be defined as in
the lemma above. we now see that Nk = bλ2mk −1 c for k ≥ 1 has ternary expansion
that omits the digit 2. This is true because we either have
Mk = 2Nk
or
Mk = 2Nk + 1,
but since each digit in the ternary expansion of Mk is either 0 or 2, Mk must be an
even number, so
Mk = 2Nk ,
and hence must the ternary expansion of Nk omit the digit 2. This proves that
Σ̃ ⊆ {λ > 0 | (bλ2mk −1 c)3 omits the digit 2 for all k ≥ 1} ⊆ E,
and so if we can lower bound the Hausdorff measure of Σ̃ instead, we are done. In
other words, we wish to prove that
H log3 2 (Σ̃) > 0.
(6.2.1)
So we need to show that for any cover {Ui } of Σ̃ we have
X
diam(Ui )log3 2 > 0.
i
It is enough to prove it for covers consisting of open intervals, and since Σ̃ is compact,
it is enough to consider finite covers. For any Ui in the covering, pick s ∈ N such
that
3−s ≤ diam(Ui ) < 3−s+1 .
Now due to the above lemmas statement on the distance between two intervals in
Xs , this means that Ui can only intersect at most two intervals in Xs , so if s0 ≥ s
0
then Ui intersects at most 2 · 2s −s intervals in Xs0 . Now
0
0
0
2 · 2s −s = 2 · 2s 3−s log3 2 ≤ 2 · 2s diam(Ui )log3 2 .
6.2. HAUSDORFF DIMENSION OF CERTAIN CANTOR-SETS
0
49
Now choose s0 so large that diam(Ui ) ≥ 3−s for all Ui in the (finite) cover. There are
0
0
at least 2s −1 intervals in Xs0 . Now since each Ui covers at least 2 · 2s diam(Ui )log3 2
intervals, we must have
X
0
0
2 · 2s diam(Ui )log3 2 ≥ 2s −1
i
which implies that
X
i
as desired.
diam Ui log3 2 ≥
1
>0
4
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