Solved exercises

1. Dynamical Systems
Solved Exercises
14 de julho de 2020
Miguel Fernandes
1. Consider the doubling map E2 : R/Z → R/Z defined by E2(x + Z) = 2x + Z.
(a) Find the number of periodic points of period n.
Remember: Given a topological dynamical system f : X → X, we say that a
point x is periodic of period p if fp
(x) = x holds.
We have:
En
2 (x + Z) = 2n
x + Z = x + Z ⇔ (2n
− 1)x = k ⇔ x =
k
2n − 1
for some integer k. Thus the periodic points of period n are those for which k
varies from 0 to 2n
− 2 and hence we have 2n
− 1 periodic points of period n.
(b) Conclude that the periodic points of E2 are dense in S1
.
Following the previous exercise, we see that the periodic points of E2 are uni-
formly distributed in S1
. Consequentely, making n → ∞ we get the desired
result.
2. Consider the dynamical system f : R → R defined by f(x) = |x − 2|.
(a) Find the fixed point of f. Is it periodic? What is its period? Comment.
The computation of the fixed point of f is the following:
f(x) = x ⇔ |x − 2| = x ⇔ x − 2 = ±x ⇔ x = 1.
Clearly, it is a periodic point of period 1, which means that, a particle starting
its trajectory at x = 1 will remain in that position for all future time.
(b) Define the first three iterates of f and sketch them.
Clearly, f(x) = |x − 2|, f2
(x) = ||x − 2| − 2| and f3
(x) = |||x − 2| − 2| − 2|, all
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2. defined in the whole real line (a piecewise definition would be too cumbersome,
a nice task for the reader!). The graphs of such iterations are shown below.
Figura 1: First three iterates of the transformation f.
(c) Find other periodic and pre–periodic points of f.
It is clear in the figure above that the points of the set [0, 2] {1} are periodic
of period 2 and the points belonging to ]−∞, 0[ ∪ ]2, +∞[ are pre–periodic.
3. Show that expanding maps of S1
are topologically mixing.
Remember: A topological continuous map f : X → X is said to be topologically
mixing if, given two non–empty open sets U and V , there exists a time N ∈ N such
that for all n ≥ N we have fn
(U) ∩ V = ∅.
Let f : S1
→ S1
be an expanding map (and hence is continuous) and F : R → R such
that π◦F = f ◦π (the existence of such function F is guaranteed by the continuity of
f), where π : R → S1
is the projection defined by π(x) = [x], for all x ∈ R. (Remark:
[x] denotes the point of S1
subject to the identification S1
= R/Z). Notice that we
can also write f ([x]) = [F(x)]. Clearly, if f([x]) = [mx], for some 1 < m ∈ Z then
F(x) = mx + n, for some n ∈ Z. Hence |F (x)| = m > 1, for all x ∈ R, and by
the mean value property it follows that |F(x) − F(y)| = |F (c)(x − y)| = m|x − y|,
where x, y ∈ R and c ∈ (x, y) (or c ∈ (y, x)). Take an open set U ⊂ R and choose an
interval I = (a, b) ⊂ U. It is straightforward that Fn
, n ∈ N, increases the length of
I by a factor of mn
. That implies that we can choose n0 such that mn0
> 1/(b − a)
and consequently the length of Fn0
(I) will exceed 1. Thus π(Fn0
(I)) = S1
and
S1
⊂ fn
(π(I)), for all n ≥ n0. This means that, given two non–empty open sets U
and V in S1
, the set fn
(U) ∩ V is non–empty, and the same for all future time.
4. Consider the rotation of the circle Rα : S1
→ S1
defined by Rα(x + Z) = x + α + Z.
(a) Show that Rα has periodic points if and only if α is rational.
If α is rational then it can be writeen in the form α = p/q, for some integers p
and q, then all points are periodic with period q. Indeed:
Rq
p/q(x + Z) = x + q ∗ p/q + Z = x + Z.
Reciprocally,
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3. Rn
α(x + Z) = x + nα + Z = x + Z ⇔ x + nα = x + p ⇔ α = p/n
for some integer p. Thus if we assume that x + Z is periodic of period n, then α
must equal p/n, for some p ∈ Z.
(b) Show that 0 is a recurrent point when α is irrational.
Remember: The orbit of a point x is said to be dense if, for all y ∈ R/Z and
all > 0, there exists a time n ∈ N such that d(Rn
α(x), y) < . In particular,
making y = x, we find that x is a recurrent point.
Let α be an irrational number. We first observe that the map n → Rn
α(0), for
each n ∈ N, is one–to–one. For that, let Rn
α(0) = Rm
α . Then nα+Z = mα+Z ⇔
(n−m)α = k, for some integer k. But, since α is irrational, we must have n = m.
Now, fix > 0. We find a time n for which we have 0 < nα+Z < +Z. The den-
sity of the irrational numbers on the real line allows us to assert that α belongs
to some interval [p/q, p/q + /2q], where p and q > 0 are integers. Consequently,
if we set n = q we have qα ∈ [p, p + /2] and thus qα + Z ≤ /2 + Z < + Z.
This implies that, for certain iterates of Rα, the trajectory is sufficiently close
to 0.
5. Consider the linear transformation T : T2
= R2
/Z2
→ R2
/Z2
defined by:
T(x, y) = (2x + y, x + y)(mod 1)1
.
(a) Show that T is well defined, i.e., if (x−x , y−y ) ∈ Z2
then L(x, y)−L(x , y ) ∈ Z2
.
It’s an immediate consequence of the linearity of T and the fact that Z2
is closed
under T.
(b) Compute the matrix of the linear transformation L : R2
→ R2
defined by
L(x, y) = (2x+y, x+y). Justify that L is invertible and compute its eigenvalues.
The matrix of L (relative to the canonical basis) is:
ML =
Å
2 1
1 1
ã
and is clearly non–singular (determinant equal to 1 and hence M−1
L has integer
entries). Furthermore, its eigenvalues are the inverses:
λ1 =
3 +
√
5
2
and λ2 = λ−1
1 =
3 −
√
5
2
.
(c) Show that the periodic points of T are dense in T2
.
Let x = a/b and y = c/b (same denominator) provided a, b, c, d ∈ Z. Then
T(x, y) = T(a/b, c/b) = 2a+c
b
, a+c
b
and so T(x, y) is also a rational point whose
coordinates have the same denominator as (x, y). But notice that we only have
b2
different choices for (x, y) in T2
(why is that) whose points shall form the set
A. Thus we have {Tn
(x, y)| n ∈ N} ⊂ A. Given T is invertible, we find that the
point (x, y) is periodic. Since b is arbitrary, such periodic points are dense in T2
(find the anology with the expanding map).
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The so–called Arnold cat map.
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4. (d) Show that the number of n−periodic points of T is given by | det(Mn
L − I2)|
(assuming this number is not zero).
The periodic points of T of period n are the solutions of:
Mn
L x = x (mod 1).
Notice that such equation is equivalent to (Mn
L −I2)x = 0 (mod 1) or more preci-
sely (Mn
L −I2)x = k, for some k ∈ Z. This makes us consider the transformation
L∗
: R2
→ R2
defined by L∗
(x) = (Mn
L −I2)x. According to the explained above,
it is clear that the periodic points of T are given by the integer points in the
set L∗
([0, 1[ × [0, 1[). Furthermore, we can easily verify that L∗
transforms the
square [0, 1[ × [0, 1[ into a paralelogram (without 2 edges). Making use of Pick’s
Theorem and regarding the fact that the area of the parallelogram is given by
| det(Mn
L − I2)|, we find the desired result.
(e) Compute the number of points of period 5.
Given the previous result, we have that the number os such periodic points is
given by | det(M5
L − I2)|. But | det(M5
L − I2)| = |(λ5
1 − 1)(λ−5
1 − 1)|, where λ1 is
the eigenvalue in (b). Since λ1 = (3 +
√
5)/2, we find |(λ5
1 − 1)(λ−5
1 − 1)| = 121.
6. Consider the alphabet β = {1, 2, 3} made of three letters. Let A be the transition
matrix defined as follows:
A =
Ñ
0 1 1
1 1 0
0 0 1
é
.
(a) Find elements of the set ΣA = x = x1x2... ∈ Σ| axnxn+1
= 1, n ≥ 1 (Σ is the
set of those infinite words whose letters belong to the alphabet β).
For instance, 221333... and 121333...
(b) How many words are there with with ’3’ as first letter?
One, 33333...
(c) Let α be a word of the form α = 1...3 in a total of 4 letters and hence finite.
Write the all the possibilities for α. How many are they?
There are only two possibilities: 1213 and 1333.
(d) Consider those finite words, of length n + 1, with first and last letters given by i
and j, respectively. Show that the number of words in such conditions is given
by the ij element of An
, where A is the transition matrix. For simplicity, denote
such number by Nn
(i, j). Verify the result for exercise (d).
We use induction on n for a proof. The case n = 1 is trivial (definition of
transition matrix). Now, using the induction hypothesis:
Nn+1
(i, j) = k Nn
(i, k)N1
(k, j) = k(An
)ikakj = (An+1
)ij.
In the case of the exercise (d), the power A3
is the matrix:
A3
=
Ñ
1 2 2
2 3 2
0 0 1
é
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5. and the result follows.
(e) Conclude that the third row of A is invariant under the power operator.
Just remember exercises (b) and (d).
(f) Conclude that A is neither irreducible nor aperiodic.
Since the third row of A is invariant under the power operator and it contains
zeros, we conclude that A is not irreducible and hence cannot be aperiodic.
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