AN INTRODUCTION TO TOPOLOGICAL ENTROPY KAREN BUTT Abstract. We develop the basic notions necessary to define topological en- tropy for topological dynamical systems. We prove a number of results which are then used to compute the entropy of some standard dynamical systems. Finally, we show topological entropy is an invariant of topological conjugacy. Contents 1. Introduction 1 2. Definition with Covers 2 3. Spanning and Separated Sets 3 4. Topological Entropy for Some Examples 6 5. Topological Conjugacy 9 Acknowledgements 10 References 10 1. Introduction A discrete-time dynamical system is a nonempty set X and a map f : X ! X, together with some additional structure, such as measure-theoretic or topological structure. The study of dynamical systems as a whole is primarily concerned with the asymptotic behavior of such systems, that is how the system evolves after repeated applications of f. In this paper, we are concerned with the study of topological dynamics, where we restrict X to be a metric space and f to be a continuous transformation. Topological entropy measures the evolution of distinguishable orbits over time, thereby providing an idea of how complex the orbit structure of a system is. Entropy distinguishes a dynamical system where points that are close together remain close from a dynamical system in which groups of points move farther. In the first section, we develop the notion of entropy by considering coverings of X. Next, we explain the notions of spanning and separated sets and prove that entropy can also be defined in these two ways. With a more flexible definition of entropy, we are able to compute it for specific examples in the subsequent section. As guiding examples, we consider the doubling map, isometric transformations and hyperbolic toral automorphisms. In the final section, we explain topological conjugacy, which is the notion of equivalence in topological dynamical systems. We then prove that topological entropy is an invariant. Date: 2014. 1 2 KAREN BUTT 2. Definition with Covers We will consider the exponential growth of distinguishable orbits with respect to the number of applications of our map f. First we will consider finite orbit segments of length n distinguishable at some finite resolution . Given a compact metric space (X; d) and map f : X ! X, we define the function dn : X × X ! R by k k dn(x; y) = max d(f (x); f (y)): 0≤k<n For each n, dn is a metric on X, since d is already defined to be a metric on X. With this new notion of distance, points are -close if they remain -close for n iterates of f. We can think of as a resolution, the smallest distance at which we can distinguish two points from one another. Next we define a quantity that counts these distinguishable orbits. We want to collect indistinguishable orbits { that is, points that are -close with respect to the dn-metric { and count how many of these collections we have. However, we want to avoid the possibility of the collections themselves being -close to one another as this would over estimate the distinguishable orbits. So we consider a covering of X that contains the smallest possible amount of collections of -close points. Definition 2.1. Fix > 0. Let cov(n; , f) denote the minimal cardinality of a covering of X by sets of dn-diameter less than , where the diameter of a set is given by the supremum of distances between pairs of points in the set. Remark 2.2. Since compactness means every open cover of X contains a finite subcover, it follows that cov(n; , f) is a finite quantity. Definition 2.3. Let 1 h(f) = lim sup log(cov(n; , f)): n!1 n This limit gives the exponential growth of cov(n; , f) with a fixed resolution as the length of orbits we consider tends to infinity. We are considering the limit superior because we do not know if the limit actually exists. To prove it does, we require the following lemma from calculus. Lemma 2.4. Let fangn≥1 be a subadditive sequence, that is am+n ≤ am +an 8m; n. Then lim an=n exists and is equal to infn an=n. n!1 The proof uses basic computation to show the limit inferior is equal to the limit superior. For the complete proof, see [4, Theorem 4.9]. 1 Proposition 2.5. The limit lim log(cov(n; , f)) = h(f) exists and is finite. n!1 n Proof. Given the previous lemma, we need to show that the sequence log(cov(n; , f)) is subadditive. Then, since log(cov(n; , f)) is bounded below by 0, inf log cov(n; , f)=n must be finite. Fix > 0. Suppose A is a cover of X with sets of dn-diameter less than and B is a cover of X with sets of dm-diameter less than . Then cov(n; , f) ≤ jAj and cov(m; , f) ≤ jBj, because cov refers to the minimum cardinality. Let U 2 A and V 2 B. If U \ f −n(V ) 6= ;, consider x; y 2 U \ f −n(V ). We have x; y 2 U, so max d(f i(x); f i(y)) < . 0≤i<n AN INTRODUCTION TO TOPOLOGICAL ENTROPY 3 −n n n Also x; y 2 f (V ) which means f (x); f (y) 2 V . Since V has dm-diameter less than , we have max d(f i(f n(x)); f i(f n(y))) < . 0≤i<m The above two inequalities give max d(f i(x); f i(y)) < . 0≤i<m+n −n So the set U \ f (V ) has dn+m-diameter less than . Let C = fU \ f −n(V ): U 2 A; V 2 Bg. Then jCj ≤ jAjjBj; equality holds when each intersection is non-empty. Additionally, we have already shown that C has dn+m-diameter less than , so cov(m+n; , f) ≤ jCj. Recalling the initial bounding of cov(n; , f) and cov(m; , f) we have cov(m + n; , f) ≤ jCj ≤ jAjjBj ≤ cov(n; , f)cov(m; , f): Recalling basic properties of logarithms, we have that the sequence log(cov(n; , f) is subadditive, which completes the proof. Next, we want to decrease the resolution. Clearly if we decrease , the amount of sets of diameter less than needed to cover X can only increase. Therefore, cov(n; , f) is monotonic, and so is h(f), which means it has a limit as goes to 0. Definition 2.6. The topological entropy of f is given by h(f) = lim h(f). !0+ This definition can be used to prove a number of properties about topological entropy, but before we proceed to examples, we develop more useful definitions in the next section. 3. Spanning and Separated Sets We will again consider the metric dn to count distinguishable orbit segments at a fixed resolution. Definition 3.1. Fix > 0. Let n 2 N. A set A ⊂ X is an (n; )-spanning set if 8x 2 X, 9y 2 A such that dn(x; y) < . Remark 3.2. Recall that the definition of dn depends on f. This is a useful characterization because we can easily construct such sets for known dynamical systems. The first dynamical system we will consider is the doubling map f : S1 ! S1 where f(x) = 2x mod 1. Here we are taking S1 to be [0; 1] with endpoints identified. The natural choice of metric for the unit circle is d(x; y) = min(jx − yj; 1 − jx − yj) with respect to which f is continuous. We will always consider this metric when discussing the doubling map. Before we can construct an (n; )-spanning set for the doubling map, we require the following lemma. Lemma 3.3. Let f be the doubling map on S1. Then we have 1 d(x; y) ≤ ) d(f(x); f(y)) = 2d(x; y): 4 4 KAREN BUTT Proof. Clearly, d(x; y) = jx − yj when jx − yj ≤ 1=2: Let x; y be such that d(x; y) ≤ 1=4. This means jx − yj ≤ 1=4. Using the definition of f, we have d(f(x); f(y)) = d(2x mod 1; 2y mod 1) = min(j2x − 2y mod 1j; 1 − j2x − 2y mod 1j): Note that j2x − 2yj ≤ 1=2, so 2x − 2y mod 1 = 2x − 2y. Therefore, d(f(x); f(y)) = 2jx − yj = 2d(x; y): k Notation 3.4. Let Sk denote the set of dyadic rationals with denominator 2 , that is i S = ; 0 ≤ i < 2k − 1 : k 2k Proposition 3.5. The set of fractions Sn+k is an (n; )-spanning set for the dou- bling map. Proof. Fix > 0. Choose k ≥ 2 such that 1=2k+1 ≤ < 1=2k. Note that for any x 2 S1, there exists i 2 f0; :::; 2n+k − 1g such that h i i + 1 x 2 ; : 2n+k 2n+k Then, choose y 2 Sn+k to be either of the endpoints of this dyadic interval. Then, d(x; y) ≤ 1=2n+k < 1=4. Using Lemma 3.3, this implies d(f(x); f(y)) = 2d(x; y) ≤ 2=2n+k < 1=4: So we can apply the lemma again to get d(f 2(x); f 2(y)) = 2d(f(x); f(y)) ≤ 22=2n+k Applying the lemma j consecutive times, for any j satisfying 0 ≤ j < n, gives 2j 2n − 1 1 d(f j(x); f j(y)) = 2jd(x; y) ≤ ≤ < ≤ . 2n−k 2n−k 2k+1 So for any x 2 S1 we have, j j max (d(f (x); f (y)) = dn(x; y) < , 0≤j<n for some y 2 Sn+k, which concludes the proof.