The C*-Algebras Associated with Irrational Time Homeomorphisms Of

Home , Interior (topology)

THE C*-ALGEBRAS ASSOCIATED WITH IRRATIONAL TIME

HOMEOMORPHISMS OF SUSPENSIONS

BENJAM´IN A. ITZA-ORTIZ´

Department of Mathematics University of Oregon June 2003 ii

An Abstract of the Dissertation of

Benjam´ınA. Itz´a-Ortiz for the degree of Doctor of Philosophy

in the Department of Mathematics to be taken June 2003

Title: THE C*-ALGEBRAS ASSOCIATED WITH IRRATIONAL TIME

HOMEOMORPHISMS OF SUSPENSIONS

Approved: Dr. N. Christopher Phillips

Let (X, h) be a dynamical system and let (X, ϕ) be the suspension ﬂow of (X, h). e

Let α be a real number. In this dissertation we study the crossed product C(X)oϕα Z e 1 associated to the dynamical system (X, ϕα). If X is ﬁnite dimensional, K (X) = 0 e and ϕα is minimal and uniquely ergodic, we ﬁnd the Elliott invariant of C(X) oϕα e Z. Conditions on h for the existence of α such that ϕα is minimal and uniquely ergodic are given as well as a formula to compute the topological entropy of ϕα in terms of the topological entropy of h. Several examples derived from this study are provided, including the existence of non orbit equivalent minimal dynamical systems on connected compact metric 1-dimensional spaces arising from time α maps, for the same α, of suspensions of strong orbit equivalent Cantor systems. iii

TABLE OF CONTENTS

Chapter Page

INTRODUCTION ...... 1

I. RANGE OF TRACE ...... 9

II. MINIMALITY ...... 25

III. UNIQUE ERGODICITY ...... 36

IV. ENTROPY ...... 46

V. EXAMPLES ...... 49

APPENDIX: ALMOST ONE-TO-ONE EXTENSIONS ...... 61

BIBLIOGRAPHY ...... 68 1

INTRODUCTION

Continuous and discrete ﬂows are two of the most fundamental concepts in topo-

logical dynamics (see e.g. [24, Chapter II]). By a continuous ﬂow we mean a pair

(X, ϕ) where X is a topological Hausdorﬀ space and ϕ: X R X is a continuous × →

map satisfying ϕ0 = IdX and ϕα+β = ϕα ϕβ for all α, β R. Notice that we follow ◦ ∈ the convention of writing ϕα(x) rather than ϕ(x, α). A discrete flow (X, h) is a pair with X a topological Hausdorff space and h: X X a homeomorphism. When → the space X is compact metric, we say that the discrete flow (X, h) is a (classical) dynamical system.

Given a discrete ﬂow (X, h) where X is a locally compact Hausdorﬀ space, there is a group homomorphism, denoted again by h, from Z to the group of automorphisms

Aut(C0(X)) of (the commutative C*-algebra) C0(X) given by

n (hn(f))(x) = (f h− )(x) ◦

for n Z, f C0(X) and x X. In other words, given a discrete ﬂow (X, h) with ∈ ∈ ∈ X a locally compact Hausdorﬀ space it induces an action of Z on the C*-algebra

C0(X). In general, given a locally compact group G, a C*-algebra A and an action

τ of G on A, one may construct a new C*-algebra, namely, the crossed product of A by G which is denoted by A oτ G, cf. [16]. Hence every discrete ﬂow (X, h) 2

induces a crossed product C0(X) oh Z. This crossed product is also known as the transformation group C*-algebra of (X, h) and sometimes denoted by C∗(Z, X, h).

Similarly, a continuous flow (X, ϕ) with X a locally compact Hausdorff space induces the group homomorphism R Aut(C0(X)), denoted again by ϕ, defined by →

(ϕα(f))(x) = (f ϕ α)(x) ◦ − for α R, f C0(X) and x X. Hence, we may consider the crossed product ∈ ∈ ∈

C0(X) oϕ R.

When one wants to study a continuous flow, it is useful to study an associated discrete flow (see e.g. [24, Chapter I]). Conversely, one way to obtain information about a discrete flow is by studying some associated continuous flow. A standard way to construct a continuous flow from a discrete flow is by means of the so called suspension construction which we describe next, cf. [24, II.5.5]. Let (X, h) be a dynamical system. The construction could be carried out for general discrete flows; however we restrict our attention to dynamical systems for technical reasons. Consider the actions of R and Z on the space X R given by ×

((x, s), α) (x, s + α) 7→ and

((x, s), n) (hn(x), s n), 7→ − respectively, where (x, s) X R, n Z and α R. The suspension X of (X, h) is ∈ × ∈ ∈ e the quotient of X R by the Z-action. Let [x, s] denote the image of (x, s) in X. Then × e 3

an action ϕ of R on X is given by ϕα([x, s]) = [x, s + α] which is well defined since e the Z and R actions on X R defined above commute. Then (X, ϕ) is a continuous × e flow which is called the suspension flow of (X, h), cf. [24, Lemma II.5.7]. Given a real number α we will refer to ϕα as the time α map on the suspension X of (X, h) or e just as the time α map when the spaces involved are understood. The suspension X e of (X, h) has the following attractive properties.

If X is compact metric then X is compact metric, cf. [24, Corollary II.5.9]. • e If h is minimal then X is connected, cf. remark on page 205 of [3]. • e If X is has (covering) dimension n then X has dimension n + 1, because X is • e e locally the product of X with the 1-dimensional space R.

In this thesis we study the crossed product C(X) oϕα Z induced by the dynamical e system (X, ϕα). As a motivation, observe that when h is the identity map on the e 1 one point space X, the dynamical system (X, ϕα) is conjugate to (S ,Rα) where e 1 1 S = R/Z is the unit circle and Rα is the rigid rotation x x + α on S . Thus ∼ 7→ when α is irrational, the crossed product C(X) oϕα Z is isomorphic to the irrational e rotation algebra Aα, cf. [23]. Therefore one of our main results, Theorem I.18, gives a diﬀerent formula to compute the range of the trace on the K0 group of irrational rotation algebras, a result originally obtained by Rieﬀel, Pimsner and Voiculescu

[23, 21]. Here we must mention that our formula for the range of the trace on K0

of C(X) oϕα Z builds upon the work of [6] and hence it works for more general e 4

(noncommutative) C*-algebras.

It has been proved by Giordano, Putnam and Skau [9, Theorem 2.1] that the

(topological) orbit structure of Cantor minimal systems is related to the isomorphism class of the associated crossed products. More precisely, they showed that K-theory yields complete information about the orbit structure of these dynamical systems and combined their discovery with the fact that simple direct limits of circle algebras with real rank zero and with the same scaled ordered K-theory are necessarily isomorphic [5]. For a larger class of minimal dynamical systems, it is natural to ask whether a similar result is true provided we assume the Elliott conjecture. The Elliott conjecture states that a complete isomorphism invariant for the associated (simple) crossed products is of (ordered) K-theoretic nature. Such invariants are known as the Elliott invariants. In [10], an example is given of a Cantor minimal system and a non-homogeneous system which are not strong orbit equivalent but have the same

Elliott invariants. More examples are known by now of dynamical systems which are not strong orbit equivalent but have the same Elliott invariants (for a survey, see

[19]). These examples involve spaces which are not connected or have dimension at least 2. A well known result about the circle states that two minimal homeomorphism of the circle are ﬂip conjugate if and only if their associated crossed products are isomorphic. However, dynamical systems on 1-dimensional connected compact metric spaces which are not homeomorphic to the circle have not been studied yet in this context. As a result of our study, we have succeeded in ﬁnding non orbit equiv- 5 alent minimal dynamical systems on connected compact metric 1-dimensional spaces arising from time α maps, for the same α, of suspensions of strong orbit equivalent

Cantor systems. We expect these dynamical systems to have the same Elliott invariants. Our example relies in the fact proved in [11] that there are two Toeplitz flows with different entropy which are simultaneously the maximal equicontinuous factors of and strong orbit equivalent to the 2-odometer. The existence of α R such that ∈ the time α map on the suspension of those Toeplitz flows are minimal gives us two minimal dynamical systems on connected compact metric 1-dimensional spaces which are not orbit equivalent because they have different topological entropy. If those time

α maps were in addition uniquely ergodic, our Theorem I.18 would give us that the

Elliott invariants of their associated crossed products are the same. This would prove the existence of non orbit equivalent dynamical systems on connected compact metric

1-dimensional spaces having associated crossed products with the same Elliott invariants. Hence our study suggests that conditions to extend [9, Theorem 2.1] to more general minimal dynamical systems might need to take into account the entropy of the systems involved.

We have divided this work into 5 chapters. Chapter 1 will give a formula to

compute the range of a trace on the K0 group of C(X) oϕα Z. Chapters 2 and 3 e will analyze conditions to ensure the existence of a number α such that the time

α map on the suspension of a dynamical system is minimal and uniquely ergodic, respectively. In Chapter 4 we will provide a formula to compute the entropy of a time 6

α map on the suspension of a dynamical system (X, h) in terms of the entropy of h.

Finally, Chapter 5 will present several examples derived from our study, including the

existence of non orbit equivalent minimal dynamical systems on connected compact

metric 1-dimensional spaces arising from time α maps, for the same α, of suspensions

of strong orbit equivalent Cantor systems.

We now want to explain some terminology and introduce some notation. We say

that a dynamical system (x, h) (or just h) is minimal if there is no nontrivial closed

h invariant subset of X or equivalently if for all x in X the orbit hn(x): n Z { ∈ } of x is dense in X. We say that a dynamical system (X, h) (or just h) is uniquely

ergodic if there is only one h invariant Borel probability measure on X. If (X1, h1)

and (X2, h2) are two dynamical systems, we say that (X1, h1) is conjugate to (X2, h2)

if there exists a homeomorphism F : X1 X2 such that F h1 = h2 F . We say that → ◦ ◦

(X1, h1) is ﬂip conjugate to (X2, h2) if (X1, h1) is conjugate either to (X2, h2) or to

1 (X2, h2− ). The dynamical system (X1, h1) is orbit equivalent to (X2, h2) if there is a

n homeomorphism F : X1 X2, called an orbit map, such that F ( h (x): n Z ) = → { 1 ∈ } n h (F (x)): n Z for all x X1. If F : X1 X2 is an orbit map then for each { 2 ∈ } ∈ → n(x) x X1 there is an integer n(x) such that F (h1(x)) = h (F (x)). Likewise there ∈ 2 m(x) exists an integer m(x) such that F (h1 (x)) = h2(F (x)). We call n and m the orbit cocycles associated to the orbit map F . When Xi is inﬁnite and hi is minimal, for i = 1, 2, the orbit cocycles m and n are uniquely deﬁned integer-valued functions on

X1. (In general they are not.) Two minimal dynamical systems (X1, h1) and (X2, h2) 7

are strong orbit equivalent if they are orbit equivalent and there is an orbit map

F : X1 X2 such that their associated orbit cocycles m, n: X1 Z have at most → → one point of discontinuity.

If (X, h) is a minimal dynamical system then the corresponding crossed product

C(X) oh Z is simple. If (X, h) is a uniquely ergodic dynamical system then the

corresponding crossed product C(X) oh Z has a unique normalized trace.

A trace τ of a C*-algebra A is a linear map A C satisfying τ(ab) = τ(ba) and →

τ(a∗) = τ(a) for all a, b A. A trace τ of a C*-algebra A extends to the C*-algebra ∈ n M (A) of n n matrices with entries in A by the formula τ(x) = τ(x ) for all n X i,i × i=1

x Mn(A). A trace τ on A induces a group homomorphism, denoted again by τ, ∈

from K0(A) to R by the formula τ([p] [q]) = τ(p) τ(q). A trace τ on a unital − − C*-algebra A is said to be normalized if τ(1) = 1.

The crossed product A oh Z induced by an action h of Z on a unital C*-algebra

A is the universal C*-algebra generated by a copy of A and a unitary u satisfying

n n n u au− = hn(a) for all a A and n Z. It follows that elements of the form anu ∈ ∈ X n Z ∈ with an A for each n Z satisfying an < form a dense subalgebra of AohZ. ∈ ∈ X k k ∞ n Z ∈ If τ is a trace on A, invariant under the action h, it extends to a trace on the dense

n subalgebra of A oh Z just mentioned by the formula anu τ(a0). This formula X 7→ n Z ∈ extends continuously to the whole A oh Z. We will denote this new trace again by τ.

If A is a separable, unital, simple C*-algebra with unique normalized trace, the

Elliott invariant of A consists of three elements: the abstract group K1(A), the scaled 8

ordered group K0(A) (with scale [1] and order deﬁned by x > 0 if and only if there is n N and a projection p Mn(A) such that x = [p]) and the the map K0(A) R ∈ ∈ → induced by the unique normalized trace.

Let A and B be two separable, unital simple C*-algebras. Assume that A and B have unique normalized traces τA and τB, respectively. We say that A and B have the same (or isomorphic) Elliott invariants if the groups K1(A) and K1(B) are isomorphic and there exists an (order and scale preserving) isomorphism ϕ: K0(A) K0(B) such → that τB ϕ = τA. ◦ If s is a real number we denote by [s] and s the integer and fractional parts of s, { } respectively. We will use the symbols C, R, Q and Z to denote the set of complex, real, rational and integer numbers, respectively. The set of complex numbers of absolute value 1 will be denoted by S1 or by T when we wish to highlight the group structure of

1 S . We will use the symbols Z+ and R+ to denote the set of strictly positive integers and strictly positive reals, respectively. If X is a metric space with a metric d, we will denote the (open) ball with center at x and radius by B(x), that is,

B(x) = y X : d(x, y) < . { ∈ } 9

CHAPTER I

RANGE OF TRACE

Let (X, h) be a dynamical system and let (X, ϕ) be the suspension ﬂow associated e to (X, h). Fix α R and assume that the time α map ϕα is minimal and uniquely ∈ ergodic. Let τ be the trace of C(X) associated to the unique invariant measure on e e X. The main goal of this chapter will be to ﬁnd the Elliott invariants of the crossed e product C(X) oϕα Z. The hard work will be to provide a formula for the range of e

τ on the K0 group of C(X) oϕα Z. There is a number of results in the literature e e which give formulas to compute the range of the trace on the K0 group of crossed products (see e.g. [20], [14] and [6]). However, we feel that the formula we provide in this chapter is more specialized and therefore it is more useful for applications. Our formula will rely on the work of [6] and so what we get is something more general than what is needed.

We begin by providing some deﬁnitions and notation, most of which are taken from [6]. Let A be a unital C*-algebra, let τ be a trace on A and let h: A A be → an automorphism. We say that τ is h invariant (or that h preserves τ) if τ(h(a)) =

τ(a) for all a A. An automorphism h: A A on a C*-algebra A induces an ∈ → automorphism on Mn(A) by the formula h(x) = (h(xij)). We denote by U (A) the ∞ 10 usual inductive limit of the sequence of groups

U (A) / U (A) / / U (A) / U (A) / 1 2 ··· n n+1 ···

where Un(A) is the subgroup of Mn(A) consisting of unitaries, i.e. Un(A) = x = {

(xij) Mn(A):(xji)(xij) = x∗x = 1 = xx∗ = (xij)(xji) , and the map Un(A) ∈ } → x 0 Un+1(A) is given by x . 7→ 0 1

Deﬁnition I.1. Let A be a unital C*-algebra and let τ be a trace on A. We say that a group homomorphism

det: U (A) T ∞ → is a determinant with trace τ if for all selfadjoint a Mn(A) ∈

det(eia) = eiτ(a).

Deﬁnition I.2. Let A be a unital C*-algebra and let τ be a trace on A. We say that

(A, τ) is integral if τ(K0(A)) Z. ⊂

We know that A admits a determinant associated with τ if and only if (A, τ) is integral, cf. [6, Theorem II.10].

Deﬁnition I.3. Let A be a unital C*-algebra and let h: A A be an automorphism. → h We denote by K1(A) the subgroup of ﬁxed points for the action of h on K1(A), i.e.

h K1(A) = x K1(A): h (x) = x { ∈ ∗ } where h (x) = [h(u)] for u M (A) such that x = [u]. ∗ ∈ ∞ 11

Deﬁnition I.4. Suppose that (A, τ) is an integral C*-algebra and let h: A A be → a trace preserving automorphism. The rotation number map of h with respect to the trace τ is the group homomorphism

τ h ρ : K1(A) T h → deﬁned by

τ ρh(x) = det(h(u∗)u)

h for x K1(A) , where u U (A) is such that x = [u] and det is a determinant for ∈ ∈ ∞ A with trace τ.

τ Proposition IV.2 in [6] shows that there is no ambiguity in the definition of ρh, that is, it is in fact a well defined group homomorphism and it does not depend on the determinant used in its definition.

Deﬁnition I.5. Let A be a unital C*-algebra and let h: A A be an automorphism. → The suspension or mapping torus of the pair (A, h) (or just of h) is the (unital) C*- algebra

Th(A) = f C(R,A): f(s + 1) = h(f(s)) . { ∈ }

The suspension of A is the C*-algebra

SA = f C([0, 1],A): f(0) = 0 = f(1) . { ∈ }

Observe that if f is in Th(A) then f restricted to [0, 1] gives a continuous function [0, 1] A satisfying f(1) = h(f(0)). Conversely, given a continuous function → 12 f : [0, 1] A satisfying f(1) = h(f(0)), we may extend f uniquely to an element in → n Th(A) by the formula f(t) = h (f(t n)) for t [n, n + 1] and n Z. With this − ∈ ∈ in mind, it follows that SA is the subalgebra of Th(A) consisting of all elements f in Th(A) such that f(0) = 0. The following result is referred as the “mapping torus exact sequence” in [7]. It is also used in [15]. We state it here as we need it in the sequel.

Lemma I.6. Let SA and Th(A) be as above. Let ev0 : Th(A) A be given by → ev0(f) = f(0). There is a short exact sequence

j ev0 0 / SA / Th(A) / A / 0 where j is the inclusion map.

1 Proof. Given a A, the map R t th(a) + (1 t)a is in ev0− (a). Hence ev0 ∈ 3 7→ − is onto. It remains to check that SA = ker(ev0) but this is clear by the remark above.

Deﬁnition I.7. Let (A, τ) be a unital traced C*-algebra and let h: A A be a trace → preserving automorphism of A. We deﬁne a trace τ on Th(A) by e τ(f) = Z (τ f)(t) dλ(t) [0,1] ◦ e for f Th(A), where λ is the Lebesgue measure on R. ∈

Definition I.8. Let (A, τ) be a unital traced C*-algebra and let h: A A be a trace → preserving automorphism of A. Let τ be as in Definition I.7 and let α R. We define ∈ e 13

ϕα to be the τ preserving automorphism on Th(A) given by e

(ϕα(f))(t) = f(t α) − for f Th(A) and t R. ∈ ∈

Lemma I.9. Let (A, τ) be a unital traced C*-algebra and let h be a trace preserving automorphism of A. Suppose that (Th(A), τ) is integral. Let α R. Then ∈ e [α] ρτ ρτ if [α] 0 ϕ α ϕ1 τ  e{ } e ≥ ρϕ =  eα  [α] ρτ ρτ − if [α] 0.  ϕ α ϕ 1  e{ } e− ≤  [α] [α] Proof. Observe that ϕα = ϕ α (ϕ1) if [α] 0 and ϕα = ϕ α (ϕ 1)− if [α] 0. { } ◦ ≥ { } ◦ − ≤ Since the time α , 1 and 1 maps are homotopic to the identity, their rotation { } − number maps all have the same domain K1(A). Thus our lemma follows from IV.3 and IV.4 in [6].

What Lemma I.9 says is that if we want to compute the rotation number map of

ϕα with respect τ, we only need to do it when α 1. We take care of this case in | | ≤ e the next lemma.

Lemma I.10. Under the hypothesis of previous lemma, let p Mm(A) be a projec- ∈ tion. Let u Um(Th(A)) be given by the formula ∈

u(t) = e2πitp for t [0, 1]. Let α 1. Then ∈ | | ≤

τ ρϕ ([u]) = exp( 2πiατ(p)). eα − 14

τ τ Proof. By using the technique in the proof of Lemma I.9, we see that ρϕα ρϕ α = 1. e e− Therefore it will suﬃce to prove the lemma for 1 α 0. Consider the function − ≤ ≤ u : [0, 1] Th(A) given by → e exp (2πi[(t 1)(1 s)p + s(t α 1)p]) if t α [0, 1]  − − − − − ∈ (u(s))(t) =  e exp (2πi[(t 1)(1 s)p + s(t α 1)h(p)]) if t α [1, 2]  − − − − − ∈  for s, t [0, 1]. For each s [0, 1], the function u(s) is indeed in Th(A) since it is ∈ ∈ e continuous and

(u(s))(1) = exp(2πi( αsh(p))) − e = h(exp(2πi( αsp))) − = h(exp(2πi [( 1)(1 s)p + s( α 1)p])) − − − − = h((u(s))(0)). e

Furthermore u is C∞ and since e

ϕα(u(0)) = u(1) e e we obtain that u Tϕ (Th(A)). Combining Theorems V.7 and V.12 in [6] we obtain ∈ α e a commutative diagram

ϕα k1 τ K1(Tϕα (Th(A))) / K0(Th(A) oϕα Z) / R SS e SSS (ev0) SSS ∗ ∂ q SSS SSS τ ) ρϕα K1(Th(A)) e / T

where q is the map R t exp(2πit) T denoted by π in [6, Theorem V.12]. Since 3 7→ ∈ 15

ev0(u) = u(0) = u we have e e

τ τ ρϕ ([u]) = ρϕ ((ev0) ([u])) eα eα ∗ e = (q τ kϕα )([u]). ◦ ◦ 1 e e

ϕα We compute τ(k1 (u)) by using [6, Theorem V.11]. First, deﬁne a function f : e e [0, 1] A by →

 (t 1)p + (t α 1)p if t α [0, 1] − − − − − ∈ f(t) =  (t 1)p + (t α 1)h(p) if t α [1, 2]. − − − − − ∈ 

We claim that τ (u0(s)∗u(s)) = τ(f). Put a(t) = (t 1)p and rewrite u as − e e e e e (u(s))(t) = exp(2πi[a(t) + sf(t)]). e Use the power series expansion for exp to get

du d ∞ (2πi)k (s) (t) = [a(t) + sf(t)]k ds ds X k! e k=0 ∞ (2πi)k d = [a(t) + sf(t)]k X k! ds k=0 k 1 ∞ (2πi)k − = [a(t) + sf(t)]k 1 jf(t)[a(t) + sf(t)]j. X k! X − − k=1 j=0 16

Hence

k k 1 ∞ ( 2πi) − j k 1 j (u0(s)∗u(s)) (t) = − [a(t) + sf(t)] f(t)[a(t) + sf(t)] − − ! X k! X e e k=1 j=0 ∞ (2πi)l [a(t) + sf(t)]l! · X l! l=0 l 1 l k k l k 1 ∞ − ( 2πi) − (2πi) − − j l 1 j = − [a(t) + sf(t)] f(t)[a(t) + sf(t)] − − X X (l k)! k! X l=1 k=0 − j=0 l 1 l k l k 1 ∞ l − ( 1) − − − j l 1 j = (2πi) − [a(t) + sf(t)] f(t)[a(t) + sf(t)] − − X X (l k)! k! X l=1 k=0 − j=0 l l 1 l k 1 ∞ (2πi) − l k l − − j l 1 j = ( 1) − [a(t) + sf(t)] f(t)[a(t) + sf(t)] − − . X l! X − k X l=1 k=0 j=0 Thus

l l 1 ∞ (2πi) − l k l l 1 τ ((u0(s)∗u(s)) (t)) = ( 1) − (l k)τ(f(t)[a(t) + sf(t)] − ) X l! X − k − e e e l=1 k=0 e l l 1 ∞ (2πi) l 1 − l k l 1 = τ(f(t)[a(t) + sf(t)] − ) ( 1) − − X (l 1)! X − k l=1 − e k=0 = 2πiτ(f(t)) − e as wanted. So, using the h invariance of the trace τ, we get

1 ϕα 1 τ(k1 ([u])) = Z τ (u0(s)∗u(s)) ds 2πi 0 e e 2πi e e e = − τ (f) 2πi 1 e = Z τ(f(t)) dt − 0 1+α 1 = Z ατ(p) dt + Z ατ(p) dt − 0 1+α = ατ(p). −

τ Thus ρϕ ([u]) = exp( 2πiατ(p)), as wanted. eα − 17

Proposition I.11. Let (A, τ) be a unital traced C*-algebra and let h be a trace pre-

serving automorphism of A. Suppose that (Th(A), τ) is integral. Let p Mm(A) be a ∈ e projection and let u Um(Th(A)) be given by the formula ∈

u(t) = e2πitp

with t [0, 1]. Then for all α R ∈ ∈

τ ρϕ ([u]) = exp( 2πiατ(p)). eα −

Proof. Immediate from Lemmas I.9 and I.10.

j Lemma I.12. If K1(A) = 0 then the inclusion SA Th(A) induces a surjective −→ j homomorphism K1(SA) ∗ K1(Th(A)). In consequence, elements in K1(Th(A)) can −→ be represented as products of unitaries of the form

u(t) = e2πitp

where p Mn(A) is a projection and n 1. ∈ ≥

Proof. This follows from the short exact sequence in Lemma I.6 and the Bott peri-

odicity K0(A) ∼= K1(SA).

Theorem I.13. Let (A, τ) be a unital traced C*-algebra and let h be a trace preserving

automorphism of A. Suppose that (Th(A), τ) is an integral C*-algebra and K1(A) = 0. e Let α R. Then the following diagram is commutative ∈

τ ρϕα K1(Th(A)) e / T O h α k1 q τ K0(A oh Z) / R 18

where qα denotes the map R t e2πiαt T and kh is the isomorphism denoted by 3 7→ ∈ 1

k1 in [6, Theorem V.3].

Proof. By Lemma I.12, it suﬃces to prove the theorem for elements in K1(Th(A))

represented by unitaries of the form

u(t) = e2πitp

where p Mn(A) is a projection and n 1. On the one hand Proposition I.11 says ∈ ≥ that

τ 2πiατ(p) ρϕ ([u]) = e− . (I.1) eα

On the other hand, using [6, Theorem V.11] we compute

1 h 1 τ k1 ([u]) = Z τ(u0(t)∗u(t)) dt. ◦ 2πi 0 1 1 = Z τ (u(t)∗( 2πip)u(t)) dt 2πi 0 − = τ(p). −

Composing the last equation with qα and comparing the result with I.1 completes the

proof.

Corollary I.14. Assume the hypothesis of Theorem I.13. Then the range of the trace

τ on the K0 group of Th(A) oϕα Z is e

Z + ατ(K0(A oh Z)).

τ Proof. As ϕα is homotopic to the identity, ρϕ has domain the whole of K1(A). eα

Therefore, given x K0(Th(A)oϕ Z), we may combine the diagrams in Theorem I.13 ∈ α 19

and [6, Theorem V.12] to obtain

exp(2πi[τ(x) ατ(kh(∂(x)))]) = 1. (I.2) − 1 e

Thus τ(K0(Th(A) oϕ Z)) Z + ατ(K0(A oh Z)). For the reverse inclusion, observe α ⊂ e that Z τ(K0(Th(A) oϕ Z)) since τ(1) = 1. It remains to prove the inclusion ⊂ α e e ατ(K0(A oh Z)) τ(K0(Th(A) oϕ Z)). For this purpose, let y be an arbitrary ⊂ α e h element in K0(A oh Z). Since k1 is an isomorphism (cf. [6, Theorem V.3]) and

∂ : K0(Th(A) oϕ Z) K1(Th(A)) is surjective (cf. the begining of the proof of [6, α → h Theorem V.12]) it follows that k ∂ is surjective. Then there is x K0(Th(A)oϕ Z) 1 ◦ ∈ α such that kh(∂(x)) = y. Using (I.2) we get that there is k Z such that τ(x) ατ(y) = 1 ∈ − e k and so τ(x k1) = ατ(y). Thus ατ(K0(A oh Z)) τ(K0(Th(A) oϕ Z)), as was to − ⊂ α e e be proved.

We now return to the case of commutative C*-algebras. Let (X, h) be a dynamical

system and let (X, ϕ) be the suspension ﬂow of (X, h). Let α R. Let π : X R X ∈ × → e e 1 be the canonical quotient map. Consider the suspension T 1 (C(X)) of (C(X), h ) h− −

(see Deﬁnition I.5). We will regard T 1 (C(X)) as a subalgebra of C(X R). Let h− ×

ϕ be the automorphism on T 1 (C(X)) as in Definition I.8. At first there seems to α h− be a conflict in our notation, as the automorphism on C(X) induced by the time α e map ϕα : X X is also denoted by ϕα (and recall is defined by (ϕα(f))([x, s]) = → e e f(ϕ α([x, s])) = f([x, s α]) for f C(X) and [x, s] X). The following results will − − ∈ ∈ e e tell us that there is virtually no distinction, either between T 1 (C(X)) and C(X), or h− e 20

between the deﬁnitions of their ϕα automorphisms. We remark that this observation

is known as it was noticed in [3].

Lemma I.15. Let (X, h), (X, ϕα) and π : X R X be as above. Let π : C(X) × → ∗ → e e e C(X R) be the map (induced by π) deﬁned by f f π. We have the following. × 7→ ◦

1. Th 1 (C(X)) = Im π . − ∗

2. π : C(X) Th 1 (C(X)) is an isomorphism. ∗ → − e 3. The following diagram is commutative.

ϕα C(X) / C(X)

π e eπ ∗ ∗ ϕα T 1 (C(X)) / T 1 (C(X)) h− h−

where the maps involved are all isomorphisms.

Proof. We have

f T 1 (C(X)) f(h(x), s) = f(x, s + 1) x X s R ∈ h− ⇒ ∀ ∈ ∀ ∈ g : X C deﬁned by g([x, t]) = f(x, t) is continuous ⇒ → e f = g π Im π ⇒ ◦ ∈ ∗ and

f Im π there is g C(X) such that f = g π ∈ ∗ ⇒ ∈ ◦ e

f T 1 (C(X)). ⇒ ∈ h− 21

Thus (1) follows. Part (2) is a consequence of (1) and the fact that π is one-to-one. ∗ Part (3) follows from the following computation.

((π ϕα)(f))(x, t) = f([x, t α]) = ϕα(f[x, t]) = (ϕα π (f))(x, t) ∗ ◦ − ◦ ∗

for f C(X) and (x, t) X R. ∈ ∈ × e

Assume now that µ is an h invariant Borel probability measure on X. Let τµ be

the trace induced by µ on C(X). Then h preserves τµ. As noticed at the beginning

of Section 1 in [14], µ induces an R invariant (in particular, ϕα invariant) Borel

probability measure µ on X by the formula e e

1 E Z (χE π)(x, s) d(µ λ)(x, s) = (µ λ)(π− (E) (X [0, 1])) 7→ X [0,1] ◦ × × ∩ × × where λ is the Lebesgue measure on R. Let us denote by τµ the trace induced by µ e e on C(X). Let τ be the trace on T 1 (C(X)) corresponding to Deﬁnition I.7. h− e e Recall that we have agreed to denote again by τµ, τµ and τ the dual traces induced e e by τ , τ and τ on the crossed products C(X)o Z, C(X)o Z and T 1 (C(X))o Z, µ µ h ϕα h− ϕα e e e respectively.

Lemma I.16. Let τµ and τ be the traces described above. Then the diagram e e

τµ C(X) e / C t: tt e τ tt π tt ∗ tte tt T 1 (C(X)) h−

is commutative. In particular 22

1. τ (K (C(X))) = τ(K (T 1 (C(X)))) and µ 0 0 h− e e e

2. τ (K (C(X) o Z)) = τ(K (T 1 (C(X)) o Z)). µ 0 ϕα 0 h− ϕα e e e Proof. We have that

τ(π (f)) = τ(f π) ∗ ◦ e e = Z τ(f π( , t)) dλ(t) [0,1] ◦ ·

= Z Z f π(x, t) dµ(x) dλ(t) [0,1] X ◦

= Z f π(x, t) d(µ λ)(x, t) X [0,1] ◦ × ×

= τµ(f) e

for all f C(X). Then we conclude that τµ = τ π as wanted. Since by Lemma ∈ ◦ ∗ e e e I.15 the function π is an isomorphism onto Th 1 (C(X)), it follows that the images ∗ − of τµ and τ coincide. Hence the last two assertions of the lemma follow. e e

Proposition I.17. Let (X, h), (X, ϕ ) and (T 1 (C(X)), ϕ ) be the dynamical sys- α h− α e tems with traces τµ, τµ and τ, as described above. Then each of the following state- e e ments implies the next.

1. (X, h) is minimal.

2. (C(X), τµ) is integral. e e

3. (C(X), τ ) and (T 1 (C(X)), τ) are integral. µ h− e e e 23

Proof. If (X, h) is minimal then X is connected. Therefore (C(X), τµ) is integral e e e (cf. [6, Proposition VI.1]) and so (T 1 (C(X)), τ) is also integral by previous the h− e lemma.

We are ready to state one of the main results of this dissertation.

Theorem I.18. Let (X, h) be a dynamical system and let α be a real number. Con-

sider the dynamical system (X, ϕα) where X is the suspension of (X, h) and ϕα is e e the time α map on X. Let µ be an h invariant Borel probability measure on X and e let µ be the induced ϕα invariant Borel probability measure on X. Let τµ and τµ be e e e the traces induced by µ and µ, respectively. Suppose that (C(X), τµ) is integral and e e e K1(X) = 0. We have the following.

1. For i = 0, 1, Ki(C(X) oϕα Z) is isomorphic to e

(a) K0(C(X)) K1(C(X)) and ⊕ e e

(b) K1(C(X) oh Z) K0(C(X) oh Z). ⊕

2. The image of τµ on K0(C(X) oϕα Z) is equal to e e

Z + ατµ(K0(C(X) oh Z)).

3. Furthermore, if X is ﬁnite dimensional, ϕα is both minimal and uniquely er-

godic, τµ is injective on K0(C(X) oh Z) and Z ατµ(K0(C(X) oh Z)) = 0 ∩ { }

then τµ induces an order isomorphism e

K0(C(X) oϕ Z) Z + ατµ(K0(C(X) oh Z)) α → e 24

where the order of the right hand side is inherited from R.

Proof. Since ϕα is homotopic to the identity, we may use the corollary in [15] (or [17,

Proposition 3.2]) and standard methods for crossed products and their K-theory to

obtain

K (C(X) o Z) = K (C(X) o Z) 0 ϕα ∼ 0 Id e e 1 = K0(C(X) C(S )) ∼ ⊗ e

= K0(C(X)) K1(C(X)). ∼ ⊕ e e

Furthermore, Ki(C(X)) = Ki 1(C(X) oh Z) by Lemma I.16 and the isomorphisms ∼ − e in [6, Theorem V.3]. Hence part (1) of the theorem follows. For part (2) we combine

Corollary I.14 with Lemma I.16 to obtain

τ (K (C(X) o Z)) = τ(K (T 1 (C(X)) o Z)) µ 0 ϕα 0 h− ϕα e e e = Z + ατ (K (C(X) o 1 Z)) µ 0 h−

= Z + ατµ(K0(C(X) oh Z))

as wanted. Part (3) follows from [18, Theorem 4.5]. 25

CHAPTER II

MINIMALITY

Let (X, h) be a dynamical system and let (X, ϕ) be the suspension ﬂow associated e to (X, h). Observe that the minimality of h is a necessary condition for the existence of a minimal time α map ϕα. (If M is a nontrivial closed h invariant subset of X, then π(M R) would be a nontrivial closed ϕα invariant subset of X.) In this chapter × e we show that if we require in addition to the minimality of h the minimality of hk for some k > 1, then not only there exists α R so that ϕα is minimal, but the set ∈ consisting of all such α is a dense Gδ set in [0, 1].

Lemma II.1. Let (X, d) be a compact metric space and let h: X X be a homeo- → morphism. Suppose that there exists an open set U X and a compact set K X ⊂ ⊂ such that

K h(U). ⊂

There exists > 0 such that if g : X X is another homeomorphism with →

sup d(g(x), h(x)) < , x X ∈ then

K g(U). ⊂ 26

Proof. Let

= inf d(x, y). x K y X∈h(U) ∈ \ Since X is metric, both K and X h(U) are compact and K (X h(U)) = , it follows \ ∩ \ ∅ that > 0. Let g : X X be another homeomorphism for which sup d(g(x), h(x)) < → x X ∈ . We must show that K g(U). For this purpose, let x K. The deﬁnition of ⊂ ∈ 1 implies B(x) h(U) and so gh− (B(x)) g(U). To complete the proof we will ⊂ ⊂ 1 1 show that x gh− (B(x)) or equivalently hg− (x) B(x). This follows because ∈ ∈

1 1 1 d(x, hg− (x)) = d(g(g− (x)), h(g− (x))) sup d(g(x), h(x)) < . ≤ x X ∈

Lemma II.2. Let X be a compact Hausdorﬀ space. If K X is compact and U0,U1 ⊂ are open sets in X such that

K U0 U1, ⊂ ∪

then there are compact sets K0 and K1 such that

K = K0 K1 ∪

and Ki Ui for i = 0, 1. ⊂

Proof. Put Yi = K Ui. Then Yi U1 i for i = 0, 1, and Y0 Y1 = . Since X is \ ⊂ − ∩ ∅ normal, there are open sets Vi X such that Yi Vi for i = 0, 1 and V0 V1 = . ⊂ ⊂ ∩ ∅

Then Ki = K Vi is compact, Ki Ui for i = 0, 1 and K0 K1 = K (V0 V1) = K \ ⊂ ∪ \ ∩ as wanted. 27

Lemma II.3. Let X be a compact Hausdorﬀ space. If K X is compact and ⊂

U0,U1,..., Un are open sets such that

K U0 U1 Un ⊂ ∪ ∪ · · · ∪

then there are compact sets K0,K1,...Kn such that

K = K0 K1 Kn ∪ ∪ · · · ∪

and Ki Ui for i = 0, 1, . . . n. ⊂

Proof. By Lemma II.2, there are compact sets K0 and K0 such that K = K0 K0 1 ∪ 1

and K0 U0, K0 U1 U2 Un. ⊂ 1 ⊂ ∪ ∪ · · · ∪

Using Lemma II.2 again for K0 , there are K1 and K0 such that K0 = K1 K0 and 1 2 1 ∪ 2

K1 U1, K0 U2 U3 Un. Repeating this argument we verify the lemma. ⊂ 2 ⊂ ∪ ∪ · · · ∪

Lemma II.4. Let (X, d) be a compact metric space and let h : X X be a home- → omorphism. Suppose that there is a nonnegative integer n and an open set U X ⊂ such that n X = hi(U). [ i=0 There exists > 0 such that if g : X X is another homeomorphism and →

sup d(g(x), h(x)) < x X ∈

then n X = gi(U). [ i=0 28

Proof. To avoid triviality, assume that n Z+. Since X is compact, the sets ∈ n U, h(U), . . . , hn(U) are open and X = hi(U), Lemma II.3 yields existence of com- [ i=0

pact sets K0,K1,...,Kn such that

n X = K [ i i=0

and

i Ki h (U) ⊂

for i = 0, 1, . . . , n. Let now i 0, 1, . . . n . We use Lemma II.1 to ﬁnd a pos- ∈ { }

itive real number i such that if g : X X is another homeomorphism with → i sup d(g(x), h (x)) < i then Ki g(U). Put 0 = min 0, 1, . . . , n . Furthermore, x X ⊂ { } ∈ i i i since h is uniformly continuous, there exists δi > 0 such that d(h (x), h (y)) < 0/n whenever d(x, y) < δi.

Let = min 0/n, δ2, δ3, . . . , δn . Let g : X X be another homeomorphism with { } → n i i sup d(g(x), h(x)) < . To show that X = g (U), it will suﬃce to prove Ki g (U) x X [ ⊂ ∈ i=0 for each i = 0, 1, . . . , n. We have K0 U by construction. We prove by induction on ⊂ i i 1 < i n that sup d(g (x), h (x)) < i0/n. When i = 1 we have sup d(g(x), h(x)) < ≤ x X x X ∈ ∈ i 1 i 1 0/n. Let i 2, 3, . . . , n . Assume that sup d(g − (x), h − (x)) < (i 1)0/n. ≤ ∈ { } x X − ∈ Let x0 X. Since ∈

d(g(x0), h(x0)) sup d(g(x), h(x)) < δi 1, ≤ x X ≤ − ∈ 29

i 1 i 1 i 1 the uniform continuity of h − gives d(h − (g(x0)), h − (h(x0))) < 0/n. So

i i i 1 i 1 i 1 i 1 d(g (x0), h (x0)) d(g − (g(x0)), h − (g(x0))) + d(h − (g(x0)), h − (h(x0))) ≤

i 1 i 1 < sup d(g − (x), h − (x)) + 0/n x X ∈

< i0/n.

i i i Thus sup d(g (x), h (x)) < i0/n i and so Ki g (U), as was to be proved. x X ≤ ⊂ ∈

Lemma II.5. Let X be a compact metric space and let Ui i Z+ be a countable basis { } ∈ for the topology of X. The set

h Homeo(X): h is minimal { ∈ } is equal to

h Homeo(X): there exists n Z such that X = n hj(U ) . \ n + j=0 i o i 1 ∈ ∈ S ≥ Proof. The proof of Proposition 4.1 in [8] shows that the set

h Homeo(X): h is minimal { ∈ } is equal to

ω \ Ui i 1 ≥ where

n j ωU = f Homeo(X): there exists n Z+ such that X = h (Ui) i n ∈ ∈ Sj=0 o

The lemma follows. 30

Deﬁnition II.6. Let (X, ϕ) be a continuous ﬂow where X is a compact metric space.

For U X open deﬁne the set V (U) [0, 1] by ⊂ ⊂

n j V (U) = α [0, 1]: there exists n Z+ such that X = ϕ (U) . n ∈ ∈ Sj=0 α o

Lemma II.7. Let (X, ϕ) be a continuous ﬂow where X is a compact metric space.

Let Ui i Z+ be a countable basis for the topology of X. The set { } ∈

α [0, 1]: ϕα is minimal { ∈ } is equal to

V (U ) \ i i 1 ≥ where V (Ui) is as in Deﬁnition II.6.

Proof. Let α be a number in [0, 1] such that ϕα is minimal. Let i 1. By Lemma II.5 ≥ we obtain

n j ϕα h Homeo(X): there exists n Z+ such that X = h (Ui) . ∈ n ∈ ∈ Sj=0 o n So there exists n Z such that X = ϕi (U ). Hence α V (U ). This proves the + [ α i i ∈ j=0 ∈ inclusion α [0, 1]: ϕ is minimal V (U ). α \ i { ∈ } ⊂ i 1 ≥ To prove the opposite inclusion, let α V (U ). Let U be an open set in X. \ i ∈ i 1 ≥ Then there exists i 1 such that Ui U. Since α V (Ui), there is n Z+ such ≥ ⊂ ∈ ∈ n ∞ that X = ϕj (U ). So we get X = ϕj (U). By [25, Theorem 5.1] we conclude [ α i [ α j=0 j= −∞ that ϕα is minimal, as wanted. 31

Lemma II.8. The set V (U) of Deﬁnition II.6 is open.

Proof. Put

n j T = α R: there exists n Z+ such that X = ϕ (U) . n ∈ ∈ Sj=0 α o

Observe that

V (U) = T [0, 1]. ∩

Hence it is equivalent to show that T is open in R. Let α T . Then there exists n ∈ ∈ n Z such that X = ϕj (U). Let d be a metric on X. By Lemma II.4, there exists > + [ α j=0

0 such that if g : X X is another homeomorphism and sup d(g(x), ϕα(x)) < then → x X n ∈ X = gj(U). Since the restriction of ϕ to X [α , α + ] is uniformly continuous, [ j=0 × − there exists 0 < δ < 1 such that d(ϕs(x), ϕt(y)) < /2 whenever max d(x, y), s t < { | − |}

δ with s, t [α , α + ]. Put 0 = min δ, . We claim that B (α) T . Indeed, ∈ − { } 0 ⊂ given x X we have ∈

β B (α) α β < 0 ∈ 0 ⇒ | − |

max d(x, x), α β = α β < 0 δ and α, β [α , α + ] ⇒ { | − |} | − | ≤ ∈ −

d(ϕα(x), ϕβ(x)) < /2. ⇒

n j We have proved that if β B0 (α) then sup d(ϕα(x), ϕβ(x)) < and so X = ϕβ(U). ∈ x X [ ∈ j=0 This proves that β T . Hence B (α) T , as wanted. ∈ 0 ⊂

Lemma II.9. Let N > 0 and let P be a set consisting of inﬁnitely many powers of a 32

single prime number. Then the set

Q = p : p P, q > N and (p, q) = 1 [0, 1] { q ∈ } ∩

is dense in [0, 1].

Proof. Because Q (0, 1) is dense in [0, 1], it will suﬃce to show that Q is dense ∩ in Q (0, 1). Write P = p1, p2, . . . , pn, pn+1,... with pn < pn+1 for n Z+. Let ∩ { } ∈ > 0. Let r Q (0, 1). We will complete the proof once we ﬁnd pk Q such that ∈ ∩ q ∈ pk r < . Let q − p + 1  n pn+1 if pn, = 1,  r r xn =  p + 1 n + 1 otherwise  r  Z for n +. Then xn n Z+ is a strictly increasing sequence of positive real numbers ∈ { } ∈

since pn + r < pn+1 for all n Z+. Furthermore, (pn, [xn]) = 1 for all n 1 because ∈ ≥

the pn’s are powers of a single prime. Then for each n Z+ we have ∈

pn+1 pn+1 xn 1 [xn] xn 1 [xn] + 1 − ≤ ≤ ⇒ r − ≤ ≤ r r pn r . 1+(1+r)/pn [xn] 1+(1 r)/pn ⇒ ≤ ≤ −

Since the elements of P form a strictly increasing sequence, we obtain that the strictly monotone sequences of real numbers at the ends of the last inequality both converge

pk to r. Hence there exists k such that r(N +1) 1 < pk and r < . Put q = [xk] [xk] − − and we are done. 33

Proposition II.10. Let (X, h) be a minimal dynamical system and let k > 1 be an

integer such that hk is also minimal. If p is a prime divisor of k then hpn is minimal

for all n 0. ≥

Proof. Let p be a prime divisor of k. Then hp is minimal. (If M is hp invariant then

M is hk invariant.) Let n > 1. We claim that hpn is minimal. Indeed, if hpn is

not minimal, we use [24, II.9.6(7)] to ﬁnd a clopen set Y X and a divisor l of pn ⊂ 2 l 1 l such that X is the disjoint union of Y, h(Y ), h (Y ), . . . , h − (Y ) and h (Y ) = Y . Note

that l > 1 since Y X is nontrivial because hpn is not minimal. As p must divide ⊂ p 2p l p p l, the set Y h (Y ) h (Y ) h − (Y ) is nontrivial, closed, and h invariant, ∪ ∪ ∪ · · · ∪ contradicting the minimality of hp.

Lemma II.11. Let (X, h) be a dynamical system and let (X, ϕ) be the suspension flow e of (X, h). Let U X be open. Consider the set V (U) as defined in Definition II.6. ⊂ e If h and hk are minimal where k > 1 is some integer then V (U) is dense in [0, 1].

Proof. Choose an open set A X and an interval I = (a, b) R such that W = ⊂ ⊂ π(A I) U, where π : X R X is the canonical quotient map. We may × ⊂ × → e regard A as an element of some countable basis for the topology of X. It is clear that

V (W ) V (U). To prove the lemma we will show that V (W ) is dense. ⊂ By Proposition II.10 there is a set P containing infinitely many powers of a single 1 prime such that hp is minimal for all p P . Let N > > 0 be an integer. ∈ b a − By Lemma II.9, it suffices to show that if p P and q > N satisfies (p, q) = 1 and ∈ 34 p p [0, 1] then V (W ). q ∈ q ∈ p Let p P and let q > N such that (p, q) = 1 and [0, 1]. Since hp is minimal ∈ q ∈ and A is an element of a countable basis, Lemma II.5 gives the existence of some m m Z such that hip(A) = X. Let be the image of I = (a, b) under the + [ ∈ i=0 I canonical quotient map R R/Z. Let Rp/q denote the rigid rotation x x + p/q on → 7→ q 1 1 R/Z. Then Rp/q( ) R − ( ) = R/Z since q > N > and (p, q) = 1. I ∪ I ∪ · · · ∪ p/q I b a n − Let n = mq + q 1. We claim that X = ϕj (W ). Indeed, let [x, s] X with − [ p/q ∈ e j=0 e 0 s < 1. Then s Rk ( ) for some k 0, 1, . . . , q 1 . So s = t+kp/q [t+kp/q] ≤ ∈ p/q I ∈ { − } − [t+kp/q] ip for some t I. Pick i 0, 1, . . . , m such that h− (x) h (A). Then ∈ ∈ { } ∈ [t+kp/q] ip y = h− − (x) A. ∈ Hence

[x, s] = π(x, s)

[t+kp/q] ip = π h− − (x), s + [t + kp/q] + ip = π (y, s + [t + kp/q] + ip)

= π (y, t + kp/q [t + kp/q] + [t + kp/q] + ip) − = π (y, t + kp/q + ip)

= π (y, t + (k + iq)p/q)

k+iq = ϕp/q (π(y, t)) n ϕj (W ) . [ p/q ∈ j=0 35

Proposition II.12. Let (X, h) be a minimal dynamical system and let (X, ϕ) be the e suspension ﬂow of (X, h). Assume that hk is minimal for some k > 1. Then the set

α [0, 1]: ϕα is minimal { ∈ }

is a dense Gδ set.

Proof. Let Ui i Z+ be a countable basis for the topology of X. By Lemma II.7, the { } ∈ e set

α [0, 1]: ϕα is minimal { ∈ } is equal to

V (U ) \ i i 1 ≥ where V (Ui) is as in Deﬁnition II.6. But, for each i 1, the set V (Ui) is open by ≥ Lemma II.8 and dense by Lemma II.11. The proposition now follows. 36

CHAPTER III

UNIQUE ERGODICITY

Let (X, h) be a dynamical system and let (X, ϕ) be the suspension ﬂow associated e to (X, h). Observe that the unique ergodicity of h is a necessary condition for the

existence of a uniquely ergodic time α map ϕα. (If µ and ν are two diﬀerent h invariant measures on X then, following the remark before Lemma I.16, we may construct two diﬀerent ϕα invariant measures µ and ν on X.) It was very pleasant to discover e e e that, just as for minimality (see Chapter II), if we require in addition to the unique ergodicity of h the unique ergodicity of hk for some k > 1, then not only does there exist α R such that ϕα is uniquely ergodic, but the set of such α is a dense Gδ set ∈ in [0, 1].

Lemma III.1. Let X be a compact metric space, let fi i Z+ be a dense set in C(X) { } ∈ and let x0 X be ﬁxed. The set ∈

h Homeo(X): h is uniquely ergodic { ∈ } is equal to

1 n 1 k k h Homeo(X): infn 1 k=0− fi h fi h (x0) = 0 . \ ≥ n i 1 ∈ P ◦ − ◦ ≥ Proof. The proof of Proposition 4.4 in [8] shows that the set

h Homeo(X): h is uniquely ergodic { ∈ } 37

is equal to

E 1(0), \ i− i 1 ≥

where Ei : Homeo(X) R+ is the function deﬁned by →

1 n 1 k k Ei(h) = infn 1 k=0− fi h fi h (x0) . ≥ n ◦ − ◦ P

Hence

1 E− (0) = h Homeo(X): Ei(h) = 0 i { ∈ }

1 n 1 k k = h Homeo(X): infn 1 k=0− fi h fi h (x0) = 0 . ∈ ≥ n P ◦ − ◦

The lemma follows.

Deﬁnition III.2. Let (X, ϕ) be a continuous ﬂow where X is a compact metric

space. For f C(X) and > 0, deﬁne the set V (f, ) [0, 1] by ∈ ⊂

1 n 1 k V (f, ) = α [0, 1]: n Z+ c C such that − f ϕ c < . ∈ ∃ ∈ ∃ ∈ n Pk=0 ◦ α −

Lemma III.3. Let (X, ϕ) be a continuous ﬂow where X is a compact metric space.

Let fi i Z+ be a dense set in C(X). The set { } ∈

α [0, 1]: ϕα is uniquely ergodic { ∈ } is equal to

V (f , 1 ) \ \ i j i 1 j 1 ≥ ≥ 1 where V (fi, j ) is as in Deﬁnition III.2. 38

Proof. Let α be a number in [0, 1] such that ϕα is uniquely ergodic. Let i 1 and ≥

let j 1. Fix x0 X. By Lemma III.1, there exists n > 0 such that ≥ ∈ n 1 1 − 1 k k fi ϕα fi ϕα(x0) < . n X ◦ − ◦ j k=0 n 1 − 1 k 1 Put c = fi ϕ (x0). Then α V fi, . This proves the inclusion α n X ◦ α ∈ j { ∈ k=0 [0, 1]: ϕ is uniquely ergodic V f , 1 . α \ \ i j } ⊂ i 1 j 1 ≥ ≥ For the opposite inclusion, let α V f , 1 , let i 1 and let > 0. Choose \ \ i j ∈ i 1 j 1 ≥ ≥ ≥ 1 1 j such that < . Since α V fi, , there is N > 0 and c C such that j 2 ∈ j ∈ N 1 1 − k fi ϕα c < . N X ◦ − 2 k=0

In particular, for x0 as in Lemma III.1,

N 1 1 − k fi ϕα(x0) c < . N X ◦ − 2 k=0

Thus N 1 1 − k k fi ϕα fi ϕα(x0) < . N X ◦ − ◦ k=0

Therefore n 1 1 − k k inf fi ϕα fi ϕα(x0) = 0. n 1 n X ◦ − ◦ ≥ k=0

Since i was chosen arbitrarily, Lemma III.1 yields unique ergodicity of ϕα.

Lemma III.4. The set V (f, ) of Deﬁnition III.2 is open in [0, 1].

Proof. Put

n 1 1 − ( R Z C j ) T = α : n + c such that f ϕα c < . ∈ ∃ ∈ ∃ ∈ n X ◦ − j=0

Observe that

V (f, ) = T [0, 1]. ∩

Hence it is equivalent to show that T is open in R. For this purpose, let αk be a { }

sequence of real numbers in R T such that αk α as k . Let n Z+ and let \ → → ∞ ∈ n 1 1 − c C. Then for all k, f ϕj c . Taking limits as k , we obtain αk ∈ n X ◦ − ≥ → ∞ j=0 α R T , as was to be proved. ∈ \

The following lemma is the analog of [24, II.9.6(7)] for the uniquely ergodic setting.

We refer the reader to [25] for an explanation of the terminology and notation used

in the proof and not explained here.

Lemma III.5. Let (X, h) be a uniquely ergodic dynamical system with unique h

invariant probability measure µ. Then for all k Z+ there is a divisor l of k and a ∈ 1 l l 1 measurable set E X such that µ(E) = , h (E) = E and E h(E) h − (E) ⊂ l ∪ ∪ · · · ∪ is µ almost equal to X.

k Proof. Let k Z+. Let ν be an ergodic h invariant measure. The set ∈

G = n Z: hn(ν) = ν { ∈ }

is a group and G contains k by assumption. Let l > 0 be a generator of G. Then l

divides k. Since µ is the only h invariant measure, we have

1 2 l 1 µ = ν + h(ν) + h (ν) + + h − (ν) . (III.1) l ··· 40

Furthermore, for i = 1, 2, . . . , l 1 it follows that hi(ν) is ergodic and ν = hi(ν). − 6 Hence for i = 1, 2, . . . , l 1, ν is mutually singular with respect to hi(ν), and so − i there is a measurable set Fi X such that ν(Fi) = 1 = (h (ν))(X Fi). Put ⊂ \ l 1 ∞ − E = hjl(F ). Then hl(E ) = E and ν(E ) = 1 = (hi(ν))(X E ). Set E = E . i \ i i i i i \ i j= \ i=0 −∞ 1 l It follows from III.1 that µ(E) = l and since h (E) = E we also have

j l 1 j µ ∞ h (E) = µ − h (E) . (III.2) Sj=0 Sj=0

But since µ(E) > 0 and µ is ergodic, the left hand side of III.2 is equal to one. The

proof is now complete.

We remark that the proof of Lemma III.5 shows that l can be chosen to be strictly

greater than 1 when hk is not uniquely ergodic. Indeed, if hk is not uniquely ergodic

one may choose ν an ergodic hk invariant measure diﬀerent from µ. Then equation

(III.1) forces l > 1.

Proposition III.6. Let (X, h) be a uniquely ergodic dynamical system and let k > 1

be an integer such that hk is also uniquely ergodic. If p is a prime divisor of k then

hpn is uniquely ergodic for all n 0. ≥

Proof. Let p be a prime divisor of k. Since the set of hp invariant probability measures

is contained in the set of hk invariant probability measures and hk is uniquely ergodic,

we conclude that hp is uniquely ergodic. Let µ be the unique h invariant measure.

Then µ is also the unique hp invariant measure. Let n > 1. We claim that hpn

is uniquely ergodic. Indeed, if hpn is not uniquely ergodic, we use Lemma III.5 to 41

ﬁnd a divisor l of pn and a measurable set E X such that hpn (E) = E, E has ⊂ 1 2 l 1 µ measure and E h(E) h (E) h − (E) is µ almost equal to X. Note l ∪ ∪ ∪ · · · ∪ l > 1 by the remark above since hpn is not uniquely ergodic. As p must divide l,

p 2p l p p the set E h (E) h (E) h − (E) is h invariant and has measure in (0, 1), ∪ ∪ ∪ · · · ∪ contradicting the ergodicity of hp with respect µ. Thus hpn is uniquely ergodic and

we are done.

Lemma III.7. Let (X, h) be a dynamical system and let (X, ϕ) be the suspension e flow of (X, h). Let f C(X) and let > 0. Consider the set V (f, ) as defined in ∈ e Definition III.2. If h and hk are uniquely ergodic where k > 1 is some integer then

V (f, ) is dense in [0, 1].

Proof. Let d be a metric on X and let d0 be a metric on X. Since f is uniformly e continuous, there is δ > 0 such that f([x, s]) f([y, t]) < whenever d([x, s], [y, t]) < | − | 2

δ. Since π X [0,1] is also uniformly continuous, there is δ1 such that d([x, s], [y, t]) < δ | ×

whenever max d0(x, y), s t < δ1 with 0 s, t < 1. { | − |} ≤ By Proposition III.6 there is a set P containing infinitely many powers of a single 1 prime such that hp is uniquely ergodic for all p P . Let N > . By Lemma II.9 ∈ δ1 p it suffices to show that if p P and q > N satisfies (p, q) = 1 and [0, 1] then ∈ q ∈ p V (f, ). q ∈ p Let p P and let q > N such that (p, q) = 1 and [0, 1]. Define functions ∈ q ∈ 42

Tm : X C by →

Tm(x) = f ([x, mp/q])

= f ϕm ([x, 0]) ◦ p/q

p for m Z. Since f is continuous, each Tm is continuous. Since h is uniquely ∈ n 1 1 − kp ergodic, the sequence Tm h converges uniformly as n to a constant n X ◦ → ∞ k=0

cm = Z Tm dµ, cf. [25, Theorem 6.19]. For all nonnegative integers k and i it is easy X to see that

kp Tkq+i = Ti h . ◦

p 1 The h invariance of µ then gives cq+i = ci for all integer i 0. Put c = (c0 + c1 + ≥ q

+ cq 1). Let i, r 0, 1, . . . , q 1 . Since ··· − ∈ { − }

qn 1 q 1 n 1 − − 1 1 1 − kp Ti+j = Ti+j h ! nq X q X n X ◦ j=0 j=0 k=0 and for r 1 ≥ qn+r 1 r 1 n − 1 n + 1 − 1 kp Ti+j = Ti+j h ! nq + r X nq + r X n + 1 X ◦ j=0 j=0 k=0 q 1 n 1 − n 1 − kp + Ti+j h ! , nq + r X n X ◦ j=r k=0

nq+r 1 1 − we see that the sequence T converges uniformly as n to c. We nq + r X i+j j=1 → ∞ n 1 1 − claim that this implies the uniform convergence of the sequence T as n n X i+j j=0 → ∞ to the same constant c. Indeed, let 0 > 0. For r = 0, 1, . . . , q 1, choose N 0 > 0 such − r 43

that nq+r 1 1 − Ti+j c < 0 nq + r X − j=0 whenever nq + r > N 0. Put N 0 = max N 0 : r = 0, 1, . . . , q 1 . If m > N 0, write r { r − } m = nq + r with r 0, 1, . . . , q 1 . Then m = nq + r > N 0 N 0 gives ∈ { − } ≥ r

m 1 1 − Ti+j c < 0. m X − j=0

This completes the proof of the claim. Let now i 0, 1, . . . , q 1 . Using the claim ∈ { − }

we just proved, we may choose Ni > 0 such that for all n > Ni

n 1 1 − Ti+j c < . nX − 2 j=0

Let N = max Ni + 1: i = 0, 1, . . . , q 1 . To complete the proof of the lemma we { − } will show N 1 1 − f ϕj c < . (III.3) p/q N X ◦ − j=0

For this purpose let [x, s] X with 0 s < 1. Choose i 0, 1, . . . , q 1 such ∈ ≤ ∈ { − } e 1 that ip/q s and ip/q s < < δ1. Then { } ≤ |{ } − | q 44

N 1 N 1 N 1 1 − 1 − 1 − f ϕj ([x, s]) c f ϕj ([x, s]) f ϕj ([x, ip/q ]) p/q p/q p/q N X ◦ − ≤ N X ◦ − N X ◦ { } j=0 j=0 j=0 N 1 1 − + (f ϕj ([x, ip/q ]) c) p/q N X ◦ { } − j=0 N 1 1 − = f ([x, s + jp/q]) f ([x, ip/q + jp/q]) N X j=0 | − { } | N 1 1 − [ip/q] + (Ti+j(h− (x)) c) N X − j=0 N 1 1 − < f ([x, s + jp/q]) f ([x, ip/q + jp/q]) + . N X 2 j=0 | − { } |

But since s + jp = ip + jp , we see that h q i h{ q } q i

[s+jp/q] [ ip/q +jp/q] max d0(h (x), h { } (x)), s + jp/q ip/q + jp/q { |{ } − {{ } }|} = s + jp/q ip/q + jp/q |{ } − {{ } }| = s ip/q − { }

< δ1

and so

d([x, s + jp/q], [x, ip/q + jp/q]) { }

[s+jp/q] [ ip/q +jp/q] = d([h (x), s + jp/q ], [h { } (x), ip/q + jp/q ]) { } {{ } } < δ.

Hence f([x, s + jp/q]) f([x, ip/q + jp/q]) < . Thus | − { } | 2

N 1 1 − (f ϕj ([x, s]) c) < . p/q N X ◦ − j=0

Since [x, s] X with 0 s < 1 was chosen arbitrarily, we conclude that (III.3) ∈ ≤ pe follows. Thus V (f, ) and we are done. q ∈

We are ready to prove the main result of this chapter.

Proposition III.8. Let (X, h) be a uniquely ergodic dynamical system and let (X, ϕ) e be the suspension ﬂow of (X, h). Assume that hk is uniquely ergodic for some k > 1.

Then the set

α [0, 1]: ϕα is uniquely ergodic { ∈ } is a dense Gδ set.

Proof. Let fi i Z+ be a dense set in C(X). By Lemma III.3, the set { } ∈ e

α [0, 1]: ϕα is uniquely ergodic { ∈ }

is equal to

V (f , 1 ) \ \ i j i 1 j 1 ≥ ≥ 1 where V (fi, ) is as in Deﬁnition III.2. But, for each i 1 and each j 1, the set j ≥ ≥ 1 V (fi, j ) is open by Lemma III.4 and dense by Lemma III.7. The proposition now

follows. 46

CHAPTER IV

ENTROPY

Let (X, h) be a dynamical system and let (X, ϕ) be the suspension flow associated e to (X, h). In this chapter we will find a formula to compute the (topological) entropy of the time α map ϕα in terms of the entropy of h. The entropy of a homeomorphism h on a compact metric space is either a nonnegative real number or infinity and is denoted by htop(h) (see [25, Chapter 7]). Entropy is an invariant of topological conjugacy. The following is a definition in Section 5 of Bowen [2].

Deﬁnition IV.1. A uniformly continuous ﬂow on a metric space (X, d) is a family of maps ϕα : X X α 0 with ϕα+β = ϕα ϕβ and such that for any α0 > 0 and > 0 { → } ≥ ◦

there is a δ > 0 for which d(ϕα(x), ϕα(y)) < whenever 0 α α0 and d(x, y) < δ. ≤ ≤

Lemma IV.2. Let (X, ϕ) be a continuous ﬂow where X is a compact metric space.

Then the family of maps ϕα α 0 is a uniformly continuous ﬂow. { } ≥

Proof. Let α0 > 0 and let > 0. Let d be a metric on X. Using the uniform continuity

of the action X [0, α0] X, choose δ > 0 such that d(ϕα(x), ϕβ(y)) < whenever × → max d(x, y), α β < δ with α, β [0, α0]. So if d(x, y) < δ then for all 0 α α0 { | − |} ∈ ≤ ≤

it follows that max d(x, y), α α < δ and therefore d(ϕα(x), ϕα(y)) < . { | − |} 47

Lemma IV.3. Let (X, ϕ) be a continuous ﬂow where X is a compact metric space.

Let α R. Then ∈

htop(ϕα) = α htop(ϕ1). | |

Proof. Since ϕα α 0 is a uniformly continuous ﬂow by Lemma IV.2, we may use [2, { } ≥ Proposition 21] to obtain for all α > 0

htop(ϕα) = αhtop(ϕ1).

Then

1 1 htop(ϕ α) = htop(ϕα− ) = αhtop(ϕ1− ) = αhtop(ϕ1) − where the last equality follows since X is compact (cf. [25, Theorem 7.3]). Further- e more, since ϕ0 = id, we have htop(ϕ0) = 0. Hence htop(ϕα) = α htop(ϕ1) for all | | α R, as wanted. ∈

Lemma IV.4. Let (X, h) be a dynamical system and let (X, ϕ) be the suspension e ﬂow of (X, h). Then

htop(ϕ1) = htop(h).

Proof. Consider the canonical quotient map π : X R X. Set × → e

Y = π X 0, 1 × 2 and

Z = π X 1 , 1 . × 2 48

Then Y and Z are closed ϕ1-invariant subsets of X. Furthermore, X = Y Z. ∪ e e Hence, by [13, Proposition 3.1.7(2)],

htop(ϕ1) = max htop(ϕ1 Y ), htop(ϕ1 Z ) . { | | }

1 Now observe that π 1 is a homeomorphism from X 0, to Y . Moreover X 0, 2 | ×h 2 i ×

ϕ1 π([x, t]) = [x, t + 1] = [h(x), t] = π (h id)([x, t]). ◦ ◦ ×

1 Hence, (X 0, , h id) is conjugate to (Y, ϕ1). Thus, using [13, Proposition × 2 × 3.1.7(4)],

htop(ϕ1 Y ) = htop(ϕ id) | ×

= htop(h) + htop(id)

= htop(h).

Similarly htop(ϕ1 Z ) = htop(h). Hence htop(ϕ1) = htop(h). |

We are ready to prove the main result of this chapter.

Theorem IV.5. Let (X, h) be a dynamical system and let (X, ϕ) be the suspension e ﬂow of (X, h). For every α R it follows that ∈

htop(ϕα) = α htop(h). | |

Proof. This follows immediately from Lemmas IV.3 and IV.4. 49

CHAPTER V

EXAMPLES

In this ﬁnal chapter we include examples (and counterexamples) about the theory studied in this thesis.

Example 1. Let (X, h) be a dynamical system and let (X, ϕ) be the suspension ﬂow e of (X, h). Chapters 2 and 3 gave suﬃcient conditions for the existence of minimal and uniquely ergodic time α maps. Is it possible to determine precisely for which α the

time α map is minimal and uniquely ergodic? We provide here an example when the

answer is aﬃrmative. Let (X, h) be a Denjoy system with irrational rotation number

β. We will explain brieﬂy what this means below. We refer to [22] for a detailed

account.

Let h be an orientation preserving homeomorphism of the circle S1 with no pe-

riodic points and let β be the irrational rotation number of h. By [22, Corollary

3.2] the homeomorphism h is uniquely ergodic and there exists a continuous sur-

1 1 jective map g : S S such that g h = Rβ g, where Rβ : x x + β is the → ◦ ◦ 7→ 1 R Z 1 rigid rotation on S ∼= / . We say that h is a Denjoy homeomorphism of S when the semiconjugation map g just described is not one-to-one. If h is a Den-

joy homeomorphism of S1 with irrational rotation number β then [22, Proposition

3.4] shows that the support X of the unique h invariant probability measure on S1 50 is a Cantor set, X is h invariant and the homeomorphism h: X X is minimal. → We then refer to the minimal and uniquely ergodic dynamical system (X, h) as a

Denjoy system. Following the notation in [22], suppose that the Cantor set X is

1 R Z obtained from S ∼= / by removing a countable disjoint union of open intervals ∞ ∞ I ,I ,.... To be more speciﬁc, let X = S1 I , where I is a countable 1 2 [ n [ n \ n=1 n=1 disjoint union of open intervals and the intervals In = (an, bn), for n Z+, are ∈ the components of S1 X. Observe that the semiconjugation map g collapses each \ of the intervals In = (a, b) into a single point. By continuity g(an) = g(bn). Set

T = an, bn : n Z+ and Q(h) = g(T ) = g(In): n Z+ . It follows that the set { ∈ } { ∈ } Q(h) is uniquely determined by h up to a rigid rotation and it is countable and invariant under Rβ. Let 1 n(h) 0 be the number of disjoint orbits of Q(h) under Rβ. n(h) ≤ ≤ ℵ Then Q(h) = Q , where Q ,...,Q are the n(h) disjoint R invariants orbits [ i 1 n(h) β i=1 of Q(h). So we have

Qi = γi + nβ : n Z { ∈ } for i = 1, . . . , n(h), where γi γj nβ : n Z when i = j. As pointed out in − 6∈ { ∈ } 6

Section 3 of [22], we may always assume γ1 = 0. Notice now that the set X T is \ dense in X and so

A = [(X T ) R]/Z \ × is dense in X. e Proposition V.1. Let (X, h) be a Denjoy system with irrational rotation number β

1 1 and let g : S S be the semiconjugation map satisfying g h = Rβ g, cf. [22]. → ◦ ◦ 51

Let (X, ϕ) be the suspension ﬂow of (X, h) and let α be a real number. Identify S1 e with R/Z. Then the map F : X S1 S1 given by → × e F ([x, t]) = (g(x) + βt, t) deﬁnes an almost one to one extension (cf. Appendix) between the dynamical systems

1 1 (X, ϕα) and (S S ,Rαβ Rα). × × e Proof. We have

F ([hn(x), t n]) = (ghn(x) + β(t n), t n) − − − = (Rng(x) + βt βn, t n) β − − = (g(x) + βn + βt βn, t n) − − = (g(x) + βt, t n) − = F ([x, t]).

Hence F is well deﬁned. Now, since

F ϕα([x, t]) = F ([x, t + α])

= (g(x) + βt + αβ, t + α)

= (Rαβ Rα)(g(x) + βt, t) ×

= (Rαβ Rα)F ([x, t]) × and F is clearly onto (because g is onto), it follows that F is an extension map. Let

A be the subset [(X T ) R]/Z of X with T as deﬁned above. Since A is dense and \ × e 1 A = [x, s] X : F − F ([x, s]) = [x, s] , n ∈ { }o e 52

we conclude that F is an almost one-to-one extension.

Proposition V.2. Let (X, h) be a Denjoy system with irrational rotation number

β. Let ϕα be the time α map on the suspension of (X, h). Then the following are

equivalent

1. ϕα is minimal.

2. ϕα is uniquely ergodic.

3. 1, α and αβ are linearly independent over Q.

1 2 4. α 0 :(r1, r2) Q (0, 0) . 6∈ { } ∪ r1β + r2 ∈ \

Proof. Let F be the map as in Proposition V.1 and let A = [(X T ) R]/Z X \ × ⊂ e with T as described above. Let λ be the Lebesgue measure on R and consider the

1 1 1 Rαβ Rα invariant measure λ λ on S S , where we identify S with [0, 1). We × × × claim that F ((T R)/Z) = F (X A) has measure zero. Indeed, using the notation × \ e described before Proposition V.1 we get

(λ λ)(F ((T R)/Z)) = × ×

Z λ( x:(x, t) (g(y) + βs, s):(y, s) T [0, 1] )dλ(t) [0,1) { ∈ { ∈ × }}

= Z λ( g(y) + βt: y T )dλ(t) [0,1) { ∈ }

= Z λ(Q(h) + βt)dλ(t) [0,1)

= Z λ(Q(h))dλ(t) [0,1) = 0. 53

Since Rα Rαβ is a group rotation, Proposition 1.4.1 in [13] and Theorem 6.20 in ×

[25] give that Rα Rαβ is minimal if and only if Rα Rαβ is uniquely ergodic if and × × only if 1, α, αβ are linearly independent over Q. The equivalence of (1), (2) and (3)

is now a consequence of Proposition V.1 and Theorem A.10. To complete the proof

of the proposition, we will to show the equivalence of (3) with (4). Set

T1 = α R: 1, α and αβ are linearly independent over Q { ∈ }

and

1 2 T2 = 0 :(r1, r2) Q (0, 0) . { } ∪ r1β + r2 ∈ \

We claim that T1 = R T2. Indeed, suppose that α T2. Then if α = 0 it is \ ∈ 1 trivial that α and αβ are not rationally independent. Otherwise, if α = r1β + r2 2 for some (r1, r2) Q (0, 0), then r1αβ + r2α 1 = 0 and so α and αβ are not ∈ \ −

rationally independent. This proves that T1 R T2. For the opposite inclusion, ⊂ \

assume that 0 = α T1; then α and αβ are not rationally independent. So there 6 6∈ exist (m, n) Z2 (0, 0) and k Z such that k = mα + nαβ = α(m + nβ). Since β ∈ \ ∈ k 1 is irrational, m + nβ = 0. Therefore k = 0 and so α = = n m , proving 6 6 nβ + m k β + k

T1 R T2, as wanted. ⊃ \

Example 2. Let (X, h) be a dynamical system. We say that (X, h) (or just h) is

a group rotation if X is a topological group and if h is rotation by a for some ﬁxed

a X (i.e. h(x) = ax for all x X). We say that a group rotation is abelian when ∈ ∈ X is abelian. In this example we show that the time α map on the suspension of 54

an abelian group rotation is an abelian group rotation. As consequence we will see

that the Elliott invariants of the crossed products of such minimal time α maps are

actually complete isomorphism invariants.

Lemma V.3. Let (X, h) be a dynamical system and an abelian group rotation. Let

(X, ϕ) be the suspension ﬂow of (X, h). Let α be a real number. Then the dynamical e system (X, ϕα) is an abelian group rotation. e Proof. Let e be the identity of X. Regarding R a group under addition, consider

the group X R with coordinatewise operation. Since h is a group rotation there is × a X such that h(x) = ax for all x X. Let H be the normal subgroup of X R ∈ ∈ × generated by (a, 1). Then X is equal to (X R)/H and so it is an abelian group. − × e The time α map is readily veriﬁed to be rotation by [e, α]. Hence the dynamical system (X, ϕα) is an abelian group rotation. e If (X, h) is a dynamical system and a group rotation then X is abelian when either

the Haar measure is an ergodic h invariant measure (cf. [25, Theorem 1.9]) or when

h is minimal (cf. [25, Theorem 3.4]).

Proposition V.4. Let (X1, h1) be a dynamical system and an abelian group rotation.

Let α be a real number and let S1 be the time α map on the suspension X1 of (X1, h1). e We have the following.

1. S1 is minimal if and only if S1 is uniquely ergodic. 55

2. Suppose that (X2, h2) is another dynamical system and an abelian group rota-

tion. Assume also that X2 abelian. Let S2 be the time α map on the suspension

o Z X2 of (X2, h2). Assume that Si is minimal for i = 1, 2. Then C(X1) S1 and e e o Z C(X2) S2 have the same Elliott invariants if and only if they are isomorphic. e

Proof. Since (X,S1) is a dynamical system and a group rotation by Lemma V.3, part e (1) follows from [25, Theorem 6.20]. Part (2) will follow from [10, Theorem 6] once we verify that (Xi,Si) is minimal and equicontinuous for i = 1, 2. But minimality is e part of the hypothesis and equicontinuity is a consequence of [10, Theorem 2] because

(Xi,Si) is a minimal dynamical system and a group rotation by Lemma V.3. e Example 3. Consider the trivial dynamical system (X, h) where X is the one point

space. It is obvious that (X, h) is a group rotation. Let α and β be irrational

numbers such that the time α map ϕα and the time β map ϕβ on the suspension X e of (X, h) are minimal (and so they are also uniquely ergodic by Proposition V.4). By

o Z o Z Theorem I.18, the Elliott invariants of C(X) ϕα and C(X) ϕβ are the same if e e and only if Z+αZ = Z+βZ if and only if α has the same image as β in R/Z. Hence ± o Z o Z using Proposition V.4 we conclude that C(X) ϕα and C(X) ϕβ are isomorphic e e if and only if α has the same image as β in R/Z. Since C(X)oϕ Z is isomorphic to ± α e 1 the irrational rotation algebra Aα = C(S ) oRα Z (likewise with β), we have obtained another way to distinguish irrational rotation algebras, a result obtained originally by Rieﬀel, Pimsner and Voiculescu [23, 21]. 56

∞ Example 4. Consider the 2-odometer (Y, a) [24, III.5.12]. That is, T = 0, 1 Y i=0{ } with the ordinary product topology and a is addition by 1 = (1, 0,...) with possibly inﬁnite carry over. Hence (Y, a) is a group rotation. Proposition V.4 then says that the Elliott invariants of crossed products induced by minimal time α maps on the suspension of (Y, a) are complete isomorphism invariants. Our Theorem I.18 gives us these Elliott invariants. We only need to ensure the existence of numbers α making the time α map on the suspension of (Y, a) minimal (and so uniquely ergodic). By

Proposition II.12 it suﬃces to show that a and ak are minimal for some integer k > 1.

The following lemma will take care of this.

Lemma V.5. Let (Y, a) be the 2-odometer. Then an is minimal for all odd positive

integers n.

Proof. It is well known that (Y, a) is minimal (and uniquely ergodic) [24, III.5.12].

Let n be an odd integer. We will show that (Y, an) is conjugate to (Y, a). We have

that an is addition by

n1 = 1 + 1 + ... + 1 . n | {z } Since n is odd, the ﬁrst component of n1 is equal to 1. Thus the multiplicative

1 n inverse (n1)− of n1 lies in Y (see the proof of [12, Theorem 10.10]). Hence (Y, a )

1 is conjugate to (Y, a) via the homeomorphism multiplication by (n1)− .

We remark that for more general p-odometers with p prime [24, III.5.12], results similar to Lemma V.5 might be obtained. 57

Example 5. Suppose that (X1, h1) and (X2, h2) are dynamical systems. Let α be a

real number. Let Si be the time α map on the suspension Xi of (Xi, hi) for i = 1, 2. f If (X1, h1) is conjugate to (X2, h2) it is easy to see that then (X1,S1) is conjugate to f (X2,S2). Suppose now that (X1, h1) is (strong) orbit equivalent to (X2, h2). Does it f follow that (X1,S1) is (strong) orbit equivalent to (X2,S2)? We show here an example f f where the answer is negative.

Suppose that (Y, a) is the 2-odometer (see Example 4). Let t1 = t2 be two real 6 numbers with 0 < t1, t2 < log(2). By [11, Theorem 16], there are Toeplitz ﬂows

(X1, h1) and (X2, h2) with entropy equal to t1 and t2 respectively and with maximal equicontinuous factor (Y, a). Moreover [11, Theorem 16] asserts that (Xi, hi) is strong orbit equivalent to (Y, a) for i = 1, 2. Denote by πi the extension map of (Xi, hi) onto

(Y, a) for i = 1, 2. As noticed in [11, Remark after Corollary 9], πi is actually an almost one-to-one extension for i = 1, 2 (see Appendix). Observe that the systems

(Xi, hi) and (Y, a) are minimal and so Proposition A.8 says that the three deﬁnitions of almost one-to-one extension presented in the Appendix are equivalent. Let k > 1

k be an odd integer. Since (Y, a ) is minimal by Lemma V.5 and πi is also an almost

k k one-to-one extension of (Xi, hi ) onto (Y, a ) for i = 1, 2, Theorem A.10 shows that

k hi is minimal for i = 1, 2. Using Proposition II.12 we may choose a real number α such that the time α map Si on the suspension Xi of (Xi, hi) is minimal for i = 1, 2. f By Theorem IV.5 we obtain

htop(S1) = α htop(h1) = α t1 = α t2 = α htop(h2) = htop(S2). (V.1) | | | | 6 | | | | 58

But since the suspensions Xi are connected for i = 1, 2, if S1 were orbit equivalent f to S2, they would be ﬂip conjugate [10, Remark after Theorem 6] and hence they would have the same entropy, contradicting (V.1). Hence (X1, h1) and (X2, h2) are

(strong) orbit equivalent but (X1,S1) and (X2,S2) are not. f f Example 6. A well known result about S1 states that two minimal homeomorphism of S1 are ﬂip conjugate if and only if their associated crossed products are isomorphic. One may ask whether a similar result follow for connected compact metric

1-dimensional spaces non homeomorphic to S1. We believe that using techniques emanating from this dissertation we may ﬁnd an example of two minimal dynamical systems on connected compact metric 1-dimensional spaces which are not strong orbit equivalent yet their associated crossed products have the same Elliott invariants.

Consider the dynamical systems (X1,S1) and (X2,S2) obtained in Example 5. We f f have that (X1,S1) and (X2,S2) are minimal dynamical systems on connected compact f f metric 1-dimensional spaces and are not strong orbit equivalent. One could use our

Theorem I.18 to compute the Elliott invariants of their associated crossed products if the maps S1 and S2 were uniquely ergodic. This Elliott invariants would be the same by [9, Theorem 2.1] and Theorem I.18. To be able to use Proposition III.8, one

k1 k2 needs the homeomorphisms h1 and h2 of Example 5 to be uniquely ergodic for some k1, k2 > 0.

Example 7. Let (X, h) be a dynamical system and suppose that the time α map on the suspension of (X, h) is minimal. Does it follow that the time α + 1 map is a 59

minimal? We show here an example where the answer is negative. Choose β [0, 1] ∈ β 1 2 such that − 0 :(r1, r2) Q (0, 0). For example, take β to 1 + β 6∈ { } ∪ r1β + r2 ∈ \ be a transcendental number. Suppose that (X, h) is a Denjoy system with irrational β 1 rotation number β. (See Example 1.) Put α = − . Then α + 1 = . Thus, 1 + β 1 + β

by Proposition V.2, ϕα is minimal but ϕα+1 is not.

Example 8. The suspension ﬂow of a dynamical system is a particular case of a

more general construction called the suspension under a ceiling function or generalized

suspension ﬂow, cf. [24, II.5.14]. Let (Y, a) be a Cantor minimal system and let F Y ⊂

be clopen. Consider the generalized suspension Yf of (Y, a) under the ceiling function

f = χX F + γχF , where γ > 0. Let ϕα be the time α map on Yf . We show here that \

asking 1, α, γ to be linearly independent over Q is not suﬃcient to make ϕα minimal.

Let α > 0 and β [0, 1] be irrational numbers such that 1, α, αβ are linearly ∈ 1 independent over Q and γ = αβ 1 > 0. For example, take α = √3 and β = . − √2 Let (X, h) be a Denjoy system with irrational rotation number β. (See Example 1.)

Consider the dynamical system (Y, a) where Y = X 0, 1 and a is deﬁned by × { }

(x, 1) if i = 0  a(x, i) =  (h(x), 0) if i = 1  

Then (Y, a) is a minimal Cantor system. Let F = X 0 and let f = χY F + γχF . × { } \ 1 α We have that the time = map on the suspension of (X, h) is not minimal β 1 + γ 1 1 (because 1, , β are not linearly independent over Q, cf. Proposition V.2) and is β β 60

conjugate to the time α map on the generalized suspension Yf of (Y, a) under the i + t ceiling function f, with conjugation map [(x, i), t] x, . We conclude that 7→ γ + 1 the time α map on Yf is not minimal even though 1, α, γ are linearly independent over Q (because γ = αβ 1 and 1, α, αβ are linearly independent over Q by choice). − I am thankful to T. Katsura for showing me the idea of the construction used in this example.

Example 9. Let (X, h) be a minimal and uniquely ergodic dynamical system. Let

(X, ϕ) be the suspension ﬂow of (X, h). Let τ be the unique normalized trace of e C(X) h Z. Theorem I.18 might suggest a connection between the minimality of ϕα × and the triviality of the intersection Z ατ(K0(C(X) oh Z)). That is, assuming that ∩

ϕα is minimal, does it follow that Z ατ(K0(C(X) oh Z)) = 0 ? We show here an ∩ { } example where the answer is negative.

Suppose that β [0, 1] and γ2 are irrational numbers such that 1, β, γ2 are linearly ∈ 1 independent over Q. For example, take β = and γ2 = √3. Let (X, h) be a Denjoy √2 system with irrational rotation number β and Q(h) = nβ : n Z γ2+nβ : n Z . { ∈ }∪{ ∈ } (See Example 1.) Then Theorem 5.2 in [22] gives

τ(K0(C(X) oh Z)) = Z + Zβ + Zγ2.

1 Put α = . Using Proposition V.2 we obtain that ϕα is minimal (because 1, α, αβ γ2 1 are linearly independent over Q since 1, β, γ = are linearly independent over Q by 2 α choice). But

Z ατK0((C(X) oh Z)) = Z. ∩ 61

APPENDIX A

ALMOST ONE-TO-ONE EXTENSIONS

Let (X, ϕ) and (Y, ψ) be dynamical systems. We say that (X, ϕ) is an extension of (Y, ψ) if there is a continuous map π of X onto Y such that π ϕ = ψ π. We ◦ ◦ call π an extension map. If the extension map π were also one-to-one, then π would be a homeomorphism and so (X, ϕ) would be conjugate to (Y, ψ). It is clear that if two dynamical systems are conjugate, then one is minimal (uniquely ergodic) if and only if the other one is also minimal (uniquely ergodic). If π is not one-to- one, one may hope for π to be “almost” one-to-one. Once we have a satisfactory definition of what we mean by almost one-to-one, we may proceed to ask, as in the case when π was one-to-one, whether minimality (unique ergodicity) of one of the dynamical systems implies minimality (unique ergodicity) of the other. In this appendix we propose a definition for almost one-to-one extensions, compare it with those (seemingly different) definitions of almost one-to-one extensions found in the literature and show that all of them coincide when the dynamical systems involved are minimal. We conclude with a result giving conditions for when an extension of dynamical systems preserves minimality (unique ergodicity). The results in this appendix might well be known to the specialists; however we decided to include them anyway as we could not find them stated in the literature. 62

The following is our proposed deﬁnition of an almost one-to-one extension.

Deﬁnition A.1. We say that an extension π : X Y is almost one-to-one if the set → 1 A = x X : π− π(x) = x is dense. { ∈ { }}

Other deﬁnitions found in the literature are the following.

Deﬁnition A.2 ([24, 4.6.1(2)]). An extension π : X Y is almost one-to-one if → 1 there is a point x in X with dense orbit such that π− π(x) = x . { }

Deﬁnition A.3 ([4, Page 152]). We say that an extension π : X Y is almost → 1 one-to-one if the set B = y Y : π− (y) = 1 contains a dense Gδ set. The symbol { ∈ | | } E means the cardinality of the set E. | |

Lemma A.4. Let π : X Y be an extension map. Let A be as in Deﬁnition A.1. →

Then π A is one-to-one. |

1 1 Proof. If a1, a2 A and π(a1) = π(a2), then a1 = π− π(a1) = π− π(a2) = a2 ∈ { } { } and thus a1 = a2.

Lemma A.5. Let π : X Y be an extension map. Let A be as in Deﬁnition A.1 → 1 and let B be as in Deﬁnition A.3. Then A = π− (B) and π(A) = B.

1 1 Proof. If a A then π− π(a) = a , that is, π− (π(a)) = 1, so π(a) B. Hence ∈ { } | | ∈ 1 1 1 a π− (B). Conversely, if a π− (B) then π(a) B, that is, π− (π(a)) = 1. It ∈ ∈ ∈ | | 1 1 follows that π− π(a) = a . This proves that A = π− (B). The other equality is now { } immediate. 63

Lemma A.6. Let (X, ϕ) be an extension of (Y, ψ) with extension map π : X → Y . The sets A and B in Deﬁnitions A.1 and A.3 are invariant sets in X and Y , respectively.

Proof. It suﬃces to show the next two statements.

(i) ψ(B) B and ⊂

1 (ii) ψ− (B) B. ⊂

1 Indeed, this would give us that ψ(B) = B = ψ− (B) and so

1 1 1 1 1 1 ϕ− (A) = ϕ− π− (B) = π− ψ− (B) = π− (B) = A as wanted.

1 1 1 1 1 To prove (i), let b B. Then ϕ− π− ψ(b) = π− ψ− ψ(b) = π− (b) = 1. From ∈ | | | | | | 1 this we obtain that π− ψ(b) = 1, that is, ψ(b) B, as was to be proved. | | ∈ The proof of (ii) is analogous.

Proposition A.7. Let (X, ϕ) be an extension of (Y, ψ) with extension map π : X → Y . We have the following.

(i) If π is an almost one-to-one extension in the sense of Deﬁnition A.2 then π is

an almost one-to-one extension in the sense of Deﬁnition A.1. The converse is

false. 64

(ii) If π is an almost-one-to-one extension in the sense of Deﬁnition A.1 then π is

an almost one-to-one extension in the sense of Deﬁnition A.3. The converse is

false.

Proof. To prove (i), assume π is an almost one-to-one extension in the sense of Deﬁ-

1 nition A.2. Then there is x in X with dense orbit and π− π(x) = x . Let A be as in { } A.1. Then x belongs to A. By Lemma A.6 the orbit of x is a subset of A. Hence A is dense. Thus π is one-to-one in the sense of Deﬁnition A.1. To show that the converse is false, consider the following example. Let X = Y consist of at least two points and put ϕ = ψ = π = IdX . Then π is an extension map which is almost one-to-one in the sense of Deﬁnition A.1 (because A = X) but is not one-to-one in the sense of

Deﬁnition A.2 (no point in X has dense orbit).

For (ii), assume that π is an almost one-to-one extension in the sense of Deﬁni- tion A.1. Let A be as in A.1 and let B be as in A.3. By Lemma A.5 we have B = π(A).

Hence B is dense since A is dense. Moreover, B is a Gδ set from a remark in [24,

VI.6.4(1)]. Thus π is an almost one-to-one extension in the sense of Definition A.3. To show that the converse is false, consider the following example. Let X = [0, 1] 2 ∪ { } and let Y = [0, 1]. Put ϕ = IdX , ψ = IdY and define π : X Y to be the identity → on [0, 1] and π(2) = 1. Then π is an extension map which is almost one-to-one in the sense of Definition A.3 (because B = Y ) but is not almost one-to-one in the sense of

Deﬁnition A.1 (because A = [0, 1] is not dense in X = [0, 1] 2 ). ∪ { }

Proposition A.8. If π : X Y is an extension of minimal systems then the Deﬁ- → 65

nitions A.1, A.2 and A.3 are equivalent to each other.

Proof. Suppose that π : X Y is an extension of minimal systems. By [1, Theorem → 1.15], π is semi-open, that is, if U = is open in X then π(U) has non-empty interior. 6 ∅ Thus A.3 implies A.1.

Let A be as in A.1. If A is dense, it is nonempty. Then there exists x X such ∈ 1 that π− π(x) = x . The orbit of x is dense since ϕ is minimal. Thus A.1 implies { } A.2.

Lemma A.9. Suppose that ϕ: X X is a minimal and uniquely ergodic homeo- → morphism with µ the unique invariant probability measure. Let A be a subset of X.

If µ(A) = 1 then A is dense.

Proof. If U X is open then µ(A U) = µ(A) + µ(U) µ(A U) = µ(U). But ⊂ ∩ − ∪ µ(U) > 0 by [25, Theorem 6.17]. Hence A U = . ∩ 6 ∅

Theorem A.10. Let (X, ϕ) and (Y, ψ) be two dynamical systems. Suppose that

(X, ϕ) is an extension of (Y, ψ) with extension map π : X Y . Let ν be an invariant → 1 probability measure on Y and consider the set A = x X : π− π(x) = x . { ∈ { }}

(a) If A is dense (i.e. π is an almost one-to-one extension in the sense of A.1) then

ϕ is minimal if and only if ψ is minimal.

(b) If ν(π(A)) = 1 then ϕ is uniquely ergodic if and only if ψ is uniquely ergodic.

1 Proof. If M is a nontrivial closed ψ-invariant subset of Y then π− (M) is a nontrivial

closed ϕ-invariant subset of X. This shows that if (X, ϕ) is an extension of (Y, ψ) and 66

ϕ is minimal, then ψ is minimal. For the converse, we use the hypothesis that the extension is almost one-to-one. Suppose that ψ is minimal. Let M be a nonempty closed ϕ invariant subset of X. Then M is compact and so π(M) is compact. Hence

π(M) is a nonempty closed ψ invariant subset of Y . Thus π(M) = Y . Since A is dense, the open set X M must be empty; otherwise, there is an element x in \ A (X M) so π(x) Y = π(M). This means that π(x) = π(y) for some y M ∩ \ ∈ ∈ 1 1 and hence x = π− π(x) = π− π(y). Thus x = y M, a contradiction. { } ∈ Let us prove now (b). If ϕ is uniquely ergodic then ψ is uniquely ergodic by

[24, Corollary IV.1.8]. For the converse, we use the hypothesis that ν(π(A)) = 1.

Assume ν is the only invariant probability measure on Y . Then there is an invariant

1 probability measure on X, call it µ, satisfying ν = µ π− (cf. [24, Corollary 4.1.9]). ◦ 1 Let µ0 be another invariant probability measure on X. Then µ0 π− must be ◦ equal to ν. Let E X be an arbitrary measurable set. Since π is one-to-one ⊂ 1 when restricted to A by Lemma A.4, we have E A = π− π(E A). Therefore ∩ ∩ 1 µ0(E A) = µ0π− π(E A) = νπ(E A). Furthermore, using Lemma A.5 and the ∩ ∩ ∩ hypothesis,

1 µ0(A) = µ0(π− (B))

= ν(B)

= ν(π(A))

= 1. 67

This gives µ0(X A) = 0 and so µ0(E (X A)) = 0. Furthermore, µ(A) = 1 implies \ ∩ \ µ(E (X A)) = 0. Hence ∩ \

µ0(E) = µ0((E A) (E (X A))) ∩ ∪ ∩ \

= µ0(E A) + µ0(E (X A)) ∩ ∩ \ = ν(π(E A)) ∩

1 = µπ− π(E A) ∩ = µ(E A) ∩ = µ(E A) + µ(E (X A)) ∩ ∩ \ = µ(E).

Thus µ = µ0. Hence ϕ is uniquely ergodic. 68

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