Tracial Rokhlin property for automorphisms on simple 6 0 A 0 T-algebras 2 n Huaxin Lin a J Department of Mathematics 0 University of Oregon 2 Eugene, Oregon 97403-1222 ] Hiroyuki Osaka A Department of Mathematical Sciences O Ritsumeikan University . h Kusatsu, Shiga, 525-8577, Japan t a m [ Abstract 1 LetAbeaunitalsimpleAT-algebraofrealrankzero. Givenanisomorphismγ1:K1(A)→ v K1(A), we show that there is an automorphism α : A → A such that α∗1 = γ1 which has 3 the tracial Rokhlin property. Consequently, the crossed product A⋊α Z is a simple unital 1 AH-algebrawithrealrankzero. Wealso showthatautomorphism withRokhlinpropertycan 5 beconstructed from minimal homeomorphisms on a connected compact metric space. 1 0 6 1 Introduction 0 / h AlanConnesintroducedthe Rokhlinpropertyinergodictheorytooperatoralgebras([2]). Several at versionsoftheRokhlinpropertyforautomorphismsonC∗-algebrashavebeenstudied(forexample m [8], [32], [14], [27], [11], [12] and [30], to name a few). Given a unital C∗-algebra A and an : automorphism α on A, one may view the pair (A,α) as a non-commutative dynamical system. v To study its dynamical structure, it is natural to introduce the notion of Rokhlin property. Let i X A be a unital simple AT-algebra (a C∗-algebra which is an inductive limit of those C∗-algebras r that are finite direct sums of continuous functions on the circle T) and let α be an automorphism a on A. In several cases, Kishimoto showed that if α is approximately inner (or homotopic to the identity) and has a Rokhlin type property, then the crossed product A⋊ Z is again a unital α simple AT of real rank zero ([14], [15], [16]). It is proved that if α induces the identity on K (A) 0 (or a “dense” subgroup of K (A)) and α has so-called tracial cyclic Rokhlin property then indeed 0 the crossed product is an AH-algebra of real rank zero ([25] and [24]). A natural question is when automorphisms have certain Rokhlin property. In [29], it is shown that if A is a unital separable simple C∗-algebra with tracial rank zero and with a unique tracial state and if α is an automorphism such that A⋊ Z has a unqiue trace, then α has the tracial Rokhlin property. α More recently, N.C. Phillips [31] showed that, for any unital simple separable C∗-algebra A with tracial rank zero, there is a dense G-δ set of approximately inner automorphisms such that every automorphism in the the set has the tracial Rokhlin property. 1 Let A be a unital simple AT-algebra of real rank zero. Suppose that γ : K (A) → K (A) 1 1 1 is an automorphism. In this note, we present an automorphism α on A with the tracial cyclic Rokhlin property such that α = γ . The automorphism α that we present in this note also has ∗1 1 the property that α = id . When α has the tracial cyclic Rokhlin property, then α must ∗0 K0(A) fix a large subgroup of K (A). In fact, it is shown in [24] that, at least in the case that A has a 0 unique tracial state, if α has tracial cyclic Rokhlin property, then α2 fixes a subgroup G⊂K (A) 0 so that ρ(G) is dense in Aff(T(A)), where T(A) is the tracialstate space of A, Aff(T(A)) is the space of all real affine continuous functions on T(A) and ρ : K (A) → Aff(T(A)) is the positive 0 homomorphism induced by the evaluation ρ([p])(τ)=τ(p) for projections p∈A. Let X be a connected compact metric space and ρ : K (C(X)) → Z be the dimension X 0 map. Denote by kerρ the kernel of the dimension map. Given a such X and a countable dense X subgroupD ⊂Q,there is astandardwayto constructa unitalsimpleC∗-algebra A with tracial X rank zero such that K (A ) = D⊕kerρ and K (A ) = K (C(X)) as well as there is a unital 0 X X 1 X 1 embedding j : C(X) → A so that (j )| = id| and j = id . This could be X ∗0 kerρX kerρX ∗1 K1(X) viewed as a version of the non-commutative space associated with X. Suppose that ψ : X → X is a minimal homeomorphism on X. We show that one can construct an automorphism α on A X associated with ψ such that α has the tracial cyclic Rokhlin property and such that α | =id , ∗0 D D α | = ψ | and α = ψ . This may be viewed as a non-commutative version of the ∗0 kerρX ∗0 kerρX ∗1 ∗1 minimal action associated with ψ. We also show that somewhat general construction can also be made. It appears that automorphisms with the tracial cyclic Rokhlin property occur quite often. While webelievethatmanyothertypesofconstructionofautomorphismswiththe Rokhlin property may be possible andperhaps notnecessarily difficult, we think these constructionin this note shed some light on how commutative Rokhlin tower lemma appears naturally in the study of non-commutative dynamical systems such as (A,α). Aknowledgement The first named author was partially supported by a NSF grant. The second author was partially supported by Open Research Center Project for Private Universi- ties:maching fund from MEXT,2004-2008. Much of the ground work of this reseach was done when both authors were visiting East China Normal University in the summer 2004. They would liketoacknowledgethesupportfromShanghaiPriorityAcademicDisciplinesandfromDepartment of Mathematics of East China Normal University. 2 Preliminaries The following conventions will be used in this paper. (1) Let A be a stably finite C∗-algebra. Denote by T(A) the tracial state space of A. Denote by Aff(T(A)) the normed space of all real affine continuous functions on T(A). (2) Denote by ρ :K (A)→Aff(T(A)) the positive homomorphism induced by ρ ([p])(τ)= A 0 A τ ⊗Tr(p) for any projection in M (A) (k =1,2,...,), where Tr is the standard trace on M . k k (3) Let X be a connected compact metric space. Denote by ρ : K (C(X)) →Z the positive X 0 homomorphism ρ defined in (2). It is the dimension function from K (C(X)) to Z. C(X) 0 (4) Let X be a compact metric space, let F be a subset of X and let ε>0. Put F ={x∈X :dist(x,F)<ε}. ε 2 (5) Let A be a C∗-algebra and let p and q be projections in A. We say p is equivalent to q if there is a v ∈A such that v∗v =p and vv∗ =q. (6)LetA=lim (A ,φ )beaninductivelimitofC∗-algebras. Hereφ isahomomorphism n→∞ n n n from A into A . We will use φ :A →A for the homomorphism induced by the inductive n n+1 n,∞ n limit system. We also use φ for φ ◦φ ◦···◦φ , if m > n. Suppose that each A = n,m m−1 m−2 n n ⊕r(n)B . When B are understood, a partial map φ(i,j) of φ is the homomorphism from B i=1 n,i n,i n n n,i to B given by φ . n+1,j n (7) Let A and B be two C∗-algebras and let φ,ψ :A→B be two maps. Suppose that G ⊂A is a subset. We write ψ ≈ φ on G, ε if kφ(a)−ψ(a)k<ε for all a∈G. The following Rokhlin property was introduced in [28]. Definition 2.1. Let A be a unital C∗-algebra and let α be an automorphism on A. We say that α has the tracial Rokhlin property if, for any ε>0, any finite subset F ⊂A, any positive integer n>0, and any a∈A \{0}, there are mutually orthogonal projections e ,e ,...,e such that + 1 2 n (1) ke x−xe k<ε for all x∈F, (0≤i≤n−1) i i (2) kα(e )−e k<ε, i=1,2,...,n−1 and i i+1 (3) 1− n−1e is equivalent to a projection in aAa. i=0 i P A stronger version of the Rokhlin property below was given in [25]. Definition 2.2. Let A be a unital C∗-algebra and let α be an automorphism on A. We say that α has the tracial cyclic Rokhlin property if, for any ε > 0, any finite subset F ⊂ A, any positive integer n > 0, and any a ∈A \{0}, there are mutually orthogonal projections e ,e ,e ,...,e + 0 1 2 n−1 (with e =e ) such that n 0 (1) ke x−xe k<ε for all x∈F, (0≤i≤n−1) i i (2) kα(e )−e k<ε, i=0,1,...,n−1 and i i+1 (3) 1− n−1e is equivalent to a projection in aAa. i=0 i P Remark 2.3. Note that, in Definition 2.2, one requires that kα(e )−e k < ε. This is not n−1 0 requiredinDefinition2.1. IfAisassumedtosatisfytheso-calledfundamentalcomparisonproperty, i.e., τ(p) > τ(q) for all τ ∈T(A) implies that q is equivalent to a projection e ≤p for any pair of projections, then condition (3) above can be replaced by τ(1− n−1e )<ε for all τ ∈T(A) and i=0 i the referenceto the elementa canbe removed. We note that if APhas tracialrank zerothen Ahas the fundamental comparison property. If A has tracial rank zero, then A has real rank zero, stable rank one, and ρ (K (A)) is dense A 0 in Aff(T(A)). One of the important consequences that a given automorphismsatisfies the tracial cyclic Rokhlin property is the following result: Theorem 2.4. (see 3.4 and 4.5 of [24]) If A is a unital separable amenable simple C∗-algebra with tracial rank zero satisfying the UCT and α is an automorphism on A with the tracial (cyclic) Rokhlin property such that (α) = id ∗0 G for some subgroup G ⊂ K (A) for which ρ (G) is dense in ρ (K (A)), then A⋊ Z has tracial 0 A A 0 α rank zero. Consequently, A⋊ Z is an AH-algebra. α 3 3 Automorphisms on simple AT-algebras The purpose of this section is to present Theorem 3.5. We start with the following construction which is certainly familiar to experts. Lemma 3.1. Let A be a unital simple infinite dimensional AF-algebra. Then, there exists a sequence of finite dimensional C∗-subalgebras F = M ⊕B and monomorphisms φ : F → n k(n) n n n F satisfying the following: n+1 (1) Let d be the identity of M . Then n k(n) lim sup {τ(φ (d ))}=0; n,∞ n n→∞τ∈T(A) (2) The partial map from M to M has multiplicity at least 2. k(n) k(n+1) (3) A=lim (F ,φ ). n→∞ n n Proof. We firstwrite A=lim (A ,ψ ), where eachA is a finite dimensionalC∗-algebra and n→∞ n n n ψ is a monomorphism. n Write A =M ⊕B′. Since A is assumed to be simple, we may also assume that the partial 1 n(1) 1 map ψ1,1 from M to a simple summand M of A has multiplicity m(1) which is at least n n(1) n(2) 2 4. Define F = M ⊕M ⊕···⊕M ⊕B′, where M repeats [m(1)/2] (integer part of 1 n(1) n(1) n(1) 1 n(1) m(1)/2)times. DenoteB =M ⊕···⊕M ⊕B′,whereM repeats[m(1)/2]−1times. So 1 n(1) n(1) 1 n(1) F =M ⊕B . Note there are monomorphism f :A →F and monomorphism ψ′ :F →A 1 n(1) 1 1 1 1 1 1 2 suchthatψ =ψ′ ◦f andthemultiplicity ofthepartialmapofψ′ fromM toM isatleast 1 1 1 1 n(1) n(2) 2. Let d be the identity of M in F . Note that 1 n(1) 1 τ(ψ ◦ψ′(d ))≤1/2 2,∞ 1 1 for all τ ∈T(A). We have A =M ⊕B′, where B′ is a finite dimensional C∗-algebra. 2 n(2) 2 2 We may alsoassume (by replacingA by someA )that the partialmapofψ fromM to a 3 n 2 n(2) simple summand M of A has multiplicity m(2) which is at least least 8. Define F =M ⊕ n(3) 3 2 n(2) M ⊕···M ⊕B′,whereM repeats[m(2)/2]times. DefineB =M ⊕···⊕M ⊕B′, n(2) n(2) 2 n(2) 2 n(2) n(2) 2 where M repeats [m(2)/2]−1 times. So F = M ⊕B . There exists a monomorphism n(2) 2 n(2) 2 f :A →F and there exists a monomorphism ψ′ :F →A such that 2 2 2 2 2 3 ψ =ψ′ ◦f . 2 2 2 Moreover,the multiplicity of the partial map of ψ′ from M to M is at least 2. 2 n(2) n(3) Define φ :F →F by φ =f ◦ψ′. Let d be the identity of the first summand M in F . 1 1 2 1 2 1 2 n(2) 2 Then τ(ψ ◦ψ′(d ))<1/4 3,∞ 2 2 for all τ ∈T(A). Write A = M ⊕B′, where B′ is a finite dimensional C∗-subalgebra. We may assume 3 n(3) 3 3 that the partialmap of ψ from M to a simple summand M of A has multiplicity m(3) at 3 n(3) n(4) 4 least 16. Define F =M ⊕M ⊕···⊕M ⊕B′, where M repeats [m(3)/2] times and 3 n(3) n(3) n(3) 3 n(3) B =M ⊕···⊕M ⊕B′, where M repeats [m(3)/2]−1 times. So F =M ⊕B . 3 n(3) n(3) 3 n(3) 3 n(3) 3 There are monomorphisms f :A →F and ψ′ :F →A such that ψ =ψ′ ◦f . Moreover, 3 3 3 3 3 4 3 3 3 the partial map of ψ from M (the first summand) to M has multiplicity at least 2. Define 3 n(3) n(4) φ :F →F by φ =f ◦ψ′. Let d be the identity of of the first summand M in F . Then 2 2 3 2 3 2 3 n(3) 3 τ(ψ ◦ψ′(d ))<1/8. 4,∞ 3 3 4 We continue this construction. We obtain F = M ⊕B , where B is a finite dimensional k n(k) k k C∗-subalgebra,andweobtainmonomorphismsf :A →F ,ψ′ :F →A andφ :F →F k k k k k k+1 k k k+1 such that ψ =ψ′ ◦f and φ =f ◦ψ′ (e3.1) k k k k k+1 k and such that the partial map of φ from M to M has multiplicity at least 2. Moreover, k n(k) n(k+1) τ(ψ ◦ψ′(d ))<1/2k (e3.2) k+1,∞ k k for all τ ∈T(A), where d is the identity of M in F . k n(k) k By (e3.1), A=lim (F ,φ ). By (e3.2), n→∞ k k τ(φ (d ))<1/2k. k,∞ k The lemma follows. Lemma 3.2. Let X =S1∨S1∨···∨S1 be identified with m copies of unit circle with a common point 1. For any integer n and finite subset F ⊂ C(X), there exists an integer N = Λ(n,F) satisfying the following: For any unital C∗-algebra B with real rank zero and any homomorphisms φ ,φ : C(X) → B 1 2 with (φ ) =(φ ) , there is a unitary u∈M (B) such that 1 ∗1 2 ∗1 mN+1 u∗(diag(φ (f),f(ξ ),f(ξ ),...,f(ξ ))u≈ 1 1 2 mN 1/2n diag(φ (f),f(ξ ),f(ξ ),...,f(ξ )), 2 1 2 mN where each of {ξ ,ξ ,...,ξ }, {ξ ,ξ ,...,ξ }, ..., {ξ ,ξ ,...,ξ }, divides 1 2 N N+1 N+2 2N (m−1)N+1 (m−1)N+2 mN the unit circles evenly. Proof. This follows from Theorem 1.1 in [7] and its remark. Definition 3.3. Let X =S1∨S1∨···∨S1 and Y =S1∨S1∨···∨S1 be identified with m and r copies of unit circle with a common point 1, respectively. Let π : Y → S1 (the ith copy of Y) i be defined as follows: π ((ξ ,ξ ,...,ξ ,...,ξ ))=ξ . i 1 2 i r i A continuous map s:X →Y is saidbe standard if π ◦s on eachS1 is describedas z →zk for i some integer k, where z is the identity map on the unit circle. Note that if k = 0, z → z0 is the map to the common point 1. Suppose that κ : Zr → Zm is the homomorphism associated with a m×r matrix (c ) with i,j integer entries. Then we say that s:X →Y is the standard map associated with κ if π ◦s on the j i-th copy of S1 can be described by z →zci,j. Note that, for a standard map s : X → Y, s maps the common point 1 of X to the common point 1 of Y. It should be also noted that if Z = S1 ∨S1 ∨···∨S1 is identified with k copies of the unit circle with a common point 1 and s′ : Y → Z is a standard map, then s′◦s is a standard map from X to Z. 5 Lemma 3.4. Let A be a unital simple AT-algebra with real rank zero. Suppose that K (A) = 1 lim (G ,κ ) 6= {0}, where G is direct sum of γ(n) copies of Z and κ : G → G is an n→∞ n n n n n n+1 homomorphism. Then A= lim (A ,φ ) n n n→∞ such that (1) A =C(X )⊗M ⊕B , where B is a finite dimensional C∗-subalgebras, n n k(n) n n (2) X =S1∨S1∨···∨S1, where S1 repeats γ(n) times, n (3) lim sup {τ(φ (d ))}=0, n,∞ n n→∞τ∈T(A) where d is the identity of C(X )⊗M , n n k(n) (4) each φ is injective. n Moreover, if {X } is fixed and a sequence of positive integers {a } is given, we may require n n that (5) the partial map of φ from C(X )⊗M to C(X )⊗M has the form n n k(n) n+1 k(n+1) φ (f)=diag(f ◦s ,f(ζ(n,1)),f(ζ(n,2)),...,f(ζ(n,t(n)))), n n where s is a standard map associated with κ and{ζ(n,1),ζ(n,2),...,ζ(n,t(n))} contains a subset n n that divides the j-th circle into b(n,j) even arcs with b(n,j)≥a ; n (6) each partial map φ(1,i) of φ from C(X )⊗M to a summand of B has the form n n n k(n) n+1 φ(1,i)(f)=⊕γ(n)diag(f(ξ(j,1)),f(ξ(j,2)),...,f(ξ(j,k ))), n j=1 j where {ξ(j,1),...,ξ(j,k )} is k points on the j-th circle of X . j j n Proof. NotethatAisaninfinitedimensionalsimpleC∗-algebra. LetBbeaunitalsimpleseparable AF-algebra such that (K (B),K (B) ,[1 ])=(K (A),K (A) ,[1 ]). 0 0 + B 0 0 + A ByLemma3.1,wemayassumethatB =lim (M ⊕B ,φ ),whereB isafinitedimensional n→∞ k(n) n n n C∗-algebra and (1) and (2) in Lemma 3.1 hold. By passing to a subsequence, one may assume that the partial map of φ from M to M has multiplicity at least m(n)≥γ(n)a −1. n k(n) k(n+1) n Put X =S1∨S1∨···∨S1, where S1 repeats γ(n) times. Define A =C(X )⊗M ⊕B . n n n k(n) n So K (A )=G =Zγ(n), n=1,2,.... Define s :X →X to be the standard map associated 1 n n n n n+1 with κ as defined in Definition 3.3. Define ψ(1,0) from C(X )⊗M to C(X )⊗M as n n n k(n) n+1 k(n+1) follows: ψ(1,0)(f)=diag(f ◦s ,f(ζ(n,1)),f(ζ(n,2)),...,f(ζ(n,t(n)))) n n where t(n) = m(n)−1 and {ζ(n,1),ζ(n,2),...,ζ(n,t(n))} contains a subset that divides the j-th circle into b(n,j) even arcs with b(n,j)≥a for each j. n ByidentifyingM withC⊗M ,wealsoassumethatψ(1,0)(1 )=φ(1,0)(1 ), k(n+1) k(n+1) n C(Xn)⊗Mk(n) n Mk(n) where φ(1,0) is the partial map of φ from M to M . Let m(n,1,i) be the multiplic- n n k(n) k(n+1) ity of the partial map of φ from M to the i-th summand of B . We may assume that n k(n) n+1 m(n,1,i)≥γ(n)n. We then define ψ(1,i)(f)=φ(1,i)(f)=⊕γ(n)diag(f(ξ(j,1)),f(ξ(j,2)),...,f(ξ(j,k ))), n n j=1 j 6 where {ξ(j,1),...,ξ(j,k )} is a set of k points on the j-th circle of X , from C(X ) ⊗ M j j n n k(n) into the i-th summand of B . We choose k so that γ(n)k = m(n,1,i). Moreover, we choose n j j=1 j {ξ(j,1),...,ξ(j,k )}sothatitis2π/ndensein(thej-thP)S1.Wemayalsoassumethatψ(1,i)(1 )= j n C(Xn)⊗Mk(n) φ(1,0)(1 ). n Mk(n) Now define ψ : A → A as follows: Define the partial map of ψ from C(X )⊗M n n n+1 n n k(n) to C(X )⊗M to be ψ(1,0). Define the partial map of ψ from C(X )⊗M to the n+1 k(n+1) n n n k(n) (1,i) i-th summand of B to be ψ . Define ψ | = φ | . Note, here, we identify M with n+1 n n Bn n Bn k(n+1) C⊗M ⊂C(X )⊗M . k(n+1) n+1 k(n+1) It is standardand easy to verify that C =lim (A ,ψ ) is a unital simple C∗-algebra with n→∞ n n tracial rank zero (for example, 3.7.8 and 3.7.9 of [19] Note that (K (A ),K (A ) ,[1 ])=(K (M ⊕B ),K (M ⊕B ) ,[1 ]). 0 n 0 n + An 0 k(n) n 0 k(n) n + Mk(n)⊕Bn Moreover (ψ ) =(φ ) . One also have n ∗0 n ∗0 K (A )=Gγ(n) 1 n and (ψ ) =κ . It follows that n ∗1 n (K (C),K (C) ,[1 ],K (C))=(K (A),K (A) ,[1 ],K (A)). 0 0 + C 1 0 0 + A 1 Since C has tracial rank zero and is an AH-algebra (so it satisfies the UCT), by [21], C ∼=A. The lemma follows from the construction. Theorem 3.5. Let A bea unital simple AT-algebra with realrank zero and h:K (A)→K (A) be 1 1 an isomorphism. Then there exists an automorphism α : A → A which satisfies the tracial cyclic Rokhlin property such that α =h and α =id . ∗1 ∗0 K0(A) Proof. If K (A)={0}, it suffices to show that there exists an automorphism on A which has the 1 Rokhlin property. But that follows from [31]. Therefore, for the rest of the proof, we may assume that K (A)6={0}. 1 Since K (A) is tosion free, we may write that K (A) = lim (G ,κ ), where each G is a 1 1 n→∞ n n n direct sum of finitely many copies of Z and κ are injective. n Leth:K (A)→K (A)bethe givenisomorphismandleth−1 bethe inverse. We mayassume, 1 1 by passingto a subsequenceif necessary,that there arehomomorphismsh ,h¯ :G →G such n n n n+1 that h◦κ =κ ◦h and h−1◦κ =κ ◦¯h n,∞ n+1,∞ n n,∞ n+1,∞ n on G . By passing to a subsequence if necessary, we may assume that, n h ◦κ =κ ◦h , h¯ ◦h =κ ◦κ and h ◦h¯ =κ ◦κ . (e3.3) n+1 n n+1 n n+1 n n+1 n n+1 n n+1 n Suppose that G is c(n) copies of Z. Homomorphisms h , h−1 and κ can be represented by n n n n matrices (a(n)), (b(n)) and (c(n)) where a(n), b(n) and c(n) are integers. i,j i,j i,j i,j i,j i,j Let J(n)=max{|a(n)|+|b(n) |+|c(n) |: i,j,i′,j′,i′′,j′′}. i,j i′,j′ i′′,j′′ We identify X with γ copies of the unit circle with a common point at 1. Denote u(n,i) ∈ n n C(X ) the function on X which is the identity map on the ith circle and 1 everywhere else. n n 7 Define γ(n+1) β (u(n,i))= z(n+1,j,i), n Y j=1 where z(n + 1,j,i) = u(n + 1,j)ai(n,j), if a(n) 6= 0, and z(n + 1,j,i) = u(n,i)(1) if a(n) = 0, i,j i,j i=1,2,...,γ(n). Note β gives a homomorphism from C(X )→C(X ) such that (β ) =h . n n n+1 n ∗1 n Define γ(n+1) β¯ (u(n,i))= z′(n+1,j,i), n Y j=1 where z′(n + 1,j,i) = u(n + 1,j)bi(n,j), if b(n) 6= 0, and z′(n + 1,j,i) = u(n,i)(1) if b(n) = 0, i,j i,j i=1,2,...,γ(n). We now write A=lim (A ,φ ), where A =C(X )⊗M ⊕B which satisfying (1)-(6) n→∞ n n n n k(n) n in Lemma 3.4 with a =J(n)n We will also use β :A →A for the homomorphism defined n n n n+1 by (β )| =β ⊗id and (β )| =φ | . n C(Xn)⊗Mk(n) n Mk(n) n Bn n Bn Let F ⊂F be a sequence of finite subsets of A such that the union of those finite subsets n n+1 is dense inA. Without loss ofgenerality,we may assumethat F ⊂φ (G ), where G is a finite n n,∞ n n subset of A . n It follows from [31] that there is an approximately inner automorphism σ ∈Aut(A) which has the tracial Rokhlin property. It follows from [25] that σ has tracial cyclic Rokhlin property. Since σ is approximately inner, by passing to a subsequence if necessary, by Lemma 3.4, we may assume that, there is a unitary v ∈A such that n n+1 σ◦φ (f)≈ adφ (v ) ◦φ (f) and (e3.4) n,∞ 1/2n+2 n+1,∞ n n,∞ σ−1φ (f)≈ adφ (v∗)◦φ (f) on G . (e3.5) n,∞ 1/2n+2 n+1,∞ n n,∞ n Define σ : A → A by σ (f) = adv (f) and δ : A → A by δ (f) = adv∗(f) for n n+1 n+1 n n n n+1 n+1 n n f ∈A . n Since a = J(n)n, by passing to a subsequence if necessary, we may assume that φ(1,1) : n n C(X )⊗M →C(X )⊗M has the following form: n k(n) n+1 k(n+1) φ(1,1)(f)=diag(f ◦s ,f(ζ(n,1)),f(ζ(n,2),...,f(ζ(n,t(n))), n n where s is a standard map (see Definition 3.3) and {ζ(n,1)),ζ(n,2),...,ζ(n,t(n)} contains sub- n sets {ξ ,...,ξ }, {ξ ,....ξ },...,{ξ ,...,ξ } such that {ξ ,...,ξ } evenly 1 N N+1 2N (γ(n)−1)N+1 γ(n)N (j−1)N+1 jN divide the j-th copy of S1 and N =Λ(n,G ). n Define a monomorphism ω :C(X )⊗M →A as follows: Define n n k(n) n+2 ω(1,1)(f)=β (g )+φ (g ), n n+1 1 n+1 2 where g = σ (diag(f ◦s ,0···0)) and g = σ (0,f(ζ(n,1)),f(ζ(n,2)),...,f(ζ(n,t(n))). Define 1 n n 2 n ω(1,i) =φ(1,i)◦φ | .LetE =φ (d ),whered istheidentityofC(X )⊗M . n n+1 n C(Xn)⊗Mk(n) n+2 n,n+2 n n n k(n) Denote by t(n) E′ =diag(1,0,...,0)∈C(X )⊗M . n+1 n+1 k(n+1) z }| { 8 Define a monomorphism ω¯ :C(X )⊗M →E A E as follows: n n k(n) n+2 n+2 n+2 Define ω¯(1,1)(f)=β¯ (g′)+φ (g′), n n+1 1 n+1 2 where g′ = δ (diag(f ◦ s ,0···0)) and g′ = δ (0,f(ζ(n,1)),f(ζ(n,2),...,f(ζ(n,t(n))). Define 1 n n 2 n (1,i) (1,i) ω¯ =φ ◦φ | . n n+1 n C(Xn)⊗Mk(n) For any unitary u∈A , define Φ:C(X )⊗M →A by 4n+3 4n+1 k(4n+1) 4n+5 Φ(f)=β¯ ◦δ ◦φ ◦adu◦β (g ). 4n+4 4n+3 4n+3 4n+2 1 Since (β¯ ◦δ ◦φ ◦adu◦β ◦φ ) =(φ ) , 4n+4 4n+3 4n+3 4n+2 4n+1 ∗1 4n+1,4n+5 ∗1 (Φ) =(φ ◦E′ φ E′ ) . ∗1 4n+2,4n+5 4n+1 4n+1 4n+1 ∗1 BychoiceofΛ(n,G ),by(e3.3)andbyapplyingLemma3.2,weobtainunitariesu ∈E A E n k k k k such that (with u =d ) 1 1 adu ◦ω¯ ◦adu ◦ω (f)≈ φ (f) on d G (e3.6) 4(n+1)+1 4n+3 4n+3 4n+1 1/2n 4n+1,4(n+1)+1 4n+1 4n+1 as well as adu ◦ω ◦adu ◦ω¯ (f)≈ φ (f) on d G (e3.7) 4(n+1)+3 4(n+1)+3 4n+3 4n+3 1/2n 4n+3,4(n+1)+3 4n+3 4n+3 (n=0,1,2,...) Define ω′ =adu ◦ω and ω′ =adu ◦ω¯ . 4n+1 4n+3 4n+1 4n+3 4n+5 4n+3 Now define α :A →A by n 4n+1 4n+3 (α )| =ω′ and (α )| =(φ ◦σ ◦φ )| n C(X4n+1)⊗Mk(4n+1) 4n+1 n B4n+1 4n+2 4n+1 4n+1 B4n+1 and define α¯ :A →A by n 4n+3 4n+5 (α¯ )| =ω′ and (α¯ )| =(φ ◦δ ◦φ )| n C(X4n+3)⊗Mk(4n+3) 4n+3 n B4n+3 4n+4 4n+3 4n+3 B4n+3 From (e3.4) and (e3.5), we have that α¯ ◦α (f)≈ φ (f) on f ∈(1−d )G (e3.8) n+1 n 1/2n+1 4n+1,4(n+1)+1 4n+1 4n+1 α ◦α¯ (f)≈ φ (f) on f ∈(1−d )G (e3.9) n+1 n 1/2n+1 4n+3,4(n+1)+3 4n+1 4n+1 It follows from (e3.6), (e3.7), (e3.8) and (e3.9), one has the following approximately inter- twining: A −φ1→,5 A −φ5→,9 A φ−9→,13 A ··· 1 5 9 17 ց ր ց ր ց ր α1 α¯1 α2 α¯2 α3 α¯3 A −φ3→,7 A φ−7→,11 A φ−11→,15··· 3 7 11 ItfollowsfromtheElliottapproximatelyintertwiningargumentthat{α }and{α¯ }definetwo n n automorphisms α and α−1. 9 Moreover,one checks that α =id and α =h. ∗0 K0(A) ∗1 It remains to show that α has tracial cyclic Rokhlin property. Fix ε>0, m>0 and a finite subset F ⊂A. Since σ has tracial cyclic Rokhlin property, there are mutually orthogonal projections e ,e ,...,e such that (with e =e ) 1 2 m m+1 1 ke a−ae k<ε/8, i=1,2,...,m,a∈F (e3.10) i i kσ(e )−e k<ε/8, i=1,2,...,m (e3.11) i i+1 and m τ(1− e )<ε/8 (e3.12) i X i=1 for all τ ∈T(A). Since σ is approximately inner, there is a unitary v ∈A such that kv∗e v−e k<ε/8, i=1,2,...,m. i i+1 Without loss of generality, we may assume that there is a finite subset G ⊂ A such that 4n+1 F ⊂φ (G) for some sufficiently large n. 4n+1,∞ It follows from the standard argument (See [34, Proposition L.2.2] for example.) that for any δ >0, (for sufficiently large n) there are projections q ∈A such that i 4n+1 ke −φ (q )k<min{δ,ε/8}, i=1,2,...,m. (e3.13) i 4n+1,∞ i Since φ is assumed to be injective, with sufficiently small δ, from [19, Lemma 2.5.6] we may n,∞ assumethatq ,q ,...,q aremutuallyorthogonal. Letq¯ =φ (q ).Wemayalsoassumethat 1 2 m i 4n+1,∞ i there is a unitary v′ ∈A such that φ (v′)=v. 4n+1 4n+1,∞ With even larger n, we may assume that, without loss of generality, α−i(F), α−i(q¯)∈φ (A ), i=1,2,...,m. (e3.14) i 4n+1,∞ 4n+1 Note that (1−d ) commutes with every element in A . Put p = q¯αi−1(φ (1− 4n+1 4n+1 i i 4n+1,∞ d )), i=1,2,...,m. Since, by (e3.14), 4n+1 αi−1◦φ (1−d )q¯ =αi−1(φ (1−d )α−i+1(q¯)) 4n+1,∞ 4n+1 i 4n+1,∞ 4n+1 i =αi−1(α−i+1(q¯)φ (1−d ))=q¯αi−1◦φ (1−d ), i 4n+1,∞ 4n+1 i 4n+1,∞ 4n+1 {p ,p ,...,p } are mutually orthogonalprojections. 1 2 m By (e3.14), if a∈F, p a=q¯αi−1(φ (1−d )α−i+1(a))=q¯aαi−1(φ (1−d )). i i 4n+1,∞ 4n+1 i 4n+1,∞ 4n+1 Therefore, by (e3.13) and (e3.10), we have (1)kp a−ap k=kq¯aαi−1((φ (1−d )))−ap k<ε/8+ε/8+ε/8<ε, i=1,2,...,m. i i i 4n+1,∞ 4n+1 i 10