Algebras Associated to Integer Dilation Matrices Ruy Exel Universidade Federal De Santa Catarina

University of Wollongong Research Online Faculty of Engineering and Information Sciences - Faculty of Engineering and Information Sciences Papers: Part A

2011 Purely infinite simple C*-algebras associated to integer dilation matrices Ruy Exel Universidade Federal de Santa Catarina

Astrid an Huef University of Otago

Iain Raeburn University of Wollongong, [email protected]

Publication Details Exel , R., Huef, A. an. & Raeburn, I. (2011). Purely infinite simple C*-algebras associated to integer dilation matrices. Indiana University Mathematics Journal, 60 (3), 1033-1058.

Research Online is the open access institutional repository for the University of Wollongong. For further information contact the UOW Library: [email protected] Purely infinite simple C*-algebras associated to integer dilation matrices

Abstract Given an n×n integer matrix A whose eigenvalues are strictly greater than 1 in absolute value, let σA be the transformation of the n-torus Tn = Rn/Zn defined by σA(e2πix) = e2πiAx for x ∈ Rn. We study the associated crossed-product C∗-algebra, which is defined using a certain transfer operator for σA, proving it to be simple and purely infinite and computing its K-theory groups.

Keywords dilation, integer, associated, matrices, algebras, purely, c, simple, infinite

Disciplines Engineering | Science and Technology Studies

This journal article is available at Research Online: http://ro.uow.edu.au/eispapers/862 Purely Inﬁnite Simple C∗-Algebras Associated to Integer Dilation Matrices

RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

ABSTRACT. Given an n × n integer matrix A whose eigenvalues are strictly greater than 1 in absolute value, let σA be the trans- n n n πix formation of the n-torus T = R /Z defined by σA(e2 ) = e2πiAx for x ∈ Rn. We study the associated crossed-product C∗-algebra, which is defined using a certain transfer operator for σA, proving it to be simple and purely infinite and computing its K-theory groups.

1. INTRODUCTION Exel has recently introduced a new kind of crossed product for an endomorphism α of a C∗-algebra B [6]. The crucial ingredient in his construction is a transfer operator, which is a positive linear map L : B → B satisfying L(α(a)b) = aL(b). In the motivating example, B = C(X), X is a compact Hausdorﬀ space, α is the endomorphism α : f f ◦ σ associated to a covering map σ : X → X,andL is deﬁned by L(f )(x) = 1 f(y) (1.1) |σ −1({x})| σ(y)=x

(see [3,7,8], for example; transfer operators themselves have been used in dynam- ics for many years [26]). Exel’s crossed product Bα,L N can be constructed in several ways, but here we view it as the Cuntz-Pimsner algebra O(ML) ofaright- Hilbert B-bimodule ML constructed from L, as discussed in [2] (see also Section 2.2 below). We became interested in this circle of ideas when we noticed that the bimod- N ule ML associated to the covering map σ : z z of the unit circle T plays a key

1033 Indiana University Mathematics Journal c , Vol. 60, No. 3 (2011) 1034 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN role in work of Packer and Rieffel on projective multi-resolution analyses [18]– [21].Themoduleelementsm ∈ ML such that m, m is the identity of C(X) are precisely the quadrature mirror filters arising in signal processing and wavelet theory, and orthonormal bases for ML are what engineers call “filter banks with perfect reconstruction” (as observed and exploited in [16]and[10], for example.) We then noticed further, using results from [8], that the associated crossed product C(T) N α C(T) σ αN ,L ,where N is the endomorphism of given by ,issimple,and accordingly computed its K-theory, finding that K0 = Z ⊕ (Z/(N − 1)Z) and K1 = Z.ButthenwesawthisK-theory occurring elsewhere, and we gradually realised C∗ C(T) N that the -algebra αN ,L had already been studied by many authors under other guises. (An almost certainly incomplete list includes [1, Theorem 2.12; 5,Example3;12, Example 4.1; 14, Appendix A; 29, Theorem 2.1] Multiplication by N, however, is just one of many dilations of interest in wavelet theory (see, for example, [27]). Here we consider the covering maps σA of Td = Rd/Zd induced by integer matrices A whose eigenvalues λ satisfy |λ| > 1, d and the crossed products of the associated systems (C(T ), αA,L),whereαA is d the endomorphism of C(T ) given by σA. We will show, using results from [8]and[14], that the crossed products C(Td) N αA,L are simple and purely infinite, and hence by the Kirchberg-Phillips theorem are classified by their K-theory. The computation of the K-theory groups C(Td) N of αA,L therefore has a special significance, and one of the main goals of this paper is to carry out this computation. C(Td) N Since αA,L is a Cuntz-Pimsner algebra, one should in principle be able to compute its K-theory using the exact sequence of [23, Theorem 4.8], but in practice we were not able to compute some of the homomorphisms in that sequence. So we have argued directly from the six-term exact sequence associated to the Toeplitz algebra of the bimodule ML, and we hope that our computation will be of independent interest. Our computation is based on a six-term exact sequence which is valid for any system (B,α,L) for which the bimodule ML is free as a right Hilbert B-module. Using an orthonormal basis for ML, we build a homomorphism Ω : B → MN (B) which has the property that Ω ◦ α(a) is the diagonal matrix a1N with N copies of a down the diagonal, and which we view as a K-theoretic left inverse for α. When the bimodule is obtained from an integral matrix A,asabove,thismapisclosely associated to the classical adjoint of A. We then show that there is an exact sequence

−Ω j K (B) id ∗ /K (B) B∗ /K (O(M )) 0 O 0 0 L

ojB∗ oid −Ω∗ K1(O(ML)) K1(B) K1(B) C∗-Algebras Associated to Dilation Matrices 1035 in which jB is the canonical embedding of B in the Cuntz-Pimsner algebra O(ML). d d When (B,α,L)= (C(T ), αA,L),weknowfrom[20]thatC(T )L is free, so this d ∗ d exact sequence applies; since we also know from [11]thatK∗(C(T )) = K (T ) is isomorphic to the exterior ring generated by a copy of Zd in K1(Td),wecanin this case compute Ω∗, and derive explicit formulas for Ki(O(ML)).

2. CROSSED PRODUCTS BY ENDOMORPHISMS 2.1. Cuntz-Pimsner algebras. A right-Hilbert bimodule over a C∗-algebra B,alsoknownasacorrespondence,isarightHilbertB-module M with a left action of B implemented by a homomorphism ϕ of B into the C∗-algebra L(M) of adjointable operators on M.InthispaperB is always unital, the bimodule M is always essential in the sense that 1 · m = m for m ∈ M, and the bimodule has a ﬁnite Parseval frame or quasi-basis: a ﬁnite subset {mj :0≤ j ≤ N − 1} for which we have the reconstruction formula

N−1 (2.1) m = mj ·mj,m for every m ∈ M. j=0 The reconstruction formula can be reformulated in terms of the rank-one operators Θm,n : x m ·n, x as

N−1 ϕ(a) = Θ a ∈ B (2.2) a·mj ,mj for every j=0 and hence implies that the homomorphism ϕ takesvaluesinthealgebraK(M) of compact operators on M. The obvious examples of Parseval frames are orthonormal bases: Lemma 2.1. Suppose that {mj :0≤ j ≤ N − 1} are vectors in a right-Hilbert ∗ bimodule M over a unital C -algebra B.Ifthemj generate M as a Hilbert B-module and satisfy mj,mk=δj,k1B ,then{mj :0≤ j ≤ N −1} is a ﬁnite Parseval frame N for M,andm (mj,m)j is an isomorphism of M onto B . Proof. A quick calculation gives the reconstruction formula for m of the form mk · b, and then linearity and continuity give it for arbitrary m. For the last (b ,...,b ) m · b assertion, check that 0 N−1 j j j is an inverse. ❐

Remark 2.2. If P ∈L(M) is a projection and {nj} is an orthonormal basis for M,then{Pnj} is a Parseval frame for P(M), and Frank and Larson have shown that every Parseval frame {mj} has this form because m (mj,m)j is an isomorphism of M onto a complemented submodule of BN [9, Theorem 5.8]. However, many interesting bimodules have Parseval frames but are not obviously presented as direct summands of free modules. For example, for a bimodule of the form C(X)L, one can construct a Parseval frame directly using a partition of unity (see, for example, [8, Proposition 8.2]). 1036 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

A Toeplitz representation of a right-Hilbert bimodule M in a C∗-algebra C consists of a linear map ψ : M → C and a homomorphism π : B → C satisfying ψ(m)∗ψ(n) = π(m, n) and ψ(ϕ(a)m) = π(a)ψ(m);wethenalsohave ψ(m · a) = ψ(m)π(a).TheToeplitz algebra T (M) is generated by a universal Toeplitz representation (iM ,iB) of M (either by theorem [23] or by deﬁnition [9]). The following lemma is implicit in the proof of [2, Corollary 3.3]. Lemma 2.3. Suppose M is an essential right-Hilbert bimodule over a unital C∗- algebra B and (ψ, π) is a Toeplitz representation of M on a Hilbert space H .Then the subspace π(1)H is reducing for (ψ, π),and

(ψ, π) = (ψπ(1)H ⊕ 0,ππ(1)H ⊕ 0).

Proof. It is standard that π = ππ(1)H ⊕ 0, and each ψ(m) = ψ(1 · m) = π(1)ψ(m) has range in π(1)H ,soitsuﬃces to show that h ⊥ π(1)H implies ψ(m)h = 0. Suppose h ⊥ π(1)H .Thenπ(m, m)h ∈ π(1)H ,sothat

ψ(m)h 2 = (ψ(m)h | ψ(m)h) = (ψ(m)∗ψ(m)h | h)

= (π(m | m)h | h) = 0. ❐ Remark 2.4. Lemma 2.3 implies that the Toeplitz algebra T (M) is universal for Toeplitz representations (ψ, π) in which π is unital, and we shall assume from now on that in all Toeplitz representations (ψ, π), π is unital. For every Toeplitz representation (ψ, π) of M, there is a unique representation (ψ, π)(1) of the algebra K(M) of compact operators on M such that

(1) ∗ (ψ, π) (Θm,n) = ψ(m)ψ(n) for m, n ∈ M (see, for example, [9, Proposition 1.6]). When1 ϕ : B →L(M) has range in K(M), we say that (ψ, π) is Cuntz-Pimsner covariant if π = (ψ, π)(1) ◦ ϕ,and the Cuntz-Pimsner algebra O(M) is the quotient of T (M) which is universal for Cuntz-Pimsner covariant representations. The algebra O(M) is generated by a canonical Cuntz-Pimsner covariant representation (jM ,jB). Now we investigate what this all means when M has an orthonormal basis. Compare with [7, Section 8] and [8, Proposition 7.1], which use quasi-bases. Lemma 2.5. Suppose that M is an essential right-Hilbert bimodule over a unital ∗ C -algebra B,andthat{mj :0≤ j ≤ N − 1} is an orthonormal basis for M.Let (ψ, π) beaToeplitzrepresentationofM.Then {ψ(m ) ≤ j ≤ N − } (1) j :0 1 is a Toeplitz-Cuntz family of isometries such that N−1 ψ(m )ψ(m )∗ π(a) j=0 j j commutes with every ; and (2) (ψ, π) is Cuntz-Pimsner covariant if and only if {ψ(mj) :0≤ j ≤ N −1} is a Cuntz family.

1As is always the case here; when the left action on the bimodule M contains non-compact operators, there are several competing deﬁnitions of O(M). C∗-Algebras Associated to Dilation Matrices 1037

∗ Proof. (1) The relations ψ(mj) ψ(mj) = π(mj,mj) = π(1) and our π( ) = ψ(m ) convention that 1 1(seeRemark 2.4) imply that the j are isometries. a ∈ B q = N−1 ψ(m )ψ(m )∗ Next, we ﬁx ,let : j=0 j j , and compute the following using the reconstruction formula (2.1):

N−1 ∗ ∗ (2.3) qπ(a)q = ψ(mj )ψ(mj) π(a)ψ(mk)ψ(mk) j,k=0 N−1 ∗ = ψ(mj )π(mj,a· mk)ψ(mk) j,k=0 N−1 N−1 ∗ = ψ(mj ·mj,a· mk) ψ(mk) k=0 j=0 N−1 ∗ = ψ(a · mk)ψ(mk) = π(a)q. k=0

Taking a = 1in(2.3)showsthatq2 = q, and since q is self-adjoint, it is a ∗ projection. Since each ψ(mj) is an isometry, each ψ(mj)ψ(mj ) is a projection, and since their sum is a projection, their ranges must be mutually orthogonal. Thus {ψ(mj)} is a Toeplitz-Cuntz family. Next we use (2.3)againtoseethat qπ(a) = (π(a∗)q)∗ = (qπ(a∗)q)∗ = qπ(a)q = π(a)q,andwehaveproved (1). (ψ, π) (2) Suppose that is Cuntz-Pimsner covariant. Plugging the formula (2.2) (1) ∗ for a = 1into(ψ, π) (ϕ(1)) = π(1) = 1showsthat j ψ(mj )ψ(mj) = 1, so {ψ(mj)} is a Cuntz family. On the other hand, if {ψ(mj)} is a Cuntz family, then we can deduce from (2.2)that

N−1 N−1 ( ) ∗ ∗ ∗ (ψ, π) 1 (ϕ(a)) = ψ(mj )ψ(a ·mj) = ψ(mj)ψ(mj ) π(a) = π(a), j=0 j=0

and (ψ, π) is Cuntz-Pimsner covariant. ❐ 2.2. Exel systems and crossed products. Let α be an endomorphism of aunitalC∗-algebra B.Atransfer operator L for (B, α) is a positive linear map L : B → B such that L(α(a)b) = aL(b) for all a, b ∈ B.Wecallthetriple (B,α,L)an Exel system. Given an Exel system (B,α,L), we construct a right-Hilbert B-module ML over B as in [6]and[2]. Let BL be a copy of the underlying vector space of B. Deﬁne a right action of a ∈ B on m ∈ BL by m · a = mα(a),andaB-valued pairing on BL by ∗ m, n=L(m n) for m, n ∈ BL. 1038 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

Modding out by {m : m, m=0} and completing yields a right Hilbert B- module ML. The action of B by left multiplication on BL extends to an action of B by adjointable operators on ML which is implemented by a unital homomorphism ϕ : B →L(ML),andthusmakesML into a right-Hilbert bimodule over B. Exel’s crossed product is constructed in two stages. First he forms a Toeplitz algebra T (B,α,L), which is isomorphic to T (ML) (see [2, Corollary 3.2]). Then the crossed product Bα,L N is the quotient of T (ML) by the ideal generated by the elements

i (a) − (i ,i )(1)(ϕ(a)) a ∈ K = ϕ−1(K(M )) ∩ Bα(B)B B ML B for α : L

(see [2, Lemma 3.7]). When ML has a finite Parseval frame and the projection − α(1) is full, we have ϕ 1(K(ML)) = B = Kα,andBα,L N is the Cuntz-Pimsner algebra O(ML). For us, the main examples of Exel systems come from surjective endomorphisms σ of a compact group K with finite kernel: the corresponding Exel system (C(K), α, L) has α(f ) = f ◦ σ and L defined by averaging over the fibres of σ ,as in (1.1). The next lemma is a mild generalisation of [20, Proposition 1]. Lemma 2.6. Suppose that σ : K → K is a surjective endomorphism of a compact abelian group K with N :=|ker σ | < ∞,and(C(K), α, L) is the corresponding Exel system. Then the norm on C(K)L defined by the inner product is equivalent to the usual sup-norm, and C(K)L is complete. It has an orthonormal basis {mj :0≤ j ≤ N − 1}. Proof. The assertions about the norm and the completeness are proved in ∧ [16, Lemma 3.3], for example. Since γ γ|ker σ is surjective and |(ker σ) |= | ker σ |=N,wecanfindasubset{γi :0≤ i ≤ N − 1} of Kˆ such that {γi|ker σ } is all of (ker σ)∧.Then γ ,γ (k) = 1 γ ( )γ ( ) i j L N i j σ( )=k = 1 γ (ζ )γ (ζ ) σ( ) = k N i 0 j 0 for any fixed 0 such that 0 ζ∈ker σ = 1 γ ( )γ ( ) γ γ (ζ). N i 0 j 0 i j ζ∈ker σ

−1 If i = j,then(γi γj)|ker σ is a nontrivial character of ker σ , and its range is a nontrivial subgroup of T, so the sum vanishes. If i = j, then the sum is N.So {γj} is orthonormal. We still need to see that {γj} generates C(K)L as a Hilbert module. The Stone-Weierstrass theorem implies that the characters of K span a dense ∗-sub- algebra of C(K), and hence by the equivalence of the norms, they also span a dense subspace of C(K)L.Soitsuﬃces to show that each γ ∈ Kˆ is in the submodule C∗-Algebras Associated to Dilation Matrices 1039

∧ generated by {γj}.Since(ker σ) ={γj},thereexistsj such that γ|ker σ = γj. −1 −1 Then γj γ vanishes on ker σ , and there is a character χ such that γj γ = χ ◦ σ . This equation unravels as γ = γj(χ ◦ σ) = γjα(χ) = γj · χ,soitimpliesthatγ belongs to the submodule generated by {γj}. ❐

Example 2.7. Suppose that A ∈ Md(Z) is an integer matrix with | det A| > 1, d πix πiAx d and σA is the endomorphism of T given by σA(e2 ) = e2 for x ∈ R . d d Then σA is surjective (because A : R → R is), and ker σA has N :=|det A| d elements. A function m ∈ C(T )L such that m, m(z) = 1forallz is called a d quadrature mirror filter for dilation by A, and an orthonormal basis for C(T )L is a filter bank. Lemma 2.6 says that for every A,filterbanksexist. Remark 2.8. Although the dilation matrices A are greatly relevant to wavelets, the filter banks we constructed in the proof of Lemma 2.6 are not the kind which are useful for the construction of wavelets. There one wants the first filter m0 to be 1/2 low-pass, which means roughly that m0(1) = N , m0 is smooth near 1, and m0 does not vanish on a sufficiently large neighbourhood of 1; for the basis in Lemma 2.6,wehave|m0(z)|=1forallz,andm0 is all-pass.Thematrix completion problem considered in [20] asks whether, given a low-pass filter m0, one can find afilterbank{mj} which includes the given m0. This amounts to asking that the m⊥ ={m ∈ C(Td) m, m = } submodule 0 : L : 0 0 is free. In [20, Section 4], Packer and Rieffel show by example that it need not be free if | det A| > 2and d> m⊥ 4. Of course, since 0 is a direct summand of a free module, it always has a Parseval frame. When α is the endomorphism of C(K) coming from a surjective endomorphism σ of K, we know from Lemma 2.6 that ML = C(K)L admits an orthonormal basis, and the associated endomorphism α : f f ◦ σ is unital, so α(1) = 1 is certainly full. Thus for the systems of interest to us, Exel’s crossed product Bα,L N is isomorphic to the Cuntz-Pimsner algebra O(ML). We will use this identification without comment.

3. THE SIX-TERM EXACT SEQUENCE We assume throughout this section that (B,α,L) is an Exel system and that {mj :0≤ j ≤ N − 1} is a Parseval frame for ML.WewriteQ for the quotient map from T (ML) →O(ML),and(ψ, π) for the universal Toeplitz covariant representation of ML in T (ML). To construct our exact sequence for K∗(O(M)),weanalysethesix-termexact sequence

ι Q K ( Q) ∗ /K (T (M )) ∗ /K (O(M )) (3.1) 0 kerO 0 L 0 L

δ1 δ0 oQ∗ oι∗ K1(O(ML)) K1(T (ML)) K1(ker Q). 1040 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

We begin by recalling from [23, Theorem 4.4] that the homomorphism π : B → T (ML) induces an isomorphism of Ki(B) onto Ki(T (ML)), so we can replace Ki(T (ML)) with Ki(B) provided we can identify the maps. Next we introduce our “K-theoretic left inverse” for α, and then we will work towards showing that B is a full corner in ker Q, so that we can replace Ki(ker Q) with Ki(B). Part (3) of the next result will not be used in this section; it is included here because it shows how Ω relates to α, and gives a hint of why we view it as a “K- theoretic left inverse” for α.(Rieffel used a similar construction in the proof of [24, Proposition 2.1], but the technicalities seem different.) Lemma 3.1. Define Ω : B → MN (B) by Ω(a) = (mj,a· mk)j,k.Then (1) Ω is a homomorphism of C∗-algebras; (2) Ω is unital if and only if {mj :0≤ j ≤ N − 1} is an orthonormal basis; (3) if B is commutative and {mj :0≤ j ≤ N − 1} is orthonormal, then Ω(α(a)) is the diagonal matrix a1N with diagonal entries a. Proof. For (1), we let a, b ∈ B and compute first

N−1 (Ω(a)Ω(b))j,k = mj,a· m m ,b· mk =0 N−1 = mj,a· m ·m ,b· mk =0

=mj,a· (b · mk)=Ω(ab)j,k, and then

∗ ∗ ∗ ∗ Ω(a ) = (mj,a ·mk)j,k = (a·mj ,mk)j,k = (mk,a·mj )j,k = Ω(a) .

Part (2)iseasy.For(3), we let qL : BL → ML be the quotient map, and consider m = q(b) ∈ q(BL). Then commutativity of B gives

m · a = q(b · a) = q(bα(a)) = q(α(a)b) = α(a) · q(b) = α(a) · m, and this formula extends to m ∈ ML by continuity. Thus

Ω(α(a))j,k =mj,α(a)· mk=mj,mk · a=mj,mka = δj,ka,

as required. ❐ ⊗i To describe ker Q, we need some standard notation. We write ML for the i-fold internal tensor product ML ⊗B ··· ⊗B ML, which is itself a right-Hilbert B (ψ⊗i,π) M⊗k T (M ) bimodule over . There is a Toeplitz representation of L in L ψ⊗k(ξ) = i ψ(ξ ) ξ = ξ ⊗ ··· ⊗ ξ such that i=1 i for elementary tensors 1 k in C∗-Algebras Associated to Dilation Matrices 1041

⊗k ⊗0 ML (see [9, Proposition 1.8], for example). By convention, we set ML := B and ψ⊗0 := π.Thenfrom[9, Lemma 2.4] we have ⊗k ⊗ ∗ ⊗k ⊗ (3.2) T (ML) = span ψ (ξ)ψ (η) : k, ≥ 0,ξ∈ ML ,η∈ ML . q = N−1 ψ(m )ψ(m )∗ We also recall from Lemma 2.5 (1) that the element : j=0 j j of T (ML) is a projection which commutes with every π(a). Lemma 3.2. With the preceding notation, we have − q = − N−1 ψ(m )ψ(m )∗ Q (1) 1 1 j=0 j j is a full projection in ker ; ⊗k ⊗k (2) (1 − q)ψ (ξ) = 0 for all ξ ∈ ML with k ≥ 1; and ⊗k ⊗ ∗ ⊗k ⊗ (3) ker Q = span{ψ (ξ)(1 − q)ψ (η) : k, ≥ 0,ξ∈ ML ,η∈ ML }. ϕ(a) = N−1 Θ Proof. (1) The reconstruction formula implies that j=0 a·mj ,mj , and so

N−1 (1) ∗ (3.3) (ψ, π) (ϕ(a)) = ψ(a · mj)ψ(mj ) = π(a)q. j=1

This implies in particular that

Q(1 − q) = Q(π(1) − π(1)q) = Q(π(1) − (ψ, π)(1)(ϕ(1))) = 0, so 1−q belongs to ker Q.SincekerQ is by deﬁnition the ideal in T (ML) generated by the elements π(a)− (ψ, π)(1)(ϕ(a)) for a ∈ B,(3.3) also implies that ker Q is generated by the elements π(a)(1 − q), and hence by the single element 1 − q. This says precisely that the projection 1 − q is full. ⊗1 (2) First we consider m ∈ ML = ML. The reconstruction formula gives

N−1 N−1 ∗ qψ(m) = ψ(mj)ψ(mj ) ψ(m) = ψ(mj ·mj,m) = ψ(m), j=0 j=0

( − q)ψ(m) = k> ξ = ξ ⊗···⊗ξ so 1 0. Now for 1 and an elementary tensor 1 k, ( − q)ψ⊗k(ξ) = ( − q)( i ψ(ξ )) = we have 1 1 i=1 i 0, and the result extends to arbitrary ξ ∈ M⊗k by linearity and continuity. (3) In consideration of part (2),weareabletodeducefrom(3.2)thatkerQ = T (ML)(1 − q)T (ML) is spanned by the elements of the form

ψ⊗k(ξ)π(a)∗(1 − q)π(b)ψ⊗ (η)∗ = ψ⊗k(ξ · a)(1 − q)ψ⊗l(η · b∗)∗

⊗k ⊗ for ξ ∈ M , η ∈ M and a, b ∈ B,whichgives(3). ❐ 1042 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

Lemma 3.3. There is a homomorphism ρ : B → ker Q such that ρ(a) = π(a)(1 − q),andρ is an isomomorphism of B onto (1 − q) ker Q(1 − q).

Proof. Lemma 3.2 says that π(a)(1 − q) belongs to ker Q, and Lemma 3.1 says that q commutes with every π(a), so there is a homomorphism ρ : B → (1 − q) ker Q(1 − q) ⊂ ker Q such that ρ(a) = π(a)(1 − q).Fromparts(2)and (3) of Lemma 3.2 we get

(1 − q) ker Q(1 − q) = span{(1 − q)ψ⊗k(ξ)(1 − q)ψ⊗l(η)∗(1 − q) : k, ≥ 0} = span{(1 − q)π(a)(1 − q)π(b)(1 − q) : a, b ∈ B} = span{(1 − q)π(ab) : a, b ∈ B}, which is precisely the range of ρ.Soρ is surjective. To see that ρ is injective we choose a faithful representation π0 : B → B(H ) and consider the Fock representation (ψF ,πF ) of ML induced from π0,asde- scribed in [9, Example 1.4]. The underlying space of this Fock representation is F(M ) ⊗ H = (M⊗k ⊗ H ) B M L B : k≥0 L B ; acts diagonally on the left, and L acts by ∗ creation operators. The crucial point for us is that each ψF (m) is an annihilation ⊗0 operator which vanishes on the subspace B ⊗B H=ML ⊗B H of F(ML) ⊗B H . Now suppose that a ∈ B.Then

0 = ψF × πF (ρ(a)) = ψF × πF (π(a)(1 − q)) N−1 ∗ = πF (a) 1 − ψF (mj )ψF (mj ) . j=0

∗ Since ψ(mj) vanishes on B ⊗B H ,wehave

N−1 ∗ ρ(a) = 0 =⇒ πF (a) 1 − ψF (mj )ψF (mj ) (1 ⊗B h) = 0forallh ∈H j=0 =⇒ πF (a)(1 ⊗B h) = 0forallh ∈H =⇒ a ⊗B h = 0forallh ∈H

=⇒ π0(a)h = 0forallh ∈H,

which implies that a = 0becauseπ0 is faithful. ❐

Lemma 3.3 implies that we can replace Ki(ker Q) in (3.1)byKi(B),asclaimed. Now we need to check what this replacement does to the map ι∗. C∗-Algebras Associated to Dilation Matrices 1043

Proposition 3.4. The following diagram commutes for i = 0:

id −Ω∗ / (3.4) Ki(B) Ki(B)

ρ∗ π∗ / Ki( Q) Ki(T (ML)). ker ι∗

If {mj :0≤ j ≤ N − 1} is orthonormal, then the diagram also commutes for i = 1. Since Ω : B → MN (B),theΩ∗ in the diagram is really the composition of Ω∗ : Ki(B) → Ki(MN (B)) with the isomorphism Ki(MN (B)) → Ki(B); the latter is induced by the map which views an element in Mr (MN (B)) as an element of MrN(B). The proof needs two standard lemmas. The ﬁrst says, loosely, that if we rewrite an r × r matrix of N × N blocks as an N × N matrix of r × r blocks, then the resulting rN × rN matrices are unitarily equivalent. We agree that this cannot be a surprise to anyone, and we apologise for failing to come up with more elegant notation. Lemma 3.5. Suppose that B is a C∗-algebra, r ≥ 1 and N ≥ 2 are integers, and

{bj,s;k,t :0≤ j,k ≤ N − 1and0≤ s,t < r} is a subset of B.Form, n satisfying 0 ≤ m, n ≤ rN − 1,wedeﬁne

cm,n = bj,s;k,t where m = sN + j and n = tN + k, and

dm,n = bj,s;k,t where m = jr + s and n = kr + t. Then there is a scalar unitary permutation matrix U such that the matrices C := ∗ (cm,n) and D := (dm,n) are related by C = UDU . Proof. For 0 ≤ p,q ≤ rN − 1, we deﬁne 1ifthereexistk, t such that p = tN + k and q = kr + t, up,q = 0 otherwise.

Each row and column contains exactly one 1, so U := (up,q) is a scalar permutation matrix, and we can verify that both (CU)m,q and (UD)m,q are equal to bj,s;k,t where m = sN + j and q = kr + t,soCU = UD. ❐ Lemma 3.6. Suppose that S is an isometry in a unital C∗-algebra B.Then S − SS∗ U = 1 : 0 S∗ is a unitary element of M2(B) and its class in K1(B) is the identity. 1044 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

Proof. A straightforward calculation shows that U is unitary. Let T=C∗(v) be the Toeplitz algebra. By Coburn’s Theorem [4]thereisa homomorphism πS : T→B such that πS (v) = S.SinceK1(T ) = 0 (see, for example, [28, Remark 11.2.2]), v − vv∗ 1 = [ ] K (T ), 0 v∗ 1 in 1 and hence S − SS∗ v − vv∗ 1 = (π ) 1 0 S∗ S ∗ 0 v∗ = (π ) ([ ]) = [ ] K (B). S ∗ 1 1 in 1 ❐

Proof of Proposition 3.4. We start with i = 0. Let a = (as,t) be a projection in Mr (B).Forπ : A → B,wewriteπr for the induced homomorphism of Mr (A) into Mr (B).Thenwehave

ρ∗([a]) = [(ρ(ar,s)] = [(π(as,t)(1 − q))] = [(π(as,t))(1 − q)1r )] = [πr (a)((1 − q)1r )] = [πr (a)] − [(πr (a)(q1r ))], and π∗ ◦ (id −Ω∗)([a]) = [πr (a)] − π∗ ◦ Ω∗([a]), so it suﬃces to show that [πr (a)(q1r )] = π∗◦Ω∗([a]) in K0(T (ML)).Theclass π∗ ◦ Ω∗([a]) appears as the class of the r × r block matrix πrN(Ωr (a)) whose (s, t) entry is the N × N block (π(mj,as,t · mk))j,k. In other words, with bj,s;k,t = π(mj,as,t · mk),thematrixπrN(Ωr (a)) is the matrix C = (cm,n) in Lemma 3.5. We now consider the matrix T in MN (Mr (T (ML))) deﬁned by ⎛ ⎞ ψ(m ) r ··· ψ(mN− ) r ⎜ 0 1 1 1 ⎟ ⎜ ··· ⎟ (3.5) T = ⎝ 0r 0r ⎠ . . . . ··· .

∗ Computations show that TT = (q1r ) ⊕ 0r(N−1), and since πr (a) is a projection which commutes with q1r ,wededucethat(πr (a) ⊕ 0r(N−1))T is a partial isometry which implements a Murray-von Neumann equivalence between ∗ ∗ T (πr (a) ⊕ 0r(N−1))T and (πr (a) ⊕ 0r(N−1))T T = (πr (a)(q1r )) ⊕ 0r(N−1). Thus we have

[πr (a)(q1r )] = [πr (a)(q1r ) ⊕ 0r(N−1)] ∗ = [T (πr (a) ⊕ 0r(N−1))T ]. C∗-Algebras Associated to Dilation Matrices 1045

∗ Another computation shows that the (j, k) entry of T (πr (a) ⊕ 0r(N−1))T is the r × r matrix (πr (mj,as,t · mk1r ))s,t. Thus with the same choice of ∗ bj,s;k,t = π(mj,as,t · mk), T (πr (a) ⊕ 0r(N−1))T is the matrix D = (dm,n) in Lemma 3.5. Since unitarily equivalent projections have the same class in K0, we can therefore deduce from Lemma 3.5 that

∗ (3.6) [πr (a)(q1r )] = [T (πr (a) ⊕ 0r(N−1))T ] = [πrN(Ωr (a))] = π∗ ◦ Ω∗([a]).

Thus Diagram 3.4 commutes when i = 0. Now consider i = 1, where we assume in addition that {mj} is orthonormal. Let u be a unitary in Mr (B). To compute ρ∗ : K1(B) → K1(ker Q),weobserve that ρ is the composition of a unital isomorphism of B onto (1 − q) ker Q(1 − q), which takes [u] to [ρr (u)] = [πr (u)((1 − q)1r )], with the inclusion of (1 − q) ker Q(1 − q) as a full corner in the non-unital algebra ker Q,whichtakes[πr (u)((1 − q)1r )] to [πr (u)((1 − q)1r ) + q1r ] ∈ + K1((ker Q) ) = K1(ker Q). On the other hand,

π∗ ◦ (id −Ω∗)([u]) = [πr (u)] − [πrN ◦ Ωr (u)]. So we need to show that

(3.7) [(πr (u)((1 − q)1r ) + q1r ) ⊕ 1r(N−1)]

= [πr (u) ⊕ 1r(N−1)] − [πrN ◦ Ωr (u)] in K1(T (ML)). To this end, we note that the left-hand side of (3.7) is unchanged by pre- or post-multiplying by any invertible matrix C ∈ M2rN(T (ML)) whose K1 class is 1. In particular, we can do this when C is • a unitary of the form S − SS∗ C = 1 , 0 S∗

where S ∈ MrN(T (ML)) is an isometry (see Lemma 3.6); • an upper- or lower-triangular matrix of the form A C = 1 C = 10 01 or A 1 t 1 tA (which are connected to 12rN via 01 and its transpose); • any constant invertible matrix C in M2rN(C) (because GL2rN(C) is connected); this implies that we can perform row and column operations without changing the class in K1. 1046 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

Since {mj} is an orthonormal basis, the matrix T deﬁned at (3.5)isanisom- etry in MrN(T (ML)).Thus (πr (u)((1 − q)1r ) + q1r ) ⊕ 1r(N−1) ∗ (πr (u)((1 − q)1r ) + q1r ) ⊕ 1r(N−1) 0rN T 1rN − TT = ∗ 0rN 1rN 0rN T (πr (u)((1 − q)1r ) + q1r ) ⊕ 1r(N−1) 0rN T(1 − q)1r ⊕ 1r(N−1) = ∗ 0rN 1rN 0rN T ((πr (u)((1 − q)1r ) + q1r ) ⊕ 1r(N−1))T πr (u)((1 − q)1r ) ⊕ 1r(N−1) = ∗ , 0rN T which, since (1 − q)ψ(mi) = 0 by Lemma 3.2 (2), is Tπr (u)((1 − q)1r ) ⊕ 1r(N−1) = ∗ 0rN T ∗ Tπr (u)((1 − q)1r ) ⊕ 1r(N−1) 1rN T (πr (u) ⊕ 1r(N−1)) = ∗ 0rN T 0rN 1rN Tπr (u) ⊕ 1r(N−1) = ∗ 0rN T ∗ since TT = q1r ⊕ 0r(N−1) and (q1r )πr (u) = πr (u)(q1r ).Byanelementary row operation this is πr (u) ⊕ 1r(N−1) T = ∗ T 0rN −1 πr (u) ⊕ 1r(N−1) T 1rN −(πr (u ) ⊕ 1r(N−1))T = ∗ T 0rN 0rN 1rN πr (u) ⊕ 1r(N−1) 0rN = ∗ ∗ − T −T (πr (u 1) ⊕ 1r(N− ))T 1 1rN 0rN πr (u) ⊕ 10rN = ∗ − ∗ ∗ − −T (πr (u 1) ⊕ 1r(N− )) 1rN T −T (πr (u 1) ⊕ 1)T 1 πr (u) ⊕ 1r(N−1) 0rN 1rN 0rN = ∗ − 0rN −T (πr (u 1) ⊕ 1r(N− ))T 0rN −1rN 1 ∗ −1 = πr (u) ⊕ 1r(N−1) + T (πr (u ) ⊕ 1r(N−1))T . Now we recall from the argument in the second paragraph (see (3.6)) that

∗ −1 −1 [T (πr (u ) ⊕ 1r(N−1))T ] = [πrN(Ωr (u ))] =−[πrN ◦ Ωr (u)],

and we see that we have proved what we wanted. ❐

Theorem 3.7. Let (B,α,L) be an Exel system with B unital and separable, and M {m }N−1 (j ,j ) suppose that L has an orthonormal basis j j=0 .Let ML B be the canonical C∗-Algebras Associated to Dilation Matrices 1047

Cuntz-Pimsner covariant representation of ML in O(ML). Then there is an exact sequence

−Ω j K (B) id ∗ /K (B) B∗ /K (O(M )) (3.8) 0 O 0 0 L

−1 −1 ρ∗ ◦δ1 ρ∗ ◦δ0 ojB∗ oid −Ω∗ K1(O(ML)) K1(B) K1(B).

(j ,j ) Proof. The canonical representation ML B is the composition of the universal Toeplitz representation (ψ, π) of ML in T (ML) with the quotient map Q, and in particular jB = Q ◦ π.SinceB is separable, [18, Theorem 4.4] says that the homomorphism π : B →T(ML) induces an isomorphism π∗ : Ki(B) → Ki(T (ML)), and since ρ : B → ker Q is an isomorphism onto a full corner, ρ∗ is an isomorphism. So splicing the commutative diagram of Proposition 3.4 into

(3.1)givestheresult. ❐

4. ENDOMORPHISMS ARISING FROM DILATION MATRICES

Throughout this section, d is an integer ≥ 2andA ∈ Md(Z) is an integer dilation matrix, by which we mean that all the complex eigenvalues λ of A satisfy |λ| > d πix 1. We consider the surjective endomorphism σA of T defined by σA(e2 ) = πiAx d e2 for x ∈ R , which has | ker σA|=|det A|, and the associated Exel system d d (C(T ), αA,L),whereαA is the endomorphism of C(T ) given by σA. C(Td) N =O(M ) We start by showing that αA,L L is simple and purely infinite. We deduce simplicity from results of Exel and Vershik [8] on crossed products by endomorphisms, and pure infiniteness from results of Katsura [14]onthe C∗-algebras of topological graphs. So we need to note that the map f N1/2f is an isomorphism of the bimodule ML = C(K)L onto the bimodule of the topolog- d d ical graph E with E0 = T , E1 = T , r = id and s = σA, and hence the crossed d ∗ ∗ product C(T )α,L N =O(ML) can also be viewed as the C -algebra C (E) studied in [13,14]. We need the following lemma on the operator norms of An acting on Rd;it must be well known, but we lack a reference. Lemma 4.1. We have A−n →0 as n →∞.

−n /n Proof. For each n in N,letan = A 1 . By the spectral radius formula we know that the limit of the an,asn tends to inﬁnity, coincides with the spectral radius r(A−1). By hypothesis we know that r(A−1)<1, so we may choose c with − r(A 1)

A = a ≤ c = . ❐ nlim→∞ nlim→∞ n nlim→∞ 0 1048 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

Proposition 4.2. The Cuntz-Pimsner algebra O(ML) is simple and purely inﬁ- nite.

Proof. We show that O(ML) is simple using [8, Theorem 11.2], which says d that C(T )α,L N is simple if and only if σA is irreducible. We recall from [8, Section 11] that x, y ∈ Td are trajectory-equivalent, written x ∼ y,ifthereare n m d n, m ∈ N such that σA (x) = σA (y),andasubsetY ⊆ T is invariant if x ∼ y ∈ Y implies that x ∈ Y ; σA is irreducible if the only closed invariant sets are ∅ and Td. Let Y be a non-empty closed invariant subset of Td, and pick a point e2πiy ∈ d πiz d d Y . We need√ to show that Y = T .Fixe2 ∈ T . Since the unit cube in R has d n diameter d, for every n ∈ N we can find kn ∈ Z such that |A z − (y + kn)|≤ √ n −n n 2πixn 2πiA xn 2πiy d.Thenxn := A (y + kn) has σA (e ) = e = e ∈ Y ,and invariance implies that e2πixn ∈ Y also. Lemma 4.1 implies that −n n −n |z − xn|≤ A |A z − (y + kn)|≤ A d → 0asn →∞, d πix πiz πiz so xn → z in R and e2 n → e2 .SinceY is closed, this implies that e2 ∈ Y , as required. Thus σA is irreducible, and O(ML) is simple. d To show that O(ML) is purely infinite, we realise O(ML) = C(T )α,L N as ∗ d d ∗ C (E) with E = (T , T , id,σA).SinceC (E) =O(ML) is simple, E is minimal by [14, Proposition 1.11]. So by [14,TheoremA]itsuffices to prove that E is 0 contracting at some vertex v0 ∈ E in the sense of Definition 2.3 of [14]; we will show that E is contracting at v = (1, 1,...,1). First, we need to see that the n 0 d positive orbit {z : σA (z) = v} of v is dense in E = T . The positive orbit of −n v contains all points of the form e2πiA k for n ∈ N and k ∈ Zd, and it follows from our proof of the irreducibility of σA above (with y = 0) that this positive orbit is dense in E0. Second, we fix a neighbourhood V of v; we need to show that V contains a contracting open set W (see [14, Definition 2.3]). For this, it suffices to find k an open neighbourhood W of v such that W ⊂ V and W ⊊ σA(W ) for some k ≥ 1. By Lemma 4.1 we can choose k such that A−k < 1. Then for every ε>0 and every x in the closed unit ball B(0,ε) in Rd,wehave|A−kx| <ε, so x = Ak(A−kx) belongs to Ak(B(0,ε)).ThusB(0,ε) ⊂ Ak(B(0,ε)).The inequality AkA−k ≤ Ak A−k implies that Ak > 1, so for every ε>0 there exists y ∈ B(0,ε) such that |Aky| >ε,andB(0,ε) ⊊ Ak(B(0,ε)).If ε is small enough to ensure that x e2πix is one-to-one on Ak(B(0,ε)),then 2πix k W :={e : x ∈ B(0,ε)} satisfies W ⊊ σA(W ), and by taking ε smaller still we can ensure that W ⊂ V .ThusE is contracting, and the result follows from

[14,TheoremA]. ❐ K C(Td) N =O(M ) We now want to calculate the -theory of αA,L L ,andwe d aim to use Theorem 3.7.Todothis,weneeddescriptionsofK∗(C(T )) and the map Ω∗. C∗-Algebras Associated to Dilation Matrices 1049

Lemma 4.3. Suppose that (B,α,L) is an Exel system with B commutative, that ML admits an orthonormal basis {mj :0≤ j ≤ N − 1},andthatΩ : B → MN (B) is the homomorphism described in Lemma 3.1.ThenΩ∗ ◦ α∗ is multiplication by N on both K0(B) and K1(B).

Proof. We know from Lemma 3.1 (3)thatΩ ◦ α(a) = a1N .Nowtakeb = (bij ) in Mr (B).Then(Ω ◦ α)r (b) is the r × r block matrix with (i, j) block equal to bij1N .Ifweview(Ω ◦ α)r (b) as an element of MN (Mr (B)),asin Lemma 3.5, it becomes b ⊕ b ⊕···⊕b. Whether b is a projection or a unitary, [b ⊕···⊕b] = N[b]. Thus by Lemma 3.5,wehave

Ω ◦ α ([b]) = (Ω ◦ α) ([b]) = [(Ω ◦ α) (b)] = [b ⊕···⊕b] = N[b].

∗ ∗ ∗ r ❐

Ji proved in [11] that the Chern character is a Z/2-graded ring isomorphism d ∗ d of K∗(C(T )) = K (T ) onto the integral cohomology ring

∞ d H∗(Td, Z) := Hk(Td, Z) = Hk(Td, Z), k∈Z k=0

∗ which in turn is isomorphic as a Z-graded ring to the exterior algebra Zd.Thus the ring H∗(Td, Z) is generated by H1(Td, Z), which is isomorphic to the set of homotopy classes of continuous functions from Td to T,andisthefreeabelian group generated by the coordinate functions uk : z = (z1,...,zn) zk.Since d d the homomorphism α∗ is induced by a continuous map σA : T → T ,the H∗(Td, Z) σ ∗ corresponding ring homomorphism on is the map A ,whichrespects ∗ d ∗ the Z-grading. Thus we can compute α∗ on Z by working out what σA does d on H1(T , Z) using the basis {ek := [uk] :1≤ k ≤ d}, and then taking exterior powers. Once we know what α∗ is, we can use the formula for Ω∗ ◦α∗ in Lemma 4.3 to work out what Ω∗ is. Lemma 4.4. With respect to the basis {[uk]}, α∗ :span{[uk]}→span{[uk]} is multiplication by the transpose AT of A.

Proof. We have α∗([uk]) = [α(uk)] = [uk ◦ σA].Since πix πiAx πi a x uk ◦ σA(e2 ) = uk(e2 ) = e2 j k,j j πia x πix a πix a = e2 k,j j = (e2 j ) k,j = uj(e2 ) k,j , j j j

ak,j we have uk ◦ σA = j uj .Hence[uk ◦ σA] = j ak,j [uj]. ❐ Since the 0-graded component is isomorphic to H0(Td, Z), the free abelian α group generated by the connected components, the action of ∗ on the 0-compo- 0 T nent (Z) = Z is the identity map. For n = 1, Lemma 4.4 implies that α∗ = A . 1050 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

For n>1, we use the basis E = e = e ∧···∧e J ⊂{ ,...,d}, |J|=n, J ={j < ···

Proof. Fix eK ∈En with K ={k1 < ···

Proposition 4.6. Let B0 =|det A|, Bd = sign(det A),and ⎧ ⎪ (τK τL) ⎨ (− )deg (AK,L ) A> , 1 det K,L if det 1 Bn = ⎩⎪ (τK τL) − (− )deg (AK,L ) A<− . 1 det K,L if det 1 d d (1) Then BnCn =|det A|1,where1 is the n × n identity matrix. C∗-Algebras Associated to Dilation Matrices 1051

(2) We have 1 − B0 = 1 −|det A| < 0, det(1 − Bn) ≠ 0 for 1 ≤ n1, 1 − Bd = 2ifdetA<−1. For the proof of Proposition 4.6 we need the following lemma; its ﬁrst part appears as equation (5.3.7) in [17], for example. Lemma 4.7. Fix n satisfying 1 ≤ n ≤ d − 1. (1) If eJ ∈En,then deg(τK τJ ) det A = (−1) det(AK,J) det(AK,J ). eK ∈En

(2) If eJ , eL ∈En and L ≠ J,then deg(τK τJ ) (−1) det(AK,J) det(AK,L ) = 0. eK ∈En

Proof. (1)FixeJ ∈En.Wehave deg σ det A = (−1) aσ(1),1 ...aσ(d),d σ ∈S d = (− )deg τJ (− )deg σ a ...a , 1 1 σ(1),j1 σ(d),jd σ ∈Sd which, by reordering the sum according to the image of In :={1,...,n} under σ ,is = (− )deg τJ (− )deg σ a ...a . (4.1) 1 1 σ(1),j1 σ(d),jd eK ∈En {σ :σ(In)=K}

Note that for ﬁxed σ ∈ Sn such that σ(In) = K,wehave

σ = (σK × σK ) ◦ τK, where σK(ki) := σ(i) and σK (k ) := σ( ).So = (− )deg τJ (− )deg τK (− )deg(σK ×σK )a ...a (4.1) 1 1 1 σK (k1),j1 σK (kn),jn eK ∈En {σ :σ(In)=K} · a ...a σK (kn+1),jn+1 σK (kd),jd = (− )deg τJ (− )deg τK (− )deg αa ...a 1 1 1 α(k1),j1 α(kn),jn eK ∈En α∈SK ,β∈SK · (− )deg βa ...a 1 β(kn+1),jn+1 β(kd),jd deg(τK τJ ) = (−1) det(AK,J) det(AK,J ). eK ∈En 1052 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN

(2)IfL ≠ J,thenL ≠ J and L ∩ J ≠ ∅. Consider the matrix D whose entries are those of A except that the L \ J columns of D have been replaced by copies of the J \ L columns of A.ThusdetD = 0. Note that AK,J and DK,L have the same columns up to permutation, so det(AK,J) =±det(DK,L). For every K we have DK,L = AK,L ,sousing(1)weget deg(τK τJ ) (−1) det(AK,J) det(AK,L ) e ∈E K n (τ τ ) deg K J =± (−1) det(DK,L) det(DK ,L ) = det D = 0. ❐ eK ∈En

ﬃ (− )deg(τK τJ ) Remark 4.8. In [17, page 92], it is observed that the coe cient 1 n ji+ki can be realised as the product i= (−1) . To see this, ﬁrst observe that n 1 (− )deg(τJ ) = (− )ji−i j − n 1 i=1 1 (because n , for example, is the number of trans- positions required to move jn to its correct place in J without changing the n J (− )deg(τK τJ ) = (− )(ji−i)+(ki−i) ordering of ), and then 1 i=1 1 . Proof of Proposition 4.6. Say det A>1. Then the (J, L) entry of CnBn is deg(τK τL) det(AK,J)(−1) det(AK,L ), eK ∈En which, by Lemma 4.7,equalsδJ,L(det A)1. If det A<−1, the same calculation gives −δJ,L(det A)1 = δJ,L| det A|1. Thus CnBn =|det A|1 = BnCn.Thisgives (1).

(2) The statements in (2)aboutB0 and Bd are immediate, so we suppose 1 ≤ n ≤ d d − 1. To compute det(I − Bn) we work over C, and choose a basis for C such that A is upper-triangular. We claim that if

J ={j1 < ···K={k1 < ···K, then there exists m such that ji = ki for ikm.SinceA is upper-triangular, jm >km a = j > ··· >j >j >k A implies jm,km 0. Moreover, n−m+1 m+1 m m,so J,K has the form U ∗ A = , J,K 0 V where U is an (m − 1) × (m − 1) upper-triangular matrix and V is a square matrix with the ﬁrst column consisting of zeros. Thus det(AJ,K) = 0, as claimed. So E ( (A )) if we order n with the lexicographic order, then the matrix det K,J J,K of n T −1 α∗|= A is lower-triangular. Hence its inverse (det A) Bn is also lower- triangular, and so is Bn. The diagonal entries of Bn are det(AK,K ) = k∈K ak,k; since each ak,k is an eigenvalue of A,wehave|ak,k| > 1, and each diagonal entry of Bn has absolute value greater than 1. Since Bn is lower-triangular, it follows that det(1 − Bn) ≠ 0. ❐ C∗-Algebras Associated to Dilation Matrices 1053

Theorem 4.9. Let A be a dilation matrix in GLd(Z) with d ≥ 1,anddeﬁne d Bn as in Proposition 4.6.LetML be the bimodule for the Exel system (C(T ), αA,L) C(Td) N =O(M ) and for which αA,L L . (1) If det A>1,then K0(O(ML)) = coker(1 − Bn) ⊕ Z, n even,n

(2) If det A<−1,then K0(O(ML)) = coker(1 − Bn), n even,n≤d K1(O(ML)) = coker(1 − Bn). n odd,n≤d

Proof. We identify n n d d d d K1(C(T )) Z and K0(C(T )) Z . n odd,n≤d n even,n≤d

d d on K0(C(T )) and K1(C(T )), respectively. By Proposition 4.6 (2), each 1 − Bn with n

Thus Diagram 3.8 gives K1(O(ML)) coker(1 − Bn) n odd,n≤d and an exact sequence 0 → coker(1 − Bn) → K0(O(ML)) → Z→ 0. n even,n

Since Z is free this sequence splits, and the formula for K0 follows. The proof for even d is similar, though this time the zero summand 1 − Bd d d appears in the map id −Ω∗ on K1(C(T )) rather than in the map on K0(C(T )). For part (2), we just note that Proposition 4.6 (2)impliesthat (1 − Bn) and (1 − Bn) n even,n≤d n odd,n≤d

are injective, and the result follows. ❐

For small d, we can identify the Bn in more familiar terms. Both B0 and Bd are just numbers (or rather, multiplication by those numbers on Z). Next we have the following result: d B =| A|(AT )−1 Proposition 4.10. For every ,wehave 1 det .Ifwelistthe d−1 d basis for Z as fk := e{1,...,d}\{k},thenBd−1 is the matrix with (k, ) entry k+ k+ +1 (−1) ak (if det A>0) or (−1) ak, (if det A<0).

Proof. For each singleton set {k}, the permutation τ{k} is the cycle which pulls k to the front and moves the elements 1,...,k− 1 to the right, which has degree k. The complements {k} are the sets kˆ :={1,...,k}\{k},andthenumber

(τK τL) τK + τL k+ (− )deg A = (− )deg deg A = (− ) A 1 det {k} ,{ } 1 det k,ˆ ˆ 1 det k,ˆ ˆ is the ( , k) entry in (det A)A−1,andthe(k, ) entry in (det A)(AT )−1.Theextra T −1 minussignintheformulaforB1 when det A<0showsthatB1 is (| det A|)(A ) . The (k, ) entry in the matrix of Bd−1 with respect to the basis {fk} is the ˆ (k,ˆ ) entry in the matrix with respect to the basis Ed−1.ForK = kˆ, τK is the cycle which moves k to the back and the last d − k terms one forward, which has d − k + A × a degree 1. Since (k)ˆ ,( )ˆ is the 1 1matrixwithentry k, ,wehave

(− )deg(τK τL) A = (− )(d−k+1)+(d− +1)a 1 det (k)ˆ ,( )ˆ 1 k, 2(d+1)−(k+ ) k+ = (−1) ak, = (−1) ak, .

This immediately gives the result for det A>0, and for det A<0, the extra minus k+ k+ +1 sign in the formula for Bd−1 means we need to replace (1) by (−1) . ❐ We can now sum up our results for small d: The ﬁrst statement of Corol- lary 4.11 is well-known, as we observed in the introduction, but the second statement and Corollary 4.12 were a bit of a surprise. Corollary 4.11. Suppose N is a non-zero integer, and consider the Exel system N (C(T), αN ,L)associated to the covering map z z . C∗-Algebras Associated to Dilation Matrices 1055

(1) If N>1,then

K (C(T) N) = (Z/(N − )Z) ⊕ Z 0 αN ,L 1 and K (C(T) N) = Z. 1 αN ,L

(2) If N<−1,then

K (C(T) N) = Z/(|N|− )Z 0 αN ,L 1 and K (C(T) N) = Z/ Z. 1 αN ,L 2

Corollary 4.12. Suppose that A = (aij ) ∈ M2(Z) is a dilation matrix. Then Z/(| det A|−1)Z ⊕ Z if det A>1, K (C(T2)α ,L N) = 0 A (Z/(| det A|−1)Z) ⊕ (Z/2Z) if det A<−1, and ⎧ ⎪ ⎪ 1 − a a ⎪Z ⊕ 11 12 A> , ⎨ coker a − a if det 1 K (C(T2) N) = 21 1 22 1 αA,L ⎪ + a −a ⎪ 1 11 12 ⎩⎪coker if det A<−1. −a21 1 + a22

Proof. The statement about K0 follows immediately from Theorem 4.9.For K1, we use the description of B1 = B2−1 in Proposition 4.10.(Ifwehadused T −1 the description of B1 as | det A|(A ) , we would have gotten a diﬀerent matrix, because we would then be calculating it with respect to the basis {e1,e2} rather than {f1,f2}={e2,e1}. However, the two matrices are conjugate in M2(Z),and hence have isomorphic cokernels.) ❐ We now look at the implications of these results for some concrete examples of dilation matrices. The ﬁrst two were used in [19] to provide examples of projective multi-resolution analyses. Examples 4.13. A = 01 A =− < − K (C(T2) N) (1) For 20 we have det 2 1. So 1 αA,L is the 1 −1 − Z cokernel of −21; since this matrix has determinant 1, it is invertible over , and we have

K (C(T2) N) = Z/ Z K (C(T2) N) = . 0 αA,L 2 and 1 αA,L 0

These K-groups are the same as those of O3, but the class of the identity is diﬀer- 2 ent. To see the last statement, note that the class [1] of the identity in K0(C(T )) 1056 RUY EXEL,ASTRID AN HUEF & IAIN RAEBURN 0 2 is the image of 1 ∈ Z = Z ,andwhen| det A|=2, 1 − B0 is invertible, so [1] belongs to the range of id −Ω∗. Thus the class of the identity 1C(T2)N = jC(T)(1) K (C(T2) N) O [ ] in 0 αA,L is 0. For 3, on the other hand, 1 is the generator of K0(O3). In K0(M2(O3)), however, the class of the identity is twice the generating class [ e ] K (O ) = Z/ Z 1O3 11 , which is also zero in 0 3 2 . Thus the classiﬁcation theorem C(T2) N of Kirchberg and Phillips [15,22,25]impliesthat αA,L is isomorphic to M (O ) 2 3 . (We thank the referee for pointing this out.) A = 11 A = > (2) The matrix −11 has det 2 1. So Corollary 4.12 implies that K (C(T2) N) = Z K (C(T2) N) = Z. 0 αA,L and 1 αA,L K (3) The vanishing of 1 in Example (1) is something of an accident. Consider A = 01 k ∈ Z k> A =−k< the matrix k 0 for with 2. We have det 0, the map (m, n) m+ n(mod (k − 1)) induces an isomorphism of coker(1 − B1) = 1 −1 Z/(k − )Z K (C(T2) N) = Z/(k − )Z coker −k 1 onto 1 ,andwehave 1 αA,L 1 . A = 21 A = > (4) The matrix −12 has det 5 1. Thus K (C(T2) N) = (Z/ Z) ⊕ Z K (C(T2) N) = Z ⊕ (Z/ Z). 0 αA,L 4 and 1 αA,L 2 A = 2 −1 − (5) The matrix 1 −3 has determinant 5, and

K (C(T2) N) = (Z/ Z) ⊕ (Z/ Z) K (C(T2) N) = Z/ Z. 0 αA,L 4 2 and 1 αA,L 5 A = 01 It is interesting to compare this with the matrix 50 in Example (3), which A =− K = Z/ Z K C(T2) N also has det 5, but has 1 4 .Sothe -theory of αA,L is not completely determined by det A. Remark 4.14. The referee asked us whether we could describe the range of the K-invariant on our class of algebras. We know from Proposition 4.6 that the matrices 1 − Bn appearing in the formulas in Theorem 4.9 are all invertible over Q, and hence all the cokernels appearing there are finite abelian groups. In the examples above, the cokernels happen to be cyclic, but in general they will not be, as is easily seen by considering diagonal matrices A.Forlarged, therefore, the torsion parts of K0 and K1 could be large direct sums of finite cyclic groups. The rank of K0 and K1, on the other hand, is completely determined by Theorem 4.9: it is 1 if det A>1,and0ifdetA<−1. Acknowledgement. This research was partially supported by the Australian Research Council and the National Council for Scientific and Technological De- velopment (CNPq) of Brazil.

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KEY WORDS AND PHRASES: crossed product, endomorphism, transfer operator, integer dilation matrix, purely inﬁnite, simple, K-theory. 2000 MATHEMATICS SUBJECT CLASSIFICATION: 46L55. Received: March 10, 2010; revised: August 3, 2010. Article electronically published on August 5, 2010.