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Concluded December 29, 1992

arXiv:funct-an/9212005v1 31 Dec 1992UNM–RE–004Concluded December 29, 1992Printed October 28, 2018A FREDHOLM OPERATOR APPROACH TOMORITA EQUIVALENCERuy Exel *Department of Mathematics and StatisticsUniversity of New MexicoAlbuquerque, New Mexico 87131e-mail: exel@math.unm.eduAbstract.Given C∗-algebras A and B and an imprimitivity A-B-bimodule X, we construct an explicit isomorphism X∗: Ki(A) →Ki(B)where Ki denote the complex K-theory functors for i = 0, 1. Our tech-niques do not require separability nor existence of countable approxi-mate identities.

We thus extend, to general C∗-algebras, the result ofBrown, Green and Rieffel according to which strongly Morita equivalentC∗-algebras have isomorphic K-groups. The method employed includesa study of Fredholm operators on Hilbert modules.

* On leave from the University of S˜ao Paulo.

2ruy exel1. IntroductionStrong Morita equivalence for C∗-algebras was introduced by Rieffel ([10], [11], [12]),generalizing the concept of Morita equivalence for rings (see for example, [13], Chapter 4).In [3], Brown, Green and Rieffel proved that, if the strongly Morita equivalent C∗-algebrasA and B are σ-unital (ie., possess countable approximate identities), then they are stablyisomorphic in the sense that K ⊗A is isomorphic to K ⊗B (K denoting the algebra ofcompact operators on a separable Hilbert space).

This result has, among many importantconsequences, the Corollary that strongly Morita equivalent σ-unital C∗-algebras haveisomorphic K-theory groups.In the absence of the σ-unital hypothesis there are examples of strongly Morita equiv-alent C∗-algebras which are not stably isomorphic, even if one allows tensoring by thecompact operators on a non-separable Hilbert space (see [3], Theorem 2.7). Thus, with-out assuming countable approximate identities, the question of whether the K-groups ofstrongly Morita equivalent C∗-algebras are isomorphic, remained open.Even though the argument can be made that the majority of C∗-algebras of interestare σ-unital, there is a more serious obstacle in applying the BGR Theorem to concretesituations.

The proof of that result rests largely, on the work of Brown on full hereditarysubalgebras of C∗-algebras [2], where a highly non-constructive method is used. The endresult is that, given strongly Morita equivalent, σ-unital C∗-algebras A and B, one knowsthat K ⊗A and K ⊗B are isomorphic to each other but there is generally little hope thatone can exhibit a concrete isomorphism, unless, of course, an isomorphism is known to existindependently of the Morita equivalence.

Likewise, no expression for the isomorphism ofK-groups may be described in general.The primary goal of the present work is to make explicit the isomorphism betweenK∗(A) and K∗(B) that is predicted by the BGR Theorem. Recently, Rieffel suggestedthat, if such an explicit construction was possible, then it would probably work withoutseparability, and in fact it does: in Theorems 5.3 and 5.5 we prove that strongly Moritaequivalent C∗-algebras have isomorphic K-groups, irrespective of countable approximateidentities.It should be noted that in a crucial point of our argument (see the proof of Theorem2.7 below), we have to use Kasparov’s stabilization Theorem [7] which requires a certaincountability condition, but in Lemma 2.6, we manage to conform ourselves to Kasparov’shypothesis, without having to assume any countability beforehand.Let us now briefly describe the strategy adopted to prove our main result.

Given aC∗-algebra A, we begin by studying Fredholm operators on Hilbert A-modules and wedefine the Fredholm index (3.4 and 3.10) which takes values in K0(A). We then provethat any element of K0(A) is the index of some Fredholm operator (3.14), showing thesurjectivity of the Fredholm index.

Next, we characterize in a somewhat geometric way,the pairs of Fredholm operators having identical index (3.16). This enables us to introducean equivalence relation on the set of all Fredholm operators, under which two operators areequivalent if and only if they have the same index.

The quotient by this relation thereforeprovides an alternate definition for K0(A) (3.17).

fredholm operators and morita equivalence3If one is given C∗-algebras A and B and a Hilbert A-B-bimodule X then, for everyFredholm operator T: M →N, where M and N are Hilbert A-modules, one can considerthe operator T ⊗IX: M ⊗B X →N ⊗B X. The crux of the whole matter is to prove thatthis operator is a Fredholm operator (4.3) if X is left-full (in the sense that the range ofthe A-valued inner-product generates A, see 4.1).

This tensor product construction is thenused to define (5.1) a group homomorphism X∗: K0(A) →K0(B) satisfying X∗(ind(T)) =ind(T ⊗IX). When X is also right-full we show that X∗is an isomorphism (5.3).

Thecase of K1 is treated, as usual, using suspensions (5.5).Our characterization of K0(A) in terms of Fredholm operators should be comparedto the well known fact that K0(A) is isomorphic to the Kasparov [8] group KK(C, A) (Cdenoting the algebra of complex numbers). However, since we do not assume countableapproximate identities, that fact does not seem to follow from the existing machineryof KK-theory.

Furthermore, this aspect of our work is, perhaps, an indication that KK-theory may be extended beyond the realm of σ-unital C∗-algebras. In particular, it soundsreasonable to conjecture that strongly Morita equivalent C∗-algebras are KK-equivalent.In fact, our methods seem particularly well suited to prove such a conjecture if only KK-theory were not so inextricably linked to σ-unital C∗-algebras.To some extent, one should also compare what we do here with the Fredholm modulesof Connes [5], in the sense that we explore, in more detail than usually found in theexisting literature, what happens when one plugs the algebra of complex numbers as oneof the variables of Kasparov’s KK-functor.

Connes’ Fredholm modules are literally theingredients of KK(· , C), according to the Fredholm picture of KK-theory (see [1]). Onthe other hand, the Fredholm operators we study here are related to KK(C, ·).Although Hilbert modules permeate all of our work, we believe the subject is oldenough that we do not need to spend much time presenting formal definitions from scratch.The reader will find all of the relevant definitions in [9], [10], [1], [6] and [4].

Neverthelesslet us stress that all of our Hilbert modules are supposed to be right modules except, ofcourse, when they are bimodules. The term operator, when referring to a map T: M →Nbetween the Hilbert A-modules M and N, will always mean an element of LA(M, N),that is, T should be an adjointable linear map in the sense that there exists T ∗: N →Msatisfying ⟨T(µ), ν⟩= ⟨µ, T ∗(ν)⟩for µ ∈M and ν ∈N.

See [6], 1.1.7 for more details.

4ruy exel2. Preliminaries on Hilbert ModulesIn this section we would like to present a few simple facts about Hilbert modules which weshall need in the sequel.

Throughout this section, A will denote a fixed C∗-algebra and Mand N will always refer to (right) Hilbert modules over A. As usual, An will be viewed asa Hilbert module equipped with the inner-product ⟨(ai)i, (bi)i⟩= Pni=1 a∗i bi.Let us observe that An will often stand for the set of n × 1 (column) matrices overA.

In that way, the above inner-product can be expressed, for v = (ai)i and w = (bi)i, as⟨v, w⟩= v∗w. Note that v∗refers to the conjugate-transpose matrix.For each n-tuple µ = (µi)i in M n, we denote by Ωµ the operator in LA(An, M) definedbyΩµ(ai)i=nXi=1µiai,(ai)i ∈An.It is easy to see that Ω∗µ is given byΩ∗µ(ξ) =⟨µi, ξ⟩i,ξ ∈M.If ν = (νi)i is an n-tuple of elements of N, then the operator T = ΩνΩ∗µ is in LA(M, N).More explicitly we haveT(ξ) =nXi=1νi⟨µi, ξ⟩,ξ ∈M.Maps such as T will be called A-finite rank operators and the set of all those will be denotedFA(M, N), or just FA(M) in case M = N. The closure of FA(M, N) in LA(M, N) isdenoted KA(M, N) and elements from this set will be referred to as A-compact operators.An expository treatment of operators on Hilbert modules may be found in [6].2.1.

Proposition. For each µ = (µi)i in M n one has that Ωµ is in KA(An, M) and hencealso that Ω∗µ is in KA(M, An).Proof.

It is obviously enough to consider the case n = 1. Let (uλ)λ be an approximateidentity for A (always assumed to be positive and of norm one).

It follows from [6], 1.1.4that limλ µuλ = µ. Therefore we have for all a in AΩµ(a) = µa = limλ µuλa = limλ µ⟨uλ, a⟩= limλ ΩµΩ∗uλ(a),the limit being uniform in ∥a∥≤1.⊓⊔2.2.

Definition. A Hilbert module M will be said to be an A-finite rank module if theidentity operator IM is in KA(M).Since FA(M) is an ideal in LA(M), which is dense in KA(M), it is easy to see thatIM must, in fact, be in FA(M) whenever M is A-finite rank.

We next give the completecharacterization of A-finite rank modules.

fredholm operators and morita equivalence52.3. Proposition.

M is A-finite rank if and only if there exists an idempotent matrix pin Mn(A) such that M is isomorphic, as Hilbert modules, to pAn.Proof. Initially we should observe that the use of the term “isomorphic”, when referringto Hilbert modules, is in accordance with [6], 1.1.18.

That is, there should exist a linearbijection, preserving the A-valued inner product.Assume M to be A-finite rank. Then IM = ΩνΩ∗µ where µ and ν are in M n. Observethat Ω∗µΩν is then an idempotent A-module operator on An, which therefore correspondsto left multiplication by the idempotent n × n matrix p =⟨µi, νj⟩i,j.The operatorΩ∗µ then gives an invertible operator in LA(M, pAn).

To make that map an (isometric)isomorphism one uses polar decomposition. The converse statement is trivial.⊓⊔Any A-finite rank module M clearly becomes a finitely generated projective moduleover the unitized C∗-algebra ˜A (the unitized algebra is given a new identity element, evenif A already has one).2.4.

Definition. The K-theory class [M]0 ∈K0( ˜A) of an A-finite rank module M, isobviously an element of K0(A) and will henceforth be denoted rank(M).If M is notnecessarily assumed to be A-finite rank, but if P is an idempotent operator in KA(M),then Im(P) is clearly an A-finite rank module.

In this case we let rank(P) = rank(Im(P)).If X, Y , Z and W are Hilbert A-modules and T is in LA(X ⊕Y, Z ⊕W), then T canbe represented by a matrixT =TZXTZYTW XTW Y,where TZX is in LA(X, Z) and similarly for the other matrix entries. Matrix notation isused to define our next important concept.2.5.

Definition. The Hilbert modules M and N are said to be quasi-stably-isomorphic ifthere exists a Hilbert module X and an invertible operator T in LA(M ⊕X, N ⊕X) suchthat IX −TXX is A-compact.Of course the concept just defined is meant to be a generalization of the well knownconcept of stable isomorphism for finitely generated projective A-modules, at least whenA is unital.

We shall discuss shortly, the precise sense in which that generalization takesplace. Before that we need a preparatory result.2.6.

Lemma. Assume M and N are A-finite rank modules.

If M and N are quasi-stably-isomorphic then the module X referred to in 2.5 can be taken to be countably generated.Proof. Let T =TNMTNXTXMTXXbe an invertible operator in LA(M ⊕X, N ⊕X) withIX −TXX A-compact, as in 2.5, and let S =SMNSMXSXNSXXbe the inverse of T. Choosea countable set Ξ0 = {ξi}i∈N of elements in X such that

6ruy exel(i) The images of TXM, SXN, T ∗NX, and S∗MX are contained in the submodule of Xgenerated by Ξ0. (ii) IX −TXX can be approximated by A-finite rank operators of the form ΩνΩ∗µ, wherethe components of µ = (µ1, .

. ., µn) and ν = (ν1, .

. .

, νn) belong to Ξ0.Define, inductively, Ξn+1 = Ξn ∪TXX(Ξn) ∪SXX(Ξn) ∪T ∗XX(Ξn) ∪S∗XX(Ξn). Theset Ξ = Sn∈N Ξn is then obviously countable, satisfies (i) and (ii) above and, in addition,(iii) Ξ is invariant under TXX, SXX, T ∗XX and S∗XX.Let X0 be the Hilbert submodule of X generated by Ξ.

Because of (i) and (iii) we seethat T(M ⊕X0) ⊆N ⊕X0 and T ∗(N ⊕X0) ⊆M ⊕X0. The restriction of T then givesan operator T ′ in LA(M ⊕X0, N ⊕X0).

The same reasoning applies to S providing S′ inLA(N ⊕X0, M ⊕X0) which is obviously the inverse of T ′. In virtue of (ii) it is clear thatT ′ satisfies the conditions of definition 2.5.⊓⊔2.7.

Theorem. If M and N are quasi-stably-isomorphic A-finite rank modules, thenrank(M) = rank(N).Proof.

According to 2.6 let X be a countably generated Hilbert module and T be aninvertible operator in LA(M ⊕X, N ⊕X) such that IX −TXX is in KA(X). By Kasparov’sstabilization Theorem [7], 3.2 (see also [6], 1.1.24), X ⊕HA is isomorphic to HA, whereHA is the completion of L∞1 A (see [6], 1.1.6 for a more precise definition).

This said, wemay assume, without loss of generality, that X = HA. Since M is finitely generated as anA-module, we conclude, again by Kasparov’s Theorem, that M ⊕X is isomorphic to HA.Choose, once and for all, an isomorphism ϕ: HA →M ⊕X and consider the operators Fand G on HA given by the compositionsF : HAϕ−→M ⊕XT−→N ⊕X −→X −→M ⊕Xϕ−1−→HAandG : HAϕ−→M ⊕X −→X −→N ⊕XT −1−→M ⊕Xϕ−1−→HA,where the unmarked arrows denote either the canonical inclusion or the canonical projec-tion.

It can be easily seen that both IHA −GF and IHA −FG belong to KA(HA). Considerthe exact sequence of C∗-algebras0 −→KA(HA) −→LA(HA) −→LA(HA)/KA(HA) −→0.Denoting by π the quotient map, one sees that π(F) and π(G) are each others inverse.Two facts need now be stressed.

The first one is that the K-theory index mapind : K1LA(HA)/KA(HA)→K0KA(HA)assigns to the class of π(F), the element rank(N) −rank(M), once KA(HA) is identifiedwith K ⊗A (according to [7], 2.4) and K0(K ⊗A) is identified with K0(A) as usual inK-theory.

fredholm operators and morita equivalence7The second fact to be pointed out is that, since TXX is an A-compact perturbation ofthe identity and M and N are A-finite rank modules (and so any operator having either Mor N as domain or codomain must be A-compact), one concludes that F is an A-compactperturbation of the identity. It follows that π(F) = 1 and hence that π(F) has trivialindex.

This concludes the proof.⊓⊔3. Fredholm OperatorsWe shall now study Fredholm operators between Hilbert modules.As before, A willdenote a fixed C∗-algebra and M and N, with or without subscripts, will denote HilbertA-modules.3.1.

Definition. Let T be in LA(M, N).Suppose there is S in LA(N, M) such thatIM −ST is in KA(M) and IN −TS is in KA(N).

Then T is said to be an A-Fredholmoperator. In case the algebra A is understood, we shall just say that T is a Fredholmoperator (but we should keep in mind that this notion does not coincide with the classicalnotion of Fredholm operators).As in the classical theory of Fredholm operators, it can be proved that whenever T isA-Fredholm, one can find S in LA(N, M) such that IM −ST is in FA(M) and IN −TSis in FA(N).

In all of our uses of the A-Fredholm hypothesis, below, we shall adopt thatcharacterization.In the initial part of the present section we shall concentrate on a special class ofoperators which we will call regular operators.This concept is the natural extension,to Hilbert modules, of the notion of operators on Hilbert spaces having closed image.Contrary to the Hilbert space case, not all A-Fredholm operators have a closed image.3.2. Definition.

An operator T in LA(M, N) is said to be regular if there is S in LA(N, M)such that TST = T and STS = S. Any operator S having these properties will be calleda pseudo-inverse of T.It is easy to see that for any pseudo-inverse S of T one has that IM −ST is theprojection onto Ker(T) and that TS is the projection onto Im(T). If T is assumed to beregular and Fredholm, there are, according to the above definitions, operators S and S′such that IM −S′T and IN −TS′ are A-finite rank and, on the other hand, TST = T andSTS = S.Observe that, because IM −ST = (IM −S′T)(IM −ST), any pseudo inverse S, must besuch that IM −ST is A-finite rank and similarly for IN −TS.

An immediate consequenceof the present discussion is the following:3.3. Proposition.

Let T ∈LA(M, N) be a regular A-Fredholm operator. Then bothKer(T) and Ker(T ∗) are A-finite rank modules.Let us now define the Fredholm index for regular A-Fredholm operators.

Shortly weshall extend that concept to general A-Fredholm operators.

8ruy exel3.4. Definition.

If T is a regular A-Fredholm operator, then the Fredholm index of T isdefined to be the element of K0(A) given byind(T) = rank(Ker(T)) −rank(Ker(T ∗)).We collect, in our next proposition, some of the elementary properties of the Fredholmindex.3.5. Proposition.

If T ∈LA(M, N) is a regular Fredholm operator, then(i) ind(T ∗) = −ind(T). (ii) For any pseudo-inverse S of T we have that rank(Ker(T ∗)) = rank(Ker(S)) andind(S) = −ind(T).

(iii) If X and Y are Hilbert A-modules and if U ∈LA(X, M) and V ∈LA(N, Y ) areinvertible, then ind(V TU) = ind(T). (iv) If T1 ∈LA(M1, N1) is regular and Fredholm, then ind(T ⊕T1) = ind(T) + ind(T1).Proof.

Left to the reader.⊓⊔The first fact about classical Fredholm operators whose generalization to Hilbert mod-ules requires some work is the invariance under compact perturbations which we now prove.3.6. Theorem.

If T is a regular Fredholm operator in LA(M) and if IM −T is A-compact,then ind(T) = 0.Proof. Let S be a pseudo-inverse for T and denote by X the image of the idempotent STor, equivalently, X = Im(S).

Consider the transformationU: Ker(T) ⊕X →Ker(S) ⊕Xgiven by U(ξ, η) =(IN −TS)(ξ + η), S(ξ + η). It is easy to see that U is invertiblewith inverse given by V (ξ, η) =(IM −ST)(ξ + T(η)), ST(ξ + T(η)).

The operator UXX(occurring in the matrix representation of U) coincides with S which is easily seen to bean A-compact perturbation of the identity. This shows that the A-finite rank modulesKer(T) and Ker(S) are quasi-stably-isomorphic.

By 2.7 we conclude that rank(Ker(T)) =rank(Ker(S)) and hence that ind(T) = 0.⊓⊔3.7. Corollary.

If T1, T2 ∈LA(M, N) are regular Fredholm operators such that T1 −T2is in KA(M, N), then ind(T1) = ind(T2).Proof. Let S1 and S2 be pseudo inverses for T1 and T2.

Define operators U and R inLA(M ⊕N) byU =IM −S1T1S1T1IN −T1S1andR =0S1T20.Observe that U 2 = I so, in particular, U is invertible. Therefore, by 3.5, we have thatind(UR) = ind(R) = ind(T2) −ind(T1).

But, sinceUR =S1T20T2 −T1S1T2T1S1is an A-compact perturbation of the identity, it follows that ind(UR) = 0.⊓⊔

fredholm operators and morita equivalence9We now start to treat general A-Fredholm operators. The crucial fact which allowsus to proceed, is that any Fredholm operator is “regularizable” over a unital algebra.3.8.

Lemma. Let B be a unital C∗-algebra and M and N be Hilbert B-modules.IfT ∈LB(M, N) is B-Fredholm (but not necessarily regular), then there exists an integer nand a regular B-Fredholm operator ˜T in LB(M ⊕Bn, N ⊕Bn) such that ˜TNM = T.Proof.

Let S in LB(N, M) be such that both IM −ST and IN −TS are B-finite rank.So, let ν = (νi)i and µ = (µi)i be such that IM −ST = ΩνΩ∗µ. Define˜T =T0Ω∗µ0and˜S =SΩν00.It is easy to see that ˜T and ˜S are each others pseudo-inverse hence, in particular, ˜T isregular.

The hypothesis that B have a unit implies that Bn is B-finite rank and hencethatI −˜S ˜T =000IBnandI −˜T ˜S =IN −TS−TΩν−Ω∗µSIBn −Ω∗µΩνare B-finite rank operators. We conclude that ˜T is Fredholm.⊓⊔Since we shall not assume the algebras we work with to be unital (nor σ-unital,as already stressed in the introduction), we will keep all the applications of the aboveLemma to the unitized algebra ˜A, in the following way.

Given an A-Fredholm operatorT in LA(M, N) consider M and N as Hilbert modules over the unitized algebra ˜A, asit is done in [6], remark 1.1.5. Obviously T is ˜A-Fredholm as well.

Let therefore ˜T ∈L ˜A(M ⊕˜An, N ⊕˜An) be the operator constructed as in 3.8.Since ˜T is regular andFredholm, ind( ˜T) is well defined as an element of K0( ˜A).The following result will allow us to return to the realm of non-unital (meaning non-necessarily-unital) algebras after our brief encounter with units.3.9. Proposition.

The Fredholm index of ˜T belongs to K0(A).Proof. Denote by ε: ˜A →C the augmentation homomorphism.

That is ε(a+λ) = λ for λin C and a in A. Since K0(A) is defined to be the kernel of the map ε∗: K0( ˜A) →K0(C),all we need to do is show that ε∗(ind( ˜T)) = 0.Let S, ˜S, µ and ν be as in the proof of 3.8.

Note that I −˜S ˜T =000I ˜An, sorank(Ker( ˜T)) = n. We then need to show that ε∗(rank(Ker( ˜S))) is also equal to n. Thekernel of ˜S is the image of the idempotentI −˜T ˜S =IN −TS−TΩν−Ω∗µSI ˜An −Ω∗µΩν(3.9.1)

10ruy exelwhich we shall simply denote by P.Since P is ˜A-compact, there are m-tuples φ =(φ1, . .

., φm) and ψ = (ψ1, . .

., ψm) of elements of N ⊕˜An such that P = ΩφΩ∗ψ. Writeeach φi as φi = (ξi, vi) and ψi = (ηi, wi).Replacing, if necessary, φi by P(φi), we can assume that PΩφ = Ωφ and therefore thatQ = Ω∗ψΩφ is an idempotent operator on ˜An whose image is isomorphic, as ˜A-modules, tothe image of P. As an m × m matrix, Q is given by Q =⟨ψi, φj⟩i,j.

Our goal is thento show that the trace of the complex idempotent matrix ε(Q) =ε⟨ψi, φj⟩i,j equals n.That trace is given bymXi=1ε⟨ψi, φi⟩=mXi=1ε⟨wi, vi⟩= ε mXi=1nXr=1⟨wi, er⟩⟨er, vi⟩,where {ei, . .

. , en} is the canonical basis of ˜An.

The above then equalsε mXi=1nXr=1⟨er, vi⟨wi, er⟩⟩= ε nXr=1⟨(0, er), P(0, er)⟩.Using the definition of P in 3.9.1, the term Pnr=1⟨(0, er), P(0, er)⟩can be expressed asnXr=1⟨er, (I −Ω∗µΩν)er⟩= n −nXr=1⟨µr, νr⟩.The last term above clearly maps to n under ε so the proof is complete.⊓⊔The statement of 3.9 is meant to refer to the specific construction of ˜T obtained is3.8. But note that any regular Fredholm operator in L ˜A(M ⊕˜An, N ⊕˜An), which has T inthe upper left corner, will differ from the ˜T above, by an ˜A-compact operator.

Thereforeits index will coincide with that of ˜T by 3.7, and so will be in K0(A) as well.3.10. Definition.

If T is an A-Fredholm operator in LA(M, N), then the Fredholm indexof T, denoted ind(T), is defined to be the index of the regular Fredholm operator ˜Tconstructed in 3.8.Clearly, if T is already regular, we can take n = 0 in 3.8 so that the above definitionextends the one given in 3.4. Elementary properties of the Fredholm index are collectedin the next proposition.3.11.

Proposition. If T ∈LA(M, N) is a Fredholm operator, then(i) ind(T ∗) = −ind(T)(ii) If U in LA(X, M) and V in LA(Y, N) are invertible, then ind(V TU) = ind(T).

(iii) If T ′ ∈LA(M, N) is such that T ′ −T is in KA(M, N), then T ′ is also Fredholm andind(T ′) = ind(T). (iv) If S ∈LA(N, M) is such that IM −ST is in KA(M), then ind(S) = −ind(T).

(v) If T1 ∈LA(M1, N1) is Fredholm, then ind(T ⊕T1) = ind(T) + ind(T1).

fredholm operators and morita equivalence11Proof. Left to the reader.⊓⊔Let us now briefly tackle the question of invariance of the Fredholm index under smallperturbations.

The proof, which we omit, is similar to the classical one.3.12. Proposition.

Let T in LA(M, N) be Fredholm. Then there is a positive real numberε such that any T ′ satisfying ∥T ′ −T∥< ε is also Fredholm with ind(T ′) = ind(T).

In factε can be taken to be ε = ∥S∥−1 for any S in LA(N, M) such that IM −ST is in KA(M).It is equally easy to prove:3.13. Proposition.

If T1 ∈LA(M, N) and T2 ∈LA(N, P) are Fredholm operators, thenT2T1 is Fredholm and ind(T2T1) = ind(T2) + ind(T1).It is part of our goal to find an alternate definition of K0(A) in terms of Fredholmoperators. For this reason it is important to have a sufficiently large collection of such op-erators.

Specifically we will need to exhibit Fredholm operators with an arbitrary elementof K0(A) as the index.This is quite easy if A has a unit: given an arbitrary element [M]0 −[N]0 in K0(A),where M and N are finitely generated projective A-modules, consider the zero operator0: M →N. If both M and N are given the Hilbert module structure they possess, beingisomorphic to direct summands of An, it follows that 0 is a Fredholm operator, its indexobviously being [M]0 −[N]0.

The situation is slightly more complicated in the non-unitalcase, and it is treated in our next result.3.14. Proposition.

For any α in K0(A) there is an A-Fredholm operator T such thatind(T) = α.Proof. Regarding K0(A) as the kernel of the augmentation map ε∗, in K0( ˜A), writeα = [p]0 −[q]0, where p and q are self-adjoint idempotent n × n matrices over ˜A such thatε∗([p]0 −[q]0) = 0.

It follows that ε(p) and ε(q) are similar complex matrices. Hence, afterperforming a conjugation of, say q, by a complex unitary matrix, we may assume that ε(p)and ε(q) are in fact equal.

Therefore p −q is in Mn(A).We now claim that the operator T: pAn →qAn given by T(v) = qv is Fredholm andthat its index is [p]0 −[q]0.Let S: qAn →pAn be given by S(v) = pv. Denote by pi the ith column of p, viewed asa n × 1 matrix, and let (uλ)λ be an approximate identity for A. Define ξi = p(p −q)pi andηλi = piuλ for 1 ≤i ≤n.

So, ξi and ηλi are elements of pAn. Let ξ = (ξi)i and ηλ = (ηλi )i.For ζ in pAn we haveΩξΩ∗ηλ(ζ) =nXi=1ξi⟨ηλi , ζ⟩=nXi=1p(p −q)piuλp∗i ζwhich converges, uniformly in ∥ζ∥≤1, as λ →∞, tonXi=1p(p −q)pip∗i ζ = p(p −q)pζ = ζ −pqζ = (I −ST)ζ.

12ruy exelIn a similar fashion we can show that I −TS is A-compact as well. In order to computethe index of T, consider the operators ˜T in L ˜A(pAn ⊕˜An, qAn ⊕˜An) and ˜S in L ˜A(qAn ⊕˜An, pAn ⊕˜An) given by˜T =qpq(I −p)(I −q)p(I −q)(I −p)and˜S =pqp(I −q)(I −p)q(I −p)(I −q).Direct computation shows that˜S ˜T =I00I −pand˜T ˜S =I00I −q,from which it follows that ˜S is a pseudo-inverse for ˜T and hence that ˜T is a regular˜A-Fredholm operator.

By definition we haveind(T) = ind( ˜T) = rank(I −˜S ˜T) −rank(I −˜T ˜S) = [p]0 −[q]0.⊓⊔Our last result showed that any K0 element is the index of some Fredholm operator.We would now like to discuss the question of when two Fredholm operators have the sameindex. As a first step, we classify operators with index zero.3.15.

Lemma. Let T ∈LA(M, N) be a Fredholm operator with ind(T) = 0.

Then thereexists an integer n such that T ⊕IAn: M ⊕An →N ⊕An is an A-compact perturbationof an invertible operator.Proof. Let ˜T in L ˜A(M ⊕˜An, N ⊕˜An) be constructed as in 3.8 with ˜A playing the roleof B.

If ˜S is also as in 3.8 we have I −˜S ˜T =000I ˜An. The hypothesis that ind(T) = 0then says that rank(I −˜T ˜S) = rank(I −˜S ˜T).

But, as we see above, the rank of I −˜S ˜Tis the same as the rank of the free ˜A-module ˜An. So, we have that Im(I −˜T ˜S) is stablyisomorphic, as ˜A-modules, to ˜An.

By increasing n, if necessary, we may thus assume thatIm(I −˜T ˜S) is, in fact, isomorphic to ˜An.This said, we may find a generating set {xi}ni=1 for Im(I −˜T ˜S) which is orthonormalin the sense that ⟨xi, xj⟩= δij. Let each xi be given by xi = (ξi, vi).

Each vj is in ˜An andhence we may write vj = (vij)i with vij ∈˜A. The fact that the xi form an orthonormalset translates to⟨ξi, ξj⟩+nXk=1v∗kivkj = δij.Recalling that ε: ˜A →C denotes the augmentation homomorphism, observe that thecomplex matrix u =uiji,j, where uij = ε(vij), is unitary.If we now define x′j =Pni=1 u∗jixi and write x′j = (ξ′j, v′j) with v′j = (v′ij)i one can show that ε(v′ij) = δij.

Inother words, we may assume, without loss of generality, that ε(vij) = δij. Otherwise,replace each xi by x′i.

fredholm operators and morita equivalence13Let ξ = (ξi)i and v = (vi)i so that Ωξ⊕Ωv is the isomorphism from ˜An onto Im(I−˜T ˜S)mentioned above. The operator U in L ˜A(M ⊕˜An, N ⊕˜An) given byU =TΩξΩ∗µΩvis therefore invertible (please note that µ is as in the proof of 3.8).Note that(M ⊕˜An) · A = M ⊕Anand similarly(N ⊕˜An) · A = N ⊕An.So, U gives, by restriction, an invertible operator from M ⊕An to N ⊕An and, denotingthe latter by U, from now on, we haveU −T00IAn=0ΩξΩ∗µΩv −IAn.The matrix on the right hand side represents an A-compact operator: the crucial pointbeing that Ωv−IAn is the operator on An given by multiplication by the matrixvij−δiji,jwhich is in Mn(A) since ε(vij −δij) was seen to be zero (see [7], Lemma 2.4).⊓⊔The following characterization of when two Fredholm operators have the same indexis an immediate corollary of our last Lemma.3.16.

Proposition. If Ti in LA(Mi, Ni) for i = 1, 2, are Fredholm operators such thatind(T1) = ind(T2), then for some integer n, the operatorT1 ⊕T ∗2 ⊕IAn : M1 ⊕N2 ⊕An →N1 ⊕M2 ⊕Anis an A-compact perturbation of an invertible operator.Using the machinery developed so far, we may provide an alternate definition of theK-theory group K0(A).

Choose, once and for all, a cardinal number ω which is biggerthan the cardinality of An for every integer n. We remark that the role of ω is merelyto avoid set theoretical problems arising from the careless reference to the set of all A-Fredholm operator. Any choice of ω, as long as it is sufficiently large, will result in thesame conclusions.Denote by F0(A) the set of all A-Fredholm operators whose domain and codomain areHilbert modules of cardinality no larger than ω (actually we should require these Hilbertmodules to have a subset of ω as their carrier set).

Declare two elements T1 and T2 of F0(A)equivalent, if there is an integer n such that T1 ⊕T ∗2 ⊕IAn is an A-compact perturbationof an invertible operator.The quotient F(A) of F0(A) by the above equivalence relation is obviously a groupwith the operation of direct sum of Fredholm operators. The inverse of the class of T beinggiven by that of T ∗by 3.11 and 3.15.

14ruy exel3.17. Corollary.

The Fredholm index map, viewed as a mapind: F(A) →K0(A),is a group isomorphism.Proof. Follows immediately from 3.14 and 3.16.⊓⊔We should remark that 3.17 is a generalization of the fact that KK(C, A) is isomorphicto K0(A).

The new aspect being that no separability is involved. This is one of the crucialsteps in achieving our main result as we shall see shortly.4.

Preliminaries on Hilbert BimodulesWe would now like to set the present section aside in order to present a few relevant aspectsof the theory of Hilbert bimodules which will be important for our discussion of Moritaequivalence. We adopt the definition of Hilbert bimodules given in [4], 1.8.

Namely, ifA and B are C∗-algebras, a Hilbert A-B-bimodule is a complex vector space X which isa left Hilbert A-module as well as a right Hilbert B-module, and such that the A-valuedinner product(·|·): X × X →Aand the B-valued inner product⟨· , ·⟩: X × X →Bare related by the identity(ξ|η)µ = ξ⟨η, µ⟩,ξ, η, µ ∈X.Some authors prefer to use the notation ⟨· , ·⟩A and ⟨· , ·⟩B for these inner-productsbut we believe the notation indicated above makes some formulas much more readable.In particular, it is implicit that any inner-product denoted by ⟨· , ·⟩will be linear in thesecond variable while those denoted by (·|·) are linear in the first variable. We shouldnevertheless remark that the differentiated notation is not meant to imply any asymmetryin the structure of bimodules.

With the obvious interchange of left and right, any resultthat holds on the “left” will also hold on the “right” and vice-versa.As mentioned in [4], Hilbert A-B-bimodules are nothing but Rieffel’s imprimitivitybimodules (see [10], 6.10) for which it is not assumed that the range of the inner-productsgenerate the coefficient algebras.The closed span of the set {⟨x, y⟩: x, y ∈X}, which we denote by ⟨X, X⟩, is a twosided ideal in B and similarly, the closed span of {(x|y): x, y ∈X} is the ideal (X|X) of A.4.1. Definition.

The Hilbert A-B-bimodule X is said to be left-full (resp. right-full) if(X|X) coincides with A (resp.

if ⟨X, X⟩coincides with B).

fredholm operators and morita equivalence15Using the terminology just introduced, Rieffel’s imprimitivity bimodules are preciselythe Hilbert-bimodules that are simultaneously left-full and right-full.Throughout this section we shall consider fixed two C∗-algebras A and B as well asa Hilbert A-B-bimodule X. As before, M and N will denote Hilbert A-modules.

If M isa Hilbert A-module (we remind the reader of our convention according to which modulewithout further adjectives, means right module), then the tensor product module M ⊗A Xhas a natural B-valued (possibly degenerated) inner-product specified by⟨ξ1 ⊗x1, ξ2 ⊗x2⟩= ⟨x1, ⟨ξ1, ξ2⟩x2⟩,ξ1, ξ2 ∈M, x1, x2 ∈X.After moding out the elements of norm zero and completing, we are left with a HilbertB-module which we also denote, for simplicity, by M ⊗A X. See [6], 1.2.3 for details, butplease observe that the notation used there is not the same as the one just described.

Itshould also be observed that one does not need the A-valued inner-product on X in orderto perform this construction. It is enough that X be a left A-module in such a way thatthe representation of A, as left multiplication operators on X, be a ∗-homomorphism.If T is in LA(M, N), we denote by T ⊗IX the linear transformationT ⊗IX : M ⊗A X →N ⊗A Xgiven by T ⊗IX(ξ, x) = T(ξ) ⊗x for ξ in M and x in X.

A slight modification of [6], 1.2.3shows that T ⊗IX is in LB(M ⊗A X, N ⊗A X) and that ∥T ⊗IX∥≤∥T∥.Let us now present one of our most important technical results. Although quite asimple fact, with an equally simple proof, it is a crucial ingredient in this work.

Compare[6], 1.2.8.4.2. Theorem.

If the Hilbert A-B-bimodule X is left-full and if T is in KA(M, N), thenT ⊗IX is B-compact.Proof. It obviously suffices to prove the statement in case T = ΩνΩ∗µ with µ in M and νin N. Given ξ ⊗x in M ⊗A X we haveT ⊗IX(ξ ⊗x) = ν⟨µ, ξ⟩⊗x = ν ⊗⟨µ, ξ⟩x.Observe that, since X is left-full and also by [6], Lemma 1.1.4, there is no harm in assumingthat ν = ν1(y|z) for some y, z in X and ν1 in N. SoT ⊗IX(ξ ⊗x) = ν1 ⊗(y|z)⟨µ, ξ⟩x = ν1 ⊗y⟨z, ⟨µ, ξ⟩x⟩= (ν1 ⊗y)⟨µ ⊗z, ξ ⊗x⟩= Ων1⊗yΩ∗µ⊗z(ξ ⊗x).This concludes the proof.⊓⊔One of the main uses we shall have for this result is recorded in:

16ruy exel4.3. Corollary.

If X is left-full and if T ∈LA(M, N) is an A-Fredholm operator, thenT ⊗IX is B-Fredholm.Proof. If S ∈LA(N, N) is such that IM −ST is in KA(M) and IN −TS is in KA(N),thenIM⊗AX −(S ⊗IX)(T ⊗IX) = (IM −ST) ⊗IXwhich is a B-compact operator by 4.2.

The same reasoning applies to IM⊗AX −(T ⊗IX)(S ⊗IX).⊓⊔At this point, the reader may have already anticipated our strategy of using a bimoduleto create a homomorphism on K0-groups: given an element α in K0(A), we may find, using3.14, an A-Fredholm operator T whose index is α. The index, in K0(B), of the B-Fredholmoperator T ⊗IX is the image of α under the homomorphism we have in mind.

In order tomake this picture work, we need to tackle the question of well definedness, which we nowdo.4.4. Proposition.

If T1, T2 ∈LA(M, N) are A-Fredholm operators such that ind(T1) =ind(T2), then ind(T1 ⊗IX) = ind(T2 ⊗IX).Proof. According to 3.16 there is an integer n such thatT1 ⊕T ∗2 ⊕IAn : M1 ⊕N2 ⊕An →N1 ⊕M2 ⊕Anis an A-compact perturbation of an invertible operator.

By 4.2 and by the fact that thetensor product distributes over direct sums, we have that (T1⊗IX)⊕(T ∗2 ⊗IX)⊕(IAn ⊗IX)is a B-compact perturbation of an invertible operator. By 3.11 its index is therefore zero.On the other hand, also by 3.11 we haveind(T1 ⊗IX) + ind(T ∗2 ⊗IX) + ind(IAn ⊗IX) = 0which says that ind(T1 ⊗IX) = ind(T2 ⊗IX).⊓⊔A important ingredient for the functoriality properties of left-full Hilbert bimodules isthe notion of tensor product of bimodules.

In order to avoid endless calculations that arisein an abstract treatment of tensor products, we shall provide an alternative picture forbimodules, as concrete operators between Hilbert spaces, in which the coefficient algebrasare represented. The notion of representation of bimodules is described next.

Compare[4], Definition 2.1.4.5. Definition.

A representation of the Hilbert A-B-bimodule X consists of the followingdata:(i) a representation πA of A on a Hilbert space HA,(ii) a representation πB of B on a Hilbert space HB and(iii) a bounded linear transformation πX from X into the Banach space B(HB, HA) of allbounded linear operators from HB to HA.

fredholm operators and morita equivalence17Furthermore it is required that, for a ∈A, b ∈B, and x, x1, x2 ∈X,(a) πX(ax) = πA(a)πX(x)(b) πX(xb) = πX(x)πB(b)(c) πA(x1|x2)= πX(x1)πX(x2)∗(d) πB⟨x1, x2⟩= πX(x1)∗πX(x2).At this point it is necessary to remark that for x in X, one has that ∥(x|x)∥= ∥⟨x, x⟩∥(see [4], Remark 1.9). So, when we speak of ∥x∥, we mean the square root of that commonvalue.

In particular, this is the norm we have in mind when we require, in (iii), that πXbe a bounded map on X.4.6. Proposition.

Let (πA, πB, πX) be any representation of X. If either πA or πB arefaithful, then πX is isometric.Proof.

Let x ∈X. We have ∥πX(x)∥2 = ∥πX(x)πX(x)∗∥= ∥πA(x|x)∥.

So, assumingthat πA is faithful, we have ∥πX(x)∥2 = ∥(x|x)∥= ∥x∥2. A similar argument applies if πBis assumed to be faithful, instead.⊓⊔Given a representation πB of B, it is natural to ask whether or not πB is part ofthe data forming a representation of X.

To answer this question we need to bring in theconjugate module and the linking algebra. The conjugate module of X is the bimodule oneobtains by reversing its structure so as to produce a Hilbert B-A-bimodule as explained in[10], 6.17, or [4], 1.4.

We shall denote the conjugate module by X∗(although ˜X is used in[10]). The linking algebra of X, introduced in [3], 1.1 in the special case when X is bothleft and right-full, and in [4], 2.2 in general, is the C∗-algebraL =AXX∗Bequipped with the multiplicationa1x1y∗1b1 a2x2y∗2b2=a1a2 + (x1|y2)a1x2 + x1b2y∗1a2 + b1y∗2⟨y1, x2⟩+ b1b2and involutionaxy∗b=a∗yx∗b∗for a, a1, a2 ∈A, b, b1, b2 ∈B and x, x1, x2, y, y1, y2 ∈X.

Here, x∗denotes the element xof X when it is viewed in X∗.4.7. Proposition.

Let πB be a non-degenerated representation of B on the Hilbert spaceHB. Then there exists a Hilbert space HA, a non-degenerated representation πA of A onHA and a bounded linear map πX: X →B(HB, HA) which, when put together, form arepresentation of X.

18ruy exelProof. Let L be the linking algebra of X.

Thus πB becomes a representation of a subal-gebra of L, namely B. Let π be a representation of L on a Hilbert space H which containsa copy of HB, such that HB is invariant under the operators π(b), for b in B, and, finally,such that π(b)|HB = πB(b).

The existence of π, in the case that πB is cyclic, follows fromthe Theorem on extension of states and the GNS construction. In the general case, itfollows from the fact that any representation is a direct sum of cyclic representations.Viewing X as the subset of L formed by the matrices with the only non-zero entrylying in the upper right hand corner, let HA = π(X)HB.

Here, and in the sequel, productsof sets, as in “π(X)HB”, will always mean the linear span of the set of individual products.Since X is a left A-module, it is clear that HA is invariant under π(A). Denote by πAthe sub-representation of A on HA given in this way.

Observe that AX = X, by [6], 1.1.4,so we conclude that πA is non-degenerated.For each x in X we have, by definition, that π(x)HB ⊆HA. By totally differentreasons we also have that πX(x)∗HA ⊆HB.

In fact, let ξ ∈HA. Without loss of generalitywe may assume that ξ = π(y)η where y is in X and η ∈HB.

We then haveπ(x)∗ξ = π(x)∗π(y)η = π⟨x, y⟩η ∈HB.For each x in X let πX(x) denote the element of B(HB, HA) obtained by restriction ofπ(x). The properties of Definition 4.5 may now be easily checked.If the reader doesdecide to do so, we suggest the formal definition πX(x) := i∗Aπ(x)iB where iA and iB arethe inclusion operators from HA and HB into H. This has the advantage of taking careof the subtle issue of reducing the size of the co-domain of an operator.⊓⊔4.8.

Proposition. There exists a representation (πA, πB, πX) such that both πA and πBare faithful (and hence πX is isometric by 4.6).Proof.

Let π be a faithful representation of the linking algebra L on the Hilbert spaceH. Define HA = π(A)H and HB = π(B)H and let πA and πB be the corresponding sub-representations of A and B on HA and HB, respectively.

Since AX = X and BX∗= X∗it is clear that π(X)H ⊆HA and that π(X)∗H ⊆HB. If, for each x in X, we denote byπX(x) the element of B(HB, HA) given by restriction of π(x), the proof can be completedas in 4.7.⊓⊔4.9.

Lemma. The setPni=1(xi|xi): n ∈N, xi ∈Xis dense in the positive cone of(A|A).Proof.

For a = Pni=1(xi|yi). We havea∗a =Xi,j(yi|xi)(xj|yj) =Xi,j(yi|xi)xj|yj=Xi,jyi⟨xi, xj⟩|yj.The matrix⟨xi, xj⟩i,j ∈Mn(B) is a positive matrix as observed in [6], 1.2.4.

So, thereisbiji,j in Mn(B) such that⟨xi, xj⟩=nXk=1bikb∗jk.

fredholm operators and morita equivalence19We then havea∗a =Xi,j,k(yibik|yjbjk).If we now define zk = Pni=1 yibik we havea∗a =nXk=1(zk|zk).Since the set of elements a∗a, with a as above, is clearly dense in the positive cone of(A|A), the proof is complete.⊓⊔4.10. Proposition.

Let (πA, πB, πX) be a representation of X. If πB is faithful then πAis faithful on (A|A).Proof.

Using 4.9, it is enough to show that ∥πAPni=1(xi|xi)∥= ∥Pni=1(xi|xi)∥. Thus,let h = Pni=1(xi|xi) and observe that∥πA(h)∥= ∥nXi=1πX(xi)πX(xi)∗∥=πX(x1), .

. ., πX(xn)πX(x1)∗...πX(xn)∗where, in the last term above, we mean the product of a row matrix by a column matrix.Since the identity ∥T ∗T∥= ∥TT ∗∥holds for general operators, the above equalsπX(x1)∗...πX(xn)∗πX(x1), .

. ., πX(xn)= ∥πX(xi)∗πX(xj)i,j∥= ∥πB⟨xi, xj⟩i,j∥= ∥⟨xi, xj⟩i,j∥.The exact same computations done so far can obviously be repeated, in reverse order, fora representation (ρA, ρB, ρX) in which all components are isometric, as for example, therepresentation provided by 4.8.

This shows that the last term above equals ∥ρA(h)∥= ∥h∥.So, ∥πA(h)∥= ∥h∥.⊓⊔From this point on, and until the end of this section, we shall consider fixed anotherC∗-algebra, denoted C, and a Hilbert B-C-bimodule Y . Our goal is to make sense ofX ⊗B Y as a Hilbert A-C-bimodule.

So, for the time being, let us denote by X ⊗B Y , thealgebraic tensor product of X and Y over B which provides us with an A-C-bimodule.4.11. Definition.

Let (·|·) and ⟨· , ·⟩be the sesqui-linear forms on X ⊗B Y (the first onebeing linear in the first variable and vice-versa) specified by(x1 ⊗y1|x2 ⊗y2) = (x1(y1|y2)|x2)

20ruy exeland⟨x1 ⊗y1, x2 ⊗y2⟩= ⟨y1, ⟨x1, x2⟩y2⟩.The only steps that are not entirely trivial in checking that this can be made into aHilbert B-C-bimodule are that(a) both sesqui-linear forms above are positive and(b) ∥(z|z)∥= ∥⟨z, z⟩∥for all z in X ⊗B Y .In order to check these, fix a faithful non-degenerated representation πB of B in someHilbert space HB. Using 4.7 we may find a representation πA on a space HA as well asa representation πX of X, by bounded operators from HB to HA, satisfying the axiomsdescribed in Definition 4.5.Using the symmetric version of 4.7 we can also find a Hilbert space HC as well as πCand πY satisfying the conditions of 4.5.

Note that, by 4.6, both πX and πY are isometric.By 4.10 it follows that πA is faithful on (A|A) and that πC is faithful on ⟨C, C⟩. Considerthe mapρ: X ⊗B Y →B(HC, HA)given by ρ(x ⊗y) = πX(x)πY (y) (meaning composition of operators).

Observe that, forx1, x2 in X and y1, y2 in Y , we haveπA(x1 ⊗y1|x2 ⊗y2)= πX(x1(y1|y2))πX(x2)∗= πX(x1)πY (y1)πY (y2)∗πX(x2)∗= ρ(x1 ⊗y1)ρ(x2 ⊗y2)∗.Similarly we have πC⟨x1 ⊗y1, x2 ⊗y2⟩= ρ(x1 ⊗y1)∗ρ(x2 ⊗y2). If z = Pni=1 xi ⊗yi isan arbitrary element of X ⊗B Y we then have thatπA(z|z)=Xρ(xi ⊗yi) Xρ(xi ⊗yi)∗which is clearly a positive element in B(HA).

Therefore, since πA is faithful on (X|X),we conclude that (z|z) is positive. Similarly it can be shown that ⟨z, z⟩is positive as well.This proves (a) above.

With respect to (c) we have∥(z|z)∥= ∥πA(z|z)∥=Xρ(xi ⊗yi) Xρ(xi ⊗yi)∗=Xρ(xi ⊗yi)∗Xρ(xi ⊗yi) = ∥πC⟨z, z⟩∥= ∥⟨z, z⟩∥.This said, we may define unambiguously, a semi-norm in X ⊗B Y by ∥z∥= ∥(z|z)∥=∥⟨z, z⟩∥. After moding out by the elements of norm zero and completing, we are left with aHilbert A-C-bimodule, which we denote, for simplicity, by X ⊗B Y , as well.

Observe thatthe procedure of moding out the null elements and completing is equivalent to consideringthe closure of ρ(X ⊗B Y ) in B(HC, HA), which, incidentally is the same as πX(X)πY (Y ).In light of 4.5, it is easy to see that the A-C-bimodule structure is reproduced asthe usual composition of operators. Furthermore, the A-valued inner product becomes(T|S) = TS∗while ⟨T, S⟩= T ∗S, as long as the appropriate identifications are made.

fredholm operators and morita equivalence214.12. Proposition.

If both X and Y are left-full then so is X ⊗B Y .Proof. By [6], 1.1.4, it follows that XB = X.

But, since we are assuming that (Y |Y ) = Bwe get X(Y |Y ) = X. It follows that(X ⊗B Y |X ⊗B Y ) = (X(Y |Y )|X) = (X|X) = A.⊓⊔Recall that X∗denotes the conjugate module of X.

Clearly X∗is left-full (resp. right-full) if and only if X is right-full (resp.

left-full).4.13 Proposition. The tensor product Hilbert A-A bimodule X ⊗B X∗is isomorphic to(A|A) (once (A|A) is given its obvious structure of Hilbert A-A-bimodule, as any ideal ofA).Proof.

Choosing a faithful representation as in 4.8 we may assume that A ⊆B(HA),B ⊆B(HB) and X ⊆B(HB, HA). In addition the bimodule structure is composition ofoperators and the inner products are given by (x|y) = xy∗and ⟨x, y⟩= x∗y.

The tensorproduct X⊗BX∗is moreover identified with XX∗(where the last occurrence of “∗” shouldbe interpreted as the usual operator involution). This said, X ⊗B X∗= (A|A).⊓⊔5.

K-Theory and Hilbert BimodulesThe stage is now set for the presentation of the main section of this work. Given C∗-algebras A and B as well as a left-full Hilbert A-B-bimodule X, we want to define a grouphomomorphism X∗: K0(A) →K0(B) which will be proven to be an isomorphism if X isright-full as well.5.1.

Definition. Let X be a left-full Hilbert A-B-bimodule.

If α is an element of thegroup K0(A), which we identify with F(A) under the isomorphism of 3.17, let T be aFredholm operator representing α in the sense that ind(T) = α. We denote by X∗(α) theelement of K0(B) defined byX∗(α) = ind(T ⊗IX).Clearly, by 4.4, X∗is well defined and it is not hard to see that it is a group homo-morphism.5.2.

Proposition. Let X be a left-full Hilbert A-B-bimodule and Y be a left-full HilbertB-C-bimodule, then Y∗◦X∗= (X ⊗B Y )∗.Proof.

This follows from the easy fact that tensor products are associative, even if wedrag along all the extra structure of Hilbert bimodules.⊓⊔Recall that the C∗-algebras A and B are said to be strongly Morita equivalent if thereexists a Hilbert A-B-bimodule which is simultaneously left and right-full. Such a module iscalled an imprimitivity bimodule.

Given an imprimitivity bimodule X, we may thereforeconsider the homomorphisms X∗: K0(A) →K0(B) and (X∗)∗: K0(B) →K0(A), whichcompose to the identity in either order by 5.2 and 4.13. The immediate outcome of thesefacts is our main result.

22ruy exel5.3. Theorem.

If A and B are strongly Morita equivalent and X is an imprimitivitybimodule, then X∗is an isomorphism from K0(A) onto K0(B).In order to treat K1-groups we shall use the usual argument of taking suspensions.The formalism of Hilbert bimodules developed in section 4 makes is very easy to discussHilbert bimodules over tensor product C∗-algebras. So, before we embark on a study ofK1, let us briefly deal with “external tensor products”.Let Ai and Bi be C∗-algebras and Xi be Hilbert Ai-Bi-bimodules for i = 1, 2.Under faithful representations we may assume that Ai ⊆B(HAi), B ⊆B(HBi) andXi ⊆B(HBi, HAi).Denote by X1 ⊗X2 the closed linear span of the set {x1 ⊗x2: x1 ∈X1, x2 ∈X2} ofoperators in B(HB1 ⊗HB2, HA1 ⊗HA2).

It is not hard to see that X1 ⊗X2 is a Hilbert(A1⊗A2)-(B1⊗B2)-bimodule. Here A1⊗A2 and B1⊗B2 mean the spatial tensor products.If xi, yi ∈Xi for i = 1, 2, observe that(x1 ⊗y1|x2 ⊗y2) = (x1 ⊗y1)(x2 ⊗y2)∗= (x1x∗2) ⊗(y1y∗2) = (x1|x2) ⊗(y1|y2)so we see that the concrete inner-products on X1 ⊗X2 may be defined abstractly, at leaston the algebraic tensor product of X1 by X2, without mentioning the representations.

Infact it is easy to see that the definition of X1 ⊗X2 given above does not depend (up tothe obvious notion of isomorphism) on the particular faithful representation chosen.The object so defined is called the “external” tensor product of X1 and X2 (compare[6], 1.2.4). It is readily apparent that if each Xi is left-full (resp.

right-full) then so isX1 ⊗X2. As an obvious consequence we have:5.4.

Proposition. If Ai and Bi are C∗-algebras and if Ai is strongly Morita equivalentto Bi under the imprimitivity bimodule Xi, for i = 1, 2, then A1 ⊗B1 is strongly Moritaequivalent to A2 ⊗B2under the imprimitivity bimodule X1 ⊗X2.Let C0(R) denote the C∗-algebra of all continuous complex valued functions on thereal line.

If A and B are strongly Morita equivalent under the imprimitivity bimodule X,it follows that the suspension of A, namely SA = C0(R) ⊗A and and SB are stronglyMorita equivalent to each other under the imprimitivity bimodule C0(R) ⊗X. Using thestandard isomorphism between K1(A) and K0(SA), we have:5.5.

Theorem. If A and B are strongly Morita equivalent and X is an imprimitivitybimodule, then (C0(R) ⊗X)∗is an isomorphism from K1(A) onto K1(B).References[1] B. Blackadar, “K-theory for operator algebras”, MSRI Publications, Springer–Verlag,1986..[2] L. G. Brown, “Stable isomorphism of hereditary subalgebras of C∗-algebras”, PacificJ.

Math. 71 (1977), 335–348.

fredholm operators and morita equivalence23[3] L. G. Brown, P. Green and M. A. Rieffel, “Stable isomorphism and strong Moritaequivalence of C∗-algebras”, Pacific J. Math.

71 (1977), 349–363. [4] L. G. Brown, J.

A. Mingo and N. T. Shen, “Quasi-multipliers and embeddings ofHilbert C∗-bimodules”, preprint, Queen’s University, 1992. [5] A. Connes, “Non-commutative differential geometry”, Publ.

Math. IHES 62 (1986),257–360.

[6] K. Jensen and K. Thomsen, “Elements of KK-Theory”, Birkh¨auser, 1991. [7] G. G. Kasparov, “Hilbert C∗-modules: Theorems of Stinespring and Voiculescu”, J.Operator Theory 4 (1980), 113–150.

[8] G. G. Kasparov, “The operator K-functor and extensions of C∗-algebras”, Math.USSR Izvestija 16 (1981), 513–572. [9] W. Paschke, “Inner product modules over B∗-algebras”, Trans.

Amer. Math.

Soc. 182(1973), 443–468.

[10] M. A. Rieffel, “Induced representations of C∗-algebras”, Adv. Math.

13 (1974), 176–257. [11], “Morita equivalence for C∗-algebras and W ∗-algebras”, J.

Pure Appl. Alge-bra 5 (1974), 51–96.

[12], “Strong Morita equivalence of certain transformation group C∗-algebras”,Math. Ann.

222 (1976), 7–22. [13] L. H. Rowen, “Ring theory – Student edition”, Academic Press, 1991.

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