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CIRCLE ACTIONS ON C*-ALGEBRAS, PARTIAL

arXiv:funct-an/9211001v1 22 Nov 1992CIRCLE ACTIONS ON C*-ALGEBRAS, PARTIALAUTOMORPHISMS AND A GENERALIZEDPIMSNER–VOICULESCU EXACTSEQUENCERuy Exel∗Department of Mathematics and StatisticsUniversity of New MexicoAlbuquerque, New Mexico 87131e-mail: exel@math.unm.eduAbstract. We introduce a method to study C∗-algebras possessing an actionof the circle group, from the point of view of its internal structure and its K-theory.

Under relatively mild conditions our structure Theorem shows that anyC∗-algebra, where an action of the circle is given, arises as the result of a con-struction that generalizes crossed products by the group of integers.Such a generalized crossed product construction is carried out for any partialautomorphism of a C∗-algebra, where by a partial automorphism we mean anisomorphism between two ideals of the given algebra.Our second main result is an extension to crossed products by partial auto-morphisms, of the celebrated Pimsner-Voiculescu exact sequence for K-groups.The representation theory of the algebra arising from our construction isshown to parallel the representation theory for C∗-dynamical systems. In partic-ular, we generalize several of the main results relating to regular and covariantrepresentations of crossed products.∗Partially supported by FAPESP, Brazil.

On leave from the University of S˜ao Paulo.1991 MR Subject Classification: Primary 46L05, 46L40, 46L45, 46L55, 46L80, 19K35; Secondary47B35, 34C35.

circle actions and partial automorphisms21. IntroductionGiven a group action on a manifold M, one of the main goals of the Dynamical Systemsspecialist is to describe M in terms of the elementary components singled out by thepresence of the action as, for example, orbits and fixed points.

The simplest case in whichsuch a description can be thoroughly carried out is that of a free action of a compact group:M can be then described as a principal bundle over the quotient space.While the interaction between Operator Algebras and Dynamical Systems has beenvery intense in the last several decades, little has been accomplished in addressing theabove question from the Operator Algebras point of view. The C∗-Dynamical Systemscounterpart of that program would be to describe the structure of a C∗-algebra, where agroup action is given, in terms of elementary data which, supposedly should be extractedfrom the action.Among the few cases in which the above mentioned task was successfully carried out,is a result by Paschke (Theorem 2.3 in [14]) in which it is shown that, under certaincircumstances, a C∗-algebra carrying an action of the circle group can be described as thecrossed product of its fixed point subalgebra (the counterpart of the quotient space) by anaction of the integers.

The main hypothesis in that Theorem is that the action have “largespectral subspaces” (see [14] and [11]), a condition that, at least in the case of the circle,singles out the non-commutative analog of free actions.If one goes as far as to accept that a crossed product is the non-commutative version ofa principal bundle, then Paschke’s result can be considered as an ipsis-literis generalizationof the well known fact about free actions of compact groups mentioned above.The circle of ideas around the notion of “large spectral subspaces” has now a longhistory. Without attempting a comprehensive account, we should mention that a versionof this notion appeared in a paper by Fell [8] in 1969 under the name of homogeneousactions, later renamed by Fell himself [9] to saturated actions.

Related notions were alsostudied by Phillips [16]. Recent work of Rieffel [20] contains yet another version of thisconcept, providing a far reaching generalization of the Takesaki-Takai duality (Corollary1.7 of [20], see also [11]).But, in the same way that free actions are not the rule, the conditions so far alludedto, exclude many actions of S1 which one would still like to investigate.The purpose of the present work is to introduce a method which allows for a descriptionof the structure of C∗-algebras carrying circle actions which are not supposed to have largespectral subspaces.

While our method does not include all possible circle actions, sincewe assume our actions to be semi-saturated (see below), the gain in generality is verysignificant in the sense that a wealth of new examples becomes tractable using our theory.A typical such example is the action of S1 on the Toeplitz algebra, given by conjugation bythe diagonal unitaries diag(1, z, z2, ...), for z ∈S1. The condition of having large spectralsubspaces fails for this action.Our method consists in first, introducing a construction, inspired on the crossed prod-uct construction, which produces a C∗-algebra equipped with a circle action.

We then goon to show that any C∗-algebra having an action of S1 arises as the result of our con-

circle actions and partial automorphisms3struction as long as the action satisfies some relatively mild restrictions. In other wordswe provide a means to “disassemble” a C∗-algebra possessing a circle action, reverting, ina sense, the crossed product construction.While a crossed product by Z depends on an automorphism of the given C∗-algebra A(i.e., an action of Z), our construction requires a partial automorphism.

Precisely speaking,a partial automorphism of A is a triple Θ = (θ, I, J) where I and J are ideals in A andθ: I →Jis a C∗-algebra isomorphism.Given a partial automorphism, we constructits “covariance algebra” which we denote by C∗(A, Θ). When both ideals agree with A,then our construction becomes the usual crossed product construction, that is, C∗(A, Θ)becomes A ×θ Z.

For that reason, C∗(A, Θ) should be considered as the crossed productof A by the partial automorphism Θ.Our second major objective, accomplished by Theorem (7.1), is a generalization of thecelebrated Pimsner–Voiculescu exact sequence [17]. Precisely, we get the following exactsequence of K-groupsK0(J)i∗−θ−1∗−→K0(A)i∗−→K0(C∗(A, Θ))↑↓K1(C∗(A, Θ))i∗←−K1(A)i∗−θ−1∗←−K1(J)As in most proofs of the Pimsner–Voiculescu exact sequence and related results ([4],[5], [6], [1]), we derive our exact sequence from the K-theory exact sequence for a suitableToeplitz extension.

The crucial step, as it is often the case, is to show that A has the sameK-groups as the Toeplitz algebra. We do so by showing that these algebras are, in fact,KK-equivalent.My first attempt at proving the exactness of the sequence above was, of course, bytrying to deduce it from the well known result of Pimsner and Voiculescu.

After failing indoing so, I am now tempted to believe that this cannot be done. Our proof is done fromscratch and, given a rather involving use of KK-theory, it turns out considerably longerthen the available proofs of the original result.We would like to thank Bill Paschke for bringing to our attention his paper [1] with J.Anderson, where they generalize a result, from unpublished lecture material of Arveson’s,as well as from Proposition (5.5) in [6], from which the crucial step in [17] follows.

Ourgeneralization of these ideas plays a central role in the proof of our result.One intersting aspect, central in our use of KK-theory, is worth mentioning here. IfA and B are C∗-algebras, then KK(A, B) may be described, as was shown by Cuntz [7],by the set of homotopy classes of homomorphisms from qA to the multiplier algebra ofB ⊗K.

Nevertheless the KK-theory elements that we need to introduce have no easydescription in such terms. Instead we exhibit these elements by replacing B ⊗K, above,by an algebra which contains B as a full corner and hence is stably isomorphic to B (atleast in the separable case).

circle actions and partial automorphisms4We feel that our structure Theorem, used in conjunction with the generalized Pimsner–Voiculescu exact sequence above, adds to the growing collection of powerful tools, devel-oped since the early eighties, designed to compute K-groups for C∗-algebras.Several attempts to generalize crossed products have been made by many authors.Among those we should mention Kumjian’s work on Localizations [12].Another wellknown example is the theory of crossed products by endomorphisms developed by Paschkein [13]. Paschke’s crossed products carry a circle action, which can easily be made tofall under our restrictions.

So, in a certain sense, his algebras are special cases of ourconstruction. In addition, we have recently learned that L. Brown has developed a theorywhich formalizes the concept of crossed products by imprimitivity bimodules.Our construction should be regarded as a generalization of crossed products, only aslong as the group Z, of integers, is concerned.

With some more work, it appears to methat one could attempt to widen the present methods to include a larger class of groups.It would be largely desirable, although quite likely very difficult, to be able to study alongthe present lines, actions of non-compact groups such as the group of real numbers. FlorinPop pointed out to me recently that the definition of a “partial action” of a discretegroup, as well as that of the corresponding covariance algebra, could be obtained by trivialmodifications of our definitions.As we already indicated, our motivation, rather than to produce new classes of C∗-algebras, is to attempt a description of the structure of C∗-algebras possessing a circularsymmetry, represented by an action of the circle.

A partial automorphism arises from agiven action α of S1 on a C∗-algebra B, in the following way. First, one lets B1 be thefirst spectral subspace of α, i.e., B1 = {b ∈B : αz(b) = zb, z ∈S1}.

The set B∗1B1(meaning, according to convention (2.2) adopted throughout this paper, the closed linearspan of the set of products) is an ideal of the subalgebra B0 of fixed points for α and thesame is true with respect to B1B∗1. These ideals are strongly Morita equivalent [18], [19]with B1 playing the role of the imprimitivity bimodule.

Morita equivalent C∗-algebras arequite often isomorphic to each other and indeed, under the assumption that the algebras bestable with strictly positive elements, they are forcibly isomorphic [3]. The two isomorphicideals are thus the ingredients of our partial automorphism.Our structure Theorem (4.21), states that B is (isomorphic to) the covariance algebraof that partial automorphism as long as B is generated, as a C∗-algebra, by the union of B0and B1.

This condition, which we call semi-saturation, is a weakening of the condition ofhaving large spectral subspace and is the point of departure for our theory. As opposed towhat happens to the latter, even when absent, our condition can be forced upon the actionby restricting one’s attention to the subalgebra of B that B0 ∪B1 happen to generate.The algebraic formalism developed here seems to flow with such a naturality that itis perhaps a bit surprising that it has been overlooked until now.

Nevertheless, it doesnot seem possible to extend our methods to rings not possessing a C∗-algebra structuresince we make extensive use of the existence of approximate units and facts like ideals ofideals of a C∗-algebra are, themselves, ideals of that algebra, or that the intersection oftwo ideals equals their product.

circle actions and partial automorphisms5Permeating most of our techniques is a concept we have not tried to formalize, but Ithink the effort to do so seems worthwhile. The reader is invited to compare the definitionof multipliers of C∗-algebras on one hand, and (4.11) and (4.13) on the other, and he willlikely see the rudiments of a concept which deserves the name of partial multipliers.After a short section (2), intended mainly to fix some notation, we describe in section(3) our construction of the covariance algebra associated to a partial automorphism.

Thefollowing section, numbered (4) is where our structure Theorem (4.21), is proved. The fifthsection deals with the representation theory for our covariance algebras and to some extentcan be regarded as the taming of the algebraic properties of partial isometric operators onHilbert’s space.

With surprising ease, partial isometries are made to play the role usuallyplayed by implementing unitaries for representations of crossed product algebras.In section (6) we introduce the Toeplitz algebra associated to a partial automorphismand describe it, within our theory, as a covariance algebra, as well. In doing so, we areable to obtain a crucial universal property, Lemma (6.7), characterizing representations ofthe Toeplitz algebra.

The seventh and last section is where our main K-theoretical work isdeveloped and where we prove the existence of the generalized Pimsner–Voiculescu exactsequence (Theorem 7.1).Our notation is reasonably standard except, possibly, for the definitions in (2.2) and(5.4) as well as for our use of the symbol L after (6.3). The reader is advised to go overthese immediately, in order to avoid possible surprises.This work is an extended version of a paper by the author, entitled “The Structure ofActions of the Circle Group on C∗-Algebras”.2.

Spectral SubspacesThis section is concerned with some preliminaries about C∗-dynamical systems based onthe circle group. Let B be a fixed C∗-algebra and α an action of S1 on B.2.1.

Definition. For each n ∈Z the nth spectral subspace for α is defined byBn = {b ∈B : αz(b) = znbfor z ∈S1}.It is an easy matter to verify that BnBm ⊆Bn+m and that B∗n = B−n.

Regardingthe product BnBm just mentioned, we adopt the following convention:2.2. Definition.

If X and Y are subsets of a C∗-algebra then XY denotes the closedlinear span of the set of products xy with x ∈X and y ∈Y .A simple fact we shall make extensive use of, is the following.2.3. Proposition.

For each n ∈Z one has that B∗nBn is a closed two sided ideal of thefixed point subalgebra B0.2.4. Definition.

The nth spectral projection for the action α is the transformationPn: B →B

circle actions and partial automorphisms6defined byPn(b) =ZS1 z−nαz(b)dz , b ∈B.It is well known that Pn is a contractive projection whose image is Bn.2.5. Proposition.

Let b ∈B and φ be a continuous linear functional on B. (a) If φ(Pn(b)) = 0 for all n ∈Z, then φ(b) = 0.

(b) If Pn(b) = 0 for all n ∈Z, then b = 0. (c) Ln∈Z Bn is dense in B.Proof.

(cf. [14], Proposition 2.1).

For the first statement it is enough to note that thenth Fourier coefficient of z ∈S1 →φ(αz(b)) is given by φ(Pn(b)). Hence, if φ(Pn(b)) = 0for all n we have that φ(αz(b)) is identically zero, as a function of z and, in particular,φ(b) = 0.

Finally (b) and (c) follow from (a) and the Hahn-Banach Theorem.⊓⊔2.6. Proposition.

If (eλ)λ∈Λ is an approximate identity for B∗nBn then, for each x ∈Bn,we have x = limλ xeλ.Proof. We have∥x −xeλ∥2 = ∥(x −xeλ)∗(x −xeλ)∥= ∥x∗x −x∗xeλ −eλx∗x + eλx∗xeλ∥≤∥x∗x −x∗xeλ∥+ ∥eλ∥∥x∗x −x∗xeλ∥→0.⊓⊔An immediate consequence is:2.7.

Corollary. For each n ∈Z one has BnB∗nBn = Bn.Since the product of two ideals in a C∗-algebra equals their intersection we have:2.8.

Proposition. If n and m are integers then B∗nBnB∗mBm = B∗mBmB∗nBn.The following concludes our preparations.2.9.

Proposition. Let B and B′ be C∗-algebras and let α and α′ be actions of S1 on Band B′, respectively.

Suppose ψ: B →B′ is a covariant homomorphism. If the restrictionof ψ to the fixed point subalgebra B0 is injective, then ψ itself, is injective.Proof.

Assume b ∈B is such that ψ(b) = 0. Then, denoting by Pn and P ′n the respectivespectral projections, we haveψ(Pn(b)Pn(b)∗) = ψ(Pn(b))ψ(Pn(b))∗= P ′n(ψ(b))P ′n(ψ(b))∗= 0.If one now notes that Pn(b)Pn(b)∗∈B0, the hypothesis is seen to imply that Pn(b) = 0and so, by (2.5.b), we have b = 0.⊓⊔

circle actions and partial automorphisms73. Partial Automorphisms and their Covariance AlgebrasIn this section we describe a generalization of the concept of crossed products by anautomorphism.

For that purpose, let A be a C∗-algebra considered fixed throughout thepresent section.3.1. Definition.

A partial automorphism of A is a triple Θ = (θ, I, J) where I and Jare ideals in A (always assumed closed and two sided) andθ: I →Jis a C∗-algebraisomorphism.If such a partial automorphism is given, we let, for each integer n, Dn denote thedomain of θ−n with the convention that D0 = A and θ0 is the identity automorphism ofA. The domain of θ−n is clearly the image of θn so this provides an equivalent definitionof Dn.Alternatively, we can give an inductive definition for these objects by letting D0 = A,Dn+1 = {a ∈J : θ−1(a) ∈Dn}for n ≥0 andDn−1 = {a ∈I : θ(a) ∈Dn}for n ≤0.According to this, one clearly has D1 = J and D−1 = I.

Of course, unless I and Jhave a substantial intersection, the sets Dn would be rather small. The extreme case inwhich I ∩J = {0} will see Dn being the singleton {0} for all n, except for n = −1, 0, 1.One of the simplest examples of partial automorphisms is obtained when one letsA = Cm, I = {(xi) ∈Cm : xm = 0}, J = {(xi) ∈Cm : x1 = 0} and θ be the forward shiftθ(x1, .

. ., xm−1, 0) = (0, x1, .

. ., xm−1).In this case Dn becomes the set of all m-tuples having n leading zeros when n ≥0 or|n| trailing zeros if n ≤0.As it will turn out, the covariance algebra for this example is isomorphic to the algebraof m × m complex matrices.3.2.

Proposition. For each integer n, Dn is an ideal in A.Proof.By definition the assertion is obvious for n = −1, 0, 1.

Arguing by inductionassume that n ≥0 and that Dn is and ideal in A. Then Dn+1 is clearly an ideal in J,being the inverse image of Dn under θ−1.

Since an ideal of an ideal of a C∗-algebra isalways an ideal of that C∗-algebra (by existence of approximate identities) we have thatDn+1 is an ideal in A. A symmetric argument yields the result for negative values of n. ⊓⊔Let’s agree to call by the name of chain any finite sequence (a0, a1, .

. ., an) of elementsin A such that a0 ∈I, an ∈J and ai ∈I ∩J for i = 1, .

. ., n −1, satisfying θ(ai−1) = aifor i = 1, 2, .

. ., n. The integer n will be called the length of said chain.

circle actions and partial automorphisms8The concept of chain can be used for giving yet another definition of Dn. Namely, Dnis the set of elements a in A which admit a chain of length |n|, ending in a in case n ≥0or, beginning in a if n ≤0.The following proposition can be easily proven if one thinks in terms of chains.

It isnevertheless crucial for what follows.3.3. Proposition.

If n and m are integers then θ−n(Dn ∩Dm) ⊆Dm−n. In addition, ifx is in Dn ∩Dm then θn−m(θ−n(x)) = θ−m(x).Denote by L the subspace of ℓ1(Z, A) formed by all summable sequences (a(n))n∈Zsuch that a(n) ∈Dn for each n. We propose to equip L with an involutive Banach algebrastructure.

For that purpose we define, for a and b in L,(a ∗b)(n) =∞Xk=−∞θk(θ−k(a(k))b(n −k))(a∗)(n) = θn(a(−n)∗)∥a∥=∞Xn=−∞∥a(n)∥.We next verify some of the axioms of involutive Banach algebras for the multipli-cation, involution and norm defined above. But, before that, we should note that ourmultiplication is well defined since, for each k, θ−k(a(k)) is in D−k while b(n −k) is inDn−k.

The product θ−k(a(k))b(n −k) is therefore in the intersection D−k ∩Dn−k. ByProposition (3.3) it follows that the kth summand in our definition of the product in factlies in D(n−k)−(−k) = Dn.

Similarly, note that a∗is an element of L for each a in L.3.4. Proposition.

The product defined above is associative.Proof. If a in is Dn we shall denote by aδn the element of L given by (aδn)(m) = δn,ma,where δn,m is the Kronecker symbol.It is readily seen that the associativity of our product follows from the identity(anδn ∗amδm) ∗apδp = anδn ∗(amδm ∗apδp)where ai ∈Di for i = n, m, p, which we now propose to prove.

Using Proposition (3.3),the left hand side above becomes(θn(θ−n(an)am)δn+m) ∗apδp =θn+m(θ−n−m(θn(θ−n(an)am))ap)δn+m+p =θn+m(θ−m(θ−n(an)am)ap)δn+m+p.

circle actions and partial automorphisms9On the other hand, the right hand side of our identity equalsanδn ∗(θm(θ−m(am)ap)δm+p) = θn(θ−n(an)θm(θ−m(am)ap))δn+m+p.Note that the term within the outermost parenthesis, to the right of the last equalsign, is in D−n ∩Dm, so that the coefficient of δn+m+p above is in Dm+n by (3.3). Thus,proving our identity, amounts to verify thatθ−m(θ−n(an)am)ap = θ−n−m(θn(θ−n(an)θm(θ−m(am)ap))or, again by (3.3), thatθ−m(θ−n(an)am)ap = θ−m(θ−n(an)θm(θ−m(am)ap)).Let (ui)i be an approximate identity for D−m.

So the left hand side above can bewritten aslimi θ−m(θ−n(an)am)uiap = limi θ−m(θ−n(an)amθm(uiap)) =limi θ−m(θ−n(an)θm(θ−m(am)uiap)) = θ−m(θ−n(an)θm(θ−m(am)ap)),concluding the proof.⊓⊔3.5. Proposition.

For a and b in L one has (ab)∗= b∗a∗.Proof. Arguing as in the beginning of the previous proof it is enough to verify that(anδn ∗amδm)∗= (amδm)∗∗(anδn)∗for an ∈Dn and am ∈Dm.

The left hand side equals, by definition and by (3.3)(θn(θ−n(an)am)δn+m)∗= θ−n−m(θn(θ−n(an)am)∗)δ−n−m =θ−m(θ−n(an)am)∗δ−n−m = θ−m((am)∗θ−n(an)∗)δ−n−m.On the other hand,(amδm)∗∗(anδn)∗= θ−m(am)∗δ−m ∗θ−n(an)∗δ−n =θ−m(θm(θ−m(am)∗)θ−n(an)∗)δ−m−n =θ−m((am)∗θ−n(an)∗)δ−m−n.⊓⊔The main difficulties being overcome, we now have:

circle actions and partial automorphisms103.6. Theorem.

L is an involutive Banach algebra with the above defined multiplication,involution and norm.3.7. Definition.

The covariance algebra for the partial automorphism Θ = (θ, I, J) isthe C∗-algebra C∗(A, Θ) obtained by taking the enveloping C∗-algebra of L.Elementary examples of this construction are standard crossed products by the groupof integers and the algebra of n × n complex matrices.The latter is obtained as thecovariance algebra for the partial automorphism mentioned after (3.1).A slightly more elaborate example is what one gets by taking A = c0(N), I = A, Jthe ideal formed by sequences with a leading zero and, finally, θ the forward shift. Thecovariance algebra, in this case, can be shown to be the algebra of compact operators onℓ2(N).

If, instead, we took A to be the unitization of c0(N), I = A, J the set of elementsin A with a leading zero coordinate, and θ the forward shift, then the covariance algebrabecomes the Toeplitz algebra, that is, the C∗-algebra generated by the forward shift onℓ2(N). The above statements follow as easy corollaries of our structure Theorem (4.21),once one considers the circle action given, in each case, by conjugation by diag(1, z, z2, ...),for z ∈S1.

With respect to the Toeplitz algebra, see also (6.6).The major example we would like to present is related to actions of the circle groupon C∗-algebras. As we shall see, any C∗-algebra having an action of S1, under relativelymild hypothesis on the action, is the covariance algebra for a certain partial automorphismof the algebra of fixed points under S1.

This will be the subject of section (4).3.8. Proposition.

Let (ei)i be a (bounded) approximate identity for A then (eiδ0)i isan approximate identity for L and hence also for C∗(A, Θ).Proof. For an ∈Dn we havelimi (eiδ0) ∗(anδn) = limi eianδn = anδn.By taking adjoints if follows that limi(anδn) ∗(eiδ0) = anδn.

Since supi ∥eiδ0∥< ∞theabove implies the conclusion.⊓⊔3.9.Proposition. The map E: a ∈L →a(0)δ0 is a contractive positive conditionalexpectation [18] from L onto the subalgebra Aδ0 of L.Proof.

To prove positivity let a ∈L. Then(a ∗a∗)(0) =∞Xk=−∞θk(θ−k(a(k))a∗(−k)) =Xkθk(θ−k(a(k))θ−k(a(k)∗)) =Xka(k)a(k)∗≥0.The remaining statements can be easily verified and are left to the reader.⊓⊔

circle actions and partial automorphisms113.10. Corollary.

The obvious inclusion of A into L, composed with the map from L intoC∗(A, Θ), gives an isometric *-homomorphism of A into the latter.Proof. Let a ∈A and let f be a state on A such that f(a∗a) = ∥a∥2.

Identifying A andits copy Aδ0 within L provides us with a state f on Aδ0 such that f((aδ0)∗∗(aδ0)) = ∥a∥2.If that state is composed with the conditional expectation of Proposition (3.9) we get astate on L. By (3.8) L has an approximate identity so one is allowed to use the GNSconstruction, which provides a representation π of L, and hence of the covariance algebra,having a cyclic unit vector ξ and which satisfies⟨π(aδ0)ξ, π(aδ0)ξ⟩= f(a∗a) = ∥a∥2.Thus, the norm ∥aδ0∥, computed in the covariance algebra, is no less than ∥a∥. Theconverse inequality follows from the fact that the map mentioned in the statement is aC∗-algebra homomorphism, hence contractive.⊓⊔We would now like to define the dual action, a concept closely related to dual actionsfor crossed products.Let, for every z ∈S1, αz be the transformation of L defined by(αz(a))(n) = zna(n)for a ∈L, z ∈S1.The reader can easily verify that each αz is a *-automorphism of L which, in turn,extends to a *-automorphism of C∗(A, Θ).

The resulting map z →αz becomes an actionof S1 on C∗(A, Θ) which we shall call the dual action.3.11. Proposition.

For each n, let Bn be the nth spectral subspace for the dual action.Then the map φn: x ∈Dn →xδn ∈C∗(A, Θ) is a linear isometry onto Bn.Proof. It is clear that xδn ∈Bn for every x ∈Dn.

Note that for such an x(xδn) ∗(xδn)∗= (xδn) ∗(θ−n(x∗)δ−n) = θn(θ−n(x)θ−n(x∗))δ0 = xx∗δ0.Thus, in order to show that φn is an isometry, it suffices to consider the case n = 0. Butthis is just the conclusion of (3.10).

It now remains to show that the image of φn is all ofBn. So let y ∈Bn and write y = limk yk where each yk belongs to L (or rather, the denseimage of L in C∗(A, Θ)).

Note that L is invariant under the spectral projections of thecovariance algebra and also that y = Pn(y) = limk Pn(yk). So, we may assume that theyk’s belong to the nth spectral subspace for the corresponding action of S1 on L. Thatspectral subspace is obviously Dnδn which, by our previous remarks, embeds isometricallyinto the covariance algebra.

This implies that (yk)k is a Cauchy sequence with respect tothe norm of L and hence that y = lim yk ∈Dnδn.⊓⊔

circle actions and partial automorphisms124. The Structure of Actions of the Circle GroupIn this section we intend to prove that any C∗-algebra admitting an action of S1 is iso-morphic to a covariance algebra, as described above, provided the action satisfies twoconditions which we now describe.4.1.

Definition. An action α of S1 on a C∗-algebra B is called semi-saturated if B isgenerated, as a C∗-algebra, by the union of the fixed point algebra B0 and the first spectralsubspace B1 (compare [8], [9], [11], [14], [16], [20]).4.2.

Definition. An action α of S1 on a C∗-algebra B is said to be stable if there existsan action α′ on a C∗-algebra B′ such that B ≃B′ ⊗K and α is the tensor product ofα′ by the trivial S1-action on the algebra K of compact operators on a separable, infinitedimensional Hilbert space.From now on we shall mainly be concerned with semi-saturated stable actions onseparable C∗-algebras, but we would like to argue that the above restriction are quite mildones.

First of all any action gives rise to a stable action by tensoring the old action withthe trivial action on K. Moreover, if the old action was semi-saturated then so will be thetensor product action.Of course not all actions of S1 are semi-saturated but this difficulty could be circum-vented in some cases. That is, given an action α on a C∗-algebra B, it follows from (2.5)that B is generated by the union of all its spectral subspaces.Now, if we let, for each positive integer n, B(n) be the sub-C∗-algebra of B generatedby B0 and Bn then B(n) is invariant under α and the formulaα(n)z= αz1/nfor z ∈S1provides a well defined semi-saturated action on B(n).With some luck, one could put together the information obtained about each B(n) tolearn something of interest about the original action.4.3.

Proposition. Let α be a stable action of S1 on a separable C∗-algebra B. Thenthere exists an isomorphism θ: B∗1B1 →B1B∗1 (see 2.2) and a linear isometry λ from B∗1onto B1B∗1 such that for x, y ∈B1, a ∈B∗1B1 and b ∈B1B∗1(i) λ(x∗b) = λ(x∗)b(ii) λ(ax∗) = θ(a)λ(x∗)(iii) λ(x∗)∗λ(y∗) = xy∗(iv) λ(x∗)λ(y∗)∗= θ(x∗y).Proof.

Everything will follow from [3], Theorem (3.4) after we verify that both B∗1B1and B1B∗1 are stable C∗-algebras with strictly positive elements. Now, stability follows atonce from our assumption that the action be stable while the existence of strictly positiveelements is a consequence of our separability hypothesis.⊓⊔

circle actions and partial automorphisms13The conclusions of the last proposition will be among our main tools in what follows.In fact the only reason we shall consider stable actions on separable C∗-algebras is toobtain the conclusions of (4.3).If, for any other reason, the existence of θ and λ as above, are guaranteed in a specificexample, we may discard stability of the action α and separability of the algebra B withouthurting the results of this section.In fact it is quite common to find examples in which the conclusions of (4.3) hold, butstill the action is not stable. The simplest such example is provided by the action of S1on B = M2(C) given by conjugation by the unitary matrices100zfor z in S1.

In thiscase we haveB0 =x00y: x, y ∈C,B1 =00y0: y ∈C,B∗1B1 =x000: x ∈C,andB1B∗1 =000y: y ∈C.The maps θ and λ defined byθx000=000xfor x ∈Candλ0x00=000xfor x ∈Csatisfy the conclusions of (4.3), as the reader may easily verify, although our action isobviously not stable.4.4. Definition.

An action of S1 on a C∗-algebra B will be called regular if the conclu-sions of (4.3) hold. That is, there should exist an isomorphism θ: B∗1B1 →B1B∗1 and alinear isometry λ from B∗1 onto B1B∗1 satisfying (i) – (iv) of (4.3).We obviously have:4.5.

Corollary. Stable circle actions on separable C∗-algebras are regular.We shall also note the following fact.

circle actions and partial automorphisms144.6. Proposition.

Let α be the dual action on the covariance algebra of a partial auto-morphism. Then α is regular.Proof.

The partial automorphism is built into the picture and λ is given byλ((xδ1)∗) = x∗δ0for x ∈D1.⊓⊔Also, there is no hope for an action of S1 to be equivalent to a dual action unless itis semi-saturated as we now demonstrate.4.7. Proposition.

If Θ = (Θ, I, J) is a partial automorphism of the C∗-algebra A, thenthe dual action α on the covariance algebra B = C∗(A, Θ) is semi-saturated.Proof. By (2.5) and by induction it is enough to show that, for n ≥1, Bn+1 ⊆B1Bn.

Ifx ∈Bn+1 is of the form x = aδn+1 for a ∈Dn+1, note that θ−1(a) ∈D−1 ∩Dn. Withoutloss of generality we may assume that θ−1(a) = bc where b, c ∈D−1 ∩Dn (the reductionto this case being made by an approximation argument).

Put b1 = θ(b) and observe that(b1δ1) ∗(cδn) = θ(θ−1(b1)c)δn+1 = θ(bc)δn+1 = aδn+1 = x.This shows that Bn+1, which equals Dn+1δn+1 by (3.11), is contained in B1Bn.⊓⊔One of the crucial properties of semi-saturated actions is obtained in our followingresult. It also provides an alternate characterization of semi-saturation.4.8.

Proposition. An action α of S1 on B is semi-saturated if and only if for each n > 0one has Bn = (B1)n. In this case, if n, m ≥0 then Bn+m = BnBm.Proof.

Let B′ be the dense *-subalgebra of B generated by B0 ∪B1. Any element of B′is a sum of “words” of the form w = x1x2 .

. .

xk where xi ∈B0 ∪B1 ∪B∗1.A word, such as above, represents an element of B which belongs to some spectralsubspace, the precise determination of which is obtained by subtracting the number of xi’sbelonging to B∗1 from the number of those in B1. For this reason, to apply the spectralprojection Pn to a sum of words is to eliminate all summands, but those which belong toBn.If b ∈Bn let b = limj bj with bj ∈B′.

Thus b = Pn(b) = limj Pn(bj), so we mayassume (upon replacing bj by Pn(bj)) not only that bj ∈B′ ∩Bn, but also that each bjis a sum of words each of which belongs to Bn. A minute of reflection will show that allwords in Bn are in (B1)n for n > 0.

The converse follows from (2.5).⊓⊔Let us now fix, for the remainder of this section, a C∗-algebra B where a regularsemi-saturated action α of S1 is defined.

circle actions and partial automorphisms154.9. Definition.

If λ and θ are as in (4.4) letρ: B∗1 →B∗1B1,λ†: B1 →B∗1B1andρ†: B1 →B1B∗1 be the isometries defined by ρ = θ−1 ◦λ, λ†(x) = ρ(x∗)∗and ρ†(x) =λ(x∗)∗for all x ∈B1.4.10. Proposition.

For a ∈B∗1B1, b ∈B1B∗1 and x, y ∈B1 one has(i) ρ(ax∗) = aρ(x∗)(ii) ρ(x∗b) = ρ(x∗)θ−1(b)(iii) ρ(x∗)ρ(y∗)∗= x∗y(iv) ρ(x∗)∗ρ(y∗) = θ−1(xy∗)(v) λ†(bx) = θ−1(b)λ†(x)(vi) λ†(xa) = λ†(x)a(vii) λ†(x)∗λ†(y) = x∗y(viii) λ†(x)λ†(y)∗= θ−1(xy∗)(ix) ρ†(bx) = bρ†(x)(x) ρ†(xa) = ρ†(x)θ(a)(xi) ρ†(x)ρ†(y)∗= xy∗(xii) ρ†(x)∗ρ†(y) = θ(x∗y).Proof. Follows by routine computations.⊓⊔4.11.

Proposition. The maps λ, ρ, λ† and ρ† extend to isometriesλ: B∗1B →B1Bρ: BB∗1 →BB1λ†: B1B →B∗1Bρ†: BB1 →BB∗1satisfying(i) λ(sb) = λ(s)bfor s ∈B∗1B1, b ∈B(ii) λ(s)∗λ(t) = s∗tfor s, t ∈B∗1B(iii) ρ(bs) = bρ(s)for s ∈BB∗1, b ∈B(iv) ρ(s)ρ(t)∗= st∗for s, t ∈BB∗1(v) λ†(sb) = λ†(s)bfor s ∈B1B, b ∈B

circle actions and partial automorphisms16(vi) λ†(s)∗λ†(t) = s∗tfor s, t ∈B1B(vii) ρ†(bs) = bρ†(s)for s ∈BB1, b ∈B(viii) ρ†(s)ρ†(t)∗= st∗for s, t ∈BB1.Moreover(a) If X ⊆B∗1B is a closed subspace, invariant under left multiplication by B∗1B1,then λ(X) = B1X. (b) If X ⊆BB∗1 is right B1B∗1-invariant then ρ(X) = XB1.

(c) If X ⊆B1B is left B1B∗1-invariant then λ†(X) = B∗1X. (d) If X ⊆BB1 is right B∗1B1-invariant then ρ†(X) = XB∗1.Finally, ρ† = ρ−1 and λ† = λ−1 hold for the extended maps.Proof.If s ∈B∗1B is of the form s = Pni=1 x∗i bi where xi ∈B1 and bi ∈B, putλ(s) = Pni=1 λ(x∗i )bi.

Observe that∥nXi=1λ(x∗i )bi∥2 = ∥Xijb∗jλ(x∗j)∗λ(x∗i )bi∥= ∥Xijb∗jxjx∗i bi∥= ∥nXi=1x∗i bi∥2,so that λ is well defined, clearly satisfies (i) and (ii) and is norm preserving. The extensionsof ρ, λ† and ρ† are handled in a similar way.Let X be a closed subspace of B∗1B, invariant under left multiplication by B∗1B1.Then by (2.6) X = B∗1B1X soλ(X) = λ(B∗1)B1X = B1B∗1B1X = B1X.Similar proofs apply to the next three statements.Finally, let x, y ∈B1 and note thatθ(y∗x) = λ(y∗)λ(x∗)∗= λ(y∗λ(x∗)∗) = λ(y∗ρ†(x)),so thaty∗x = θ−1(λ(y∗ρ†(x))) = ρ(y∗ρ†(x)) = y∗ρ(ρ†(x)).Therefore, by (2.6), x = ρρ†(x).For s = bx with b ∈B and x ∈B1 we have ρρ†(bx) = ρ(bρ†(x)) = bρρ†(x) = bx sothe compositionBB1ρ†→BB∗1ρ→BB1is the identity.

Since ρ is isometric, hence injective, we get ρ† = ρ−1. The proof thatλ† = λ−1 goes along identical lines.⊓⊔

circle actions and partial automorphisms174.12. Proposition.

The compositionB∗1BB1λ→B1BB1ρ†→B1BB∗1is a C∗-algebra isomorphism which extends θ and whose inverse is given by the compositionB1BB∗1ρ→B1BB1λ†→B∗1BB1.Proof. For x, y ∈B1 and a ∈B we haveρ†(λ(x∗ay)) = ρ†(λ(x∗)ay) = λ(x∗)aρ†(y).Thus if x1, y1 ∈B1 and a1 ∈Bρ†(λ(x∗ay))ρ†(λ(x∗1a1y1)) = λ(x∗)aρ†(y)λ(x∗1)a1ρ†(y) =λ(x∗)aλ(y∗)∗λ(x∗1)a1ρ†(y) = λ(x∗)ayx∗1a1ρ†(y) = ρ†(λ(x∗ayx∗1a1y1)),proving ρ† ◦λ to be multiplicative.We also haveρ†(λ(x∗ay))∗= ρ†(y)∗a∗λ(x∗)∗= λ(y∗)a∗ρ†(x) = ρ†(λ(y∗a∗x)),which shows that ρ† ◦λ is star preserving.To check the statement about extending θ note that for x, y ∈B1 we haveρ†(λ(x∗y)) = λ(x∗)ρ†(y) = λ(x∗)λ(y∗)∗= θ(x∗y).The fact that λ†ρ inverts ρ†λ follows from (4.11).⊓⊔Our next result resembles one of the main axioms for multipliers of C∗-algebras.4.13.

Proposition. If s ∈BB∗1 and t ∈B∗1B then ρ(s)t = sλ(t).Proof.

It clearly suffices to verify the case in which s = x∗and t = y∗with x, y ∈B1.We haveρ(x∗)y∗= ρ(x∗)ρ(ρ†(y))∗= x∗ρ†(y)∗= x∗λ(y∗).⊓⊔4.14. Proposition.

Regarding the partial automorphism (θ, B∗1B1, B1B∗1) of B0 we haveDom(θn) = B∗nBn for all integers n.Proof. This is clearly true for n = −1, 0, 1.

For n ≥1, using induction, we haveDom(θn+1) = {x ∈Dom(θ) : θ(x) ∈Dom(θn)} = {x ∈B∗1B1 : θ(x) ∈B∗nBn}.

circle actions and partial automorphisms18Thus, given x in Dom(θn+1)θ(x) ∈Im(θ) ∩Dom(θn) = Im(θ) · Dom(θn) = B1B∗1B∗nBnB1B∗1,sox = θ−1(θ(x)) = λ†ρ(θ(x)) ∈λ†(ρ(B1B∗1B∗nBnB1B∗1)) =B∗1B1B∗1B∗nBnB1B∗1B1 = B∗n+1Bn+1.This shows that Dom(θn+1) ⊆B∗n+1Bn+1.Conversely, if x ∈B∗n+1Bn+1 then clearlyx ∈Dom(θ) andθ(x) = ρ†λ(x) ∈ρ†(λ(B∗n+1Bn+1)) = B1B∗n+1Bn+1B∗1 ⊆B∗nBn = Dom(θn).So x ∈Dom(θn+1).⊓⊔Our main goal, as we already indicated, is to describe B as the covariance algebrafor the partial automorphism Θ = (θ, B∗1B1, B1B∗1) of the fixed point algebra B0. Theconclusions of (4.14) suggest that, if we are to succeed, then BnB∗n, being essentially thenth spectral subspace of the covariance algebra C∗(B0, Θ), should correspond to Bn.

Thisleads us to the following definitions.4.15. Definition.

Let i0: B0 →B0 denote the identity map of B0 and in: Bn →B0 bedefined, inductively, by in(x) = in−1(ρ†(x)) when n ≥1 or by in(x) = in+1(ρ(x)) whenn ≤−1.It is clear that in is an isometry for each n (although not necessarily surjective).4.16. Proposition.

For all integers m and n and for any xn ∈Bn and ym ∈Bm we havein+m(xnym) = in(xnim(ym)).Proof. Assume initially that n = 1.

If m = 0 there is nothing to prove and if m > 0,using induction, we havei1+m(x1ym) = im(ρ†(x1ym)) = im(x1ρ†(ym)) =i1(x1im−1ρ†(ym)) = i1(x1im(ym)).If m = −1, still assuming n = 1, we havei1(x1i−1(y−1)) = ρ†(x1ρ(y−1)) = ρ†ρ(x1y−1) = x1y−1 = i0(x1y−1).Now let m < −1 and observe that, by inductioni1+m(x1ym) = i1+(m+1)(ρ(x1ym)) = i1+(m+1)(x1ρ(ym)) =i1(x1im+1ρ(ym)) = i1(x1im(ym)).If n > 1 we may assume, without loss of generality, by (4.8) or (2.6) that xn = x1xn−1where x1 ∈B1 and xn−1 ∈Bn−1. So, by induction once more,in+m(xnyn) = i1+(n−1)+m(x1xn−1ym) = i1(x1in−1+m(xn−1ym)) =i1(x1in−1(xn−1im(ym))) = in(x1xn−1im(ym)).The case n = 0 is trivial and the proof for n < 0 is done in a similar way.⊓⊔

circle actions and partial automorphisms194.17. Proposition.

For every integer n one has in(Bn) = BnB∗n.Proof. The statement follows by definition when n = −1, 0, 1.

If n ≥1, by induction,(2.7), (4.8) and (4.11) one hasin+1(Bn+1) = i1(B1in(Bn)) = ρ†(B1BnB∗n) =ρ†(B1B∗1B1BnB∗n) = ρ†(B1BnB∗nB∗1B1) = B1BnB∗nB∗1B1B∗1 = Bn+1B∗n+1.As usual, the proof for negative values of n is omitted.⊓⊔4.18. Proposition.

If n ∈Z, a ∈B∗nBn and xn ∈Bn then in(xna) = in(xn)θn(a).Proof.For n = −1, 0, 1 the statement is obvious. For n > 1 assume, as before, thatxn = x1xn−1, the subscripts indicating the spectral subspace each factor lies in.

So, byinduction,in(xna) = in(x1xn−1a) = i1(x1in−1(xn−1a)) =ρ†(x1in−1(xn−1)θn−1(a)) = ρ†(x1in−1(xn−1))θn(a) = in(xn)θn(a).For n < −1 the proof is similar.⊓⊔4.19. Proposition.

For all integers n and m and for all xn ∈Bn and xm ∈Bmθ−n(in+m(xnym)) = θ−n(in(xn))im(xm).Proof. Let {ei} be an approximate identity for B∗nBn.

We haveθ−n(in+m(xnym)) = θ−n(in(xnim(ym))) = limj θ−n(in(xnejim(ym))) =limj θ−n(in(xn)θn(ejim(ym))) = limj θ−n(in(xn))ejim(ym)) = θ−n(in(xn))im(ym).⊓⊔4.20. Proposition.

For any n and for any xn ∈Bn one has θ−n(in(xn)) = i−n(x∗n)∗.Proof. This is obvious for n = 0.

If n = 1 we haveθ−1(i1(x1)) = λ†ρρ†(x1) = λ†(x1) = ρ(x∗1)∗= i−1(x∗1)∗.If n ≥1 assume, without loss of generality, that xn = x1xn−1 where x1 ∈B1 and xn−1 ∈Bn−1. So, by induction, we haveθ−n(in(xn)) = θ−(n−1)θ−1(i1(x1in−1(xn−1))) = θ−(n−1)(i−1(in−1(xn−1)∗x∗1))∗=θ−(n−1)(in−1(xn−1)∗i−1(x∗1))∗= θ−(n−1)(i−1(x∗1)in−1(xn−1)) =θ−(n−1)(in−1(i−1(x∗1)xn−1)) = i−(n−1)(x∗n−1i−1(x∗1)∗)∗=i−n(x∗n−1x∗1)∗= i−n(x∗n)∗.The proof for negative values of n is similar.⊓⊔We are now prepared to present our first main result.

circle actions and partial automorphisms204.21. Theorem.

Let α be a semi-saturated regular action of S1 on a C∗-algebra B. IfΘ = (θ, B∗1B1, B1B∗1) is the partial automorphism of the fixed point algebra B0 as in (4.4),then there exists an isomorphismφ: C∗(B0, Θ) →Bwhich is covariant with respect to the dual action.Proof. Recall that L is the Banach *-algebra formed by summable sequences (a(n))n∈Zwith a(n) ∈Dom(θ−n) = BnB∗n.

Defineφ: a ∈L →Xi−1n (a(n)) ∈B.To show that φ is multiplicative, let an ∈BnB∗n, bm ∈BmB∗m and put xn = i−1n (an) andym = i−1m (bm). We must then verify thatφ(anδn ∗bmδm) = φ(anδn)φ(bmδm).For this end, we need to check thati−1n+m(θn(θ−n(an)bm)) = i−1n (an)i−1m (bm)or thatθn(θ−n(in(xn))im(ym)) = in+m(xnym)which follows from (4.19).To check that φ preserves adjoints it is enough to show that, for any n and anya ∈BnB∗n,φ((aδn)∗) = φ(aδn)∗which translates toi−1−n(θ−n(a∗)) = i−1n (a)∗.Writing a = in(xn), for xn ∈Bn, the above becomesi−1−n(θ−n(in(xn)∗)) = x∗nwhich is equivalent to the conclusion of (4.20).If follows that φ extends to a *-homomorphism from C∗(B0, Θ) to B which is surjec-tive, because its image contains any Bn, and which is equivariant with respect to the dualaction, since spectral subspaces are mapped accordingly.It remains to show that φ is injective, but in view of (2.9) we only need to checkinjectivity on the fixed point subalgebra of C∗(B0, Θ), a fact that follows from (3.11).⊓⊔

circle actions and partial automorphisms215. Representations of Covariance AlgebrasThe parallel between our covariance algebras and crossed product algebras is reflected,also, in their representation theory which we shall now study in detail.

In particular, it ispossible to define an analog for the regular representation of crossed products which is ournext goal.Fix, for the remainder of this section, a C∗-algebra A and a partial automorphismΘ = (θ, I, J) of A.5.1. Definition.

Let π be a representation of A on the Hilbert space H. The regularrepresentation ˜π of C∗(A, Θ), associated to π, is that which is obtained by inducing π toL [18] via the conditional expectation E of (3.9) and then extending it to the envelopingC∗-algebra of L.A result which could perhaps be thought of as a new manifestation of the amenabilityof the group of integers is the following (see [15]):5.2. Theorem.

If π is a faithful representation of A then ˜π is faithful on C∗(A, Θ).Proof. The inducing process [18] leads us to consider the Hilbert space ˜H obtained bycompleting L⊗H under the semi-norm given by the (sometimes degenerated) inner product⟨y1 ⊗ξ1, y2 ⊗ξ2⟩= ⟨π(E(y∗2y1))ξ1, ξ2⟩for yi ∈L, ξi ∈H, i = 1, 2.The induced representation itself is specified by˜π(x)(y ⊗ξ) = (xy) ⊗ξfor x, y ∈L, ξ ∈H.Let Ln be the nth spectral subspace for the dual action restricted to L, that is,Ln = Dnδn.

Since, for n ̸= m, E(L∗nLm) = 0 and because Ln Ln is dense in L, wehave that the subspaces ˜Hn of ˜H, obtained by closing the image of Ln ⊗H in ˜H, form anorthogonal decomposition of ˜H.If y = dnδn for dn ∈Dn, then it is clear that ˜π(y) sends ˜Hm into ˜Hm+n for every m.Thus, if y is the finite sum y = PNn=−N dnδn, it follows thatpm+n ◦˜π(y)|Hm = ˜π(Pn(y))|Hmfor n, m ∈Z,where pn is the orthogonal projection of ˜H onto ˜Hn and Pn is, as in (2.4), the nth spectralprojection. Since the finite sums are dense in C∗(A, Θ), a continuity argument yields theabove formula for all y ∈C∗(A, Θ).From this it can easily be seen that if ˜π(y) = 0 then ˜π(Pn(y)) = 0 for any y in thecovariance algebra.

This reduces our task to the easier one of proving ˜π to be injective oneach spectral subspace of C∗(A, Θ), which we have already shown to be essentially Dnδn.So let dn ∈Dn and suppose that ˜π(dnδn) = 0. Note that for a ∈D0 = A and ξ ∈H0 = ˜π(dnδn)(aδ0 ⊗ξ) = θn(θ−n(dn)a)δn ⊗ξ.

circle actions and partial automorphisms22Therefore, the right hand side represents the zero vector in ˜H. By definition, its norm in˜H is computed by the horrible looking expression⟨π(E((θn(θ−n(dn)a)δn)∗∗(θn(θ−n(dn)a)δn)))ξ, ξ⟩1/2which, fortunately, can be simplified to ∥π(θ−n(dn)a)ξ∥.Since ξ can be any vector of H, it follows that π(θ−n(dn)a) = 0.

But π was supposedfaithful. Letting a run through an approximate unit for A we get θ−n(dn) = 0 and hencedn = 0.⊓⊔Representations of crossed product algebras are known to correspond to covariantrepresentations of the corresponding C∗-dynamical system [15].

With the necessary mod-ifications, the same is true for covariance algebras of partial automorphisms as we shallnow see.5.3. Definition.

A covariant representation of the pair (A, Θ) is a triple (H, π, u) whereπ is a *-representation of A on the Hilbert space H and u ∈B(H) is a partial isometrywhose initial space is π(I)H (meaning closed linear span) and whose final space if π(J)H.In addition it is required that, for a ∈I,π(θ(a)) = uπ(a)u∗.The rules of the game are often so tough when dealing with algebraic properties ofpartial isometries that the reader may be stricken by the naiveness of the above definition.In fact that was also the author’s first impression. It is really quite surprising that thisdefinition happens to do it’s job in a satisfactory way.A covariant representation gives rise to a representation of C∗(A, Θ) as we shall see.Before that, let us introduce some notation.5.4.Definition.

If n is a negative integer and if v is a partial isometry on a Hilbertspace, then vn stands for v−n∗(note that this will not cause any confusion since, when vis invertible, v∗= v−1 and the two possible interpretations of vn coincide).For y ∈L (cf. 3.6) put(π × u)(y) =Xn∈Zπ(y(n))un.5.5.Proposition.

π × u is a representation of L on H and therefore, extends to arepresentation of C∗(A, Θ).Proof. The proof of the multiplicativity of π × u consists in justifying all steps of thefollowing calculation, for an ∈Dn and am ∈Dmπ(an)unπ(am)um = unπ(θ−n(an)am)um =

circle actions and partial automorphisms23π(θn(θ−n(an)am))unum = π(θn(θ−n(an)am))un+m.The first two steps follow from the formulaunπ(θ−n(an)) = π(an)unfor an ∈Dn,which can be proved, using induction, separately for n > 0 and n < 0.The final step is a consequence of the formula,π(an)unum = π(an)un+mfor an ∈Dn,which is obviously true when n and m have the same sign, but which must be proved byinduction otherwise. The proof that π × u is invariant under the star operation is left tothe reader.⊓⊔5.6.

Theorem. Let σ be a *-representation of C∗(A, Θ) on the Hilbert space H. Thenthere exists a covariant representation (π, H, u) of the pair (A, Θ) such that σ = π × u.Proof.Let B = C∗(A, Θ) and let λ, ρ, λ† and ρ† be as in (4.11).Denote by π therestriction of σ to A, identified with Aδ0.

For xi ∈I and ξi ∈H, i = 1, . .

., n letu(nXi=1π(xi)ξi) =nXi=1σ(λ(xiδ0))ξi,and let us verify that this extends to a well defined isometry from π(I)H to π(J)H. Wehave∥Xiσ(λ(xiδ0))ξi∥2 =Xij⟨σ(λ(xjδ0)∗∗λ(xiδ0))ξi, ξj⟩=Xij⟨π(x∗jxi)ξi, ξj⟩= ∥Xiπ(xi)ξi∥,so u is well defined and isometric, but we still have to check that the image of u is π(J)H.That image equalsσ(λ(B∗1B1))H = σ(B1B∗1B1)H ⊆π(B1B∗1)H = π(J)H,whileπ(J)H = π(B1B∗1)H = π(B1B∗1B1B∗1)H ⊆σ(B1B∗1B1)H = σ(λ(B∗1B1))H = Im(u).We next need to show that (π, H, u) satisfies the covariance equation π(θ(a)) =uπ(a)u∗for a ∈I. Given b ∈I and ξ ∈H we haveuπ(a)π(b)ξ = uπ(ab)ξ = σ(λ(abδ0))ξ = σ((θ(a)δ0) ∗λ(bδ0))ξ = π(θ(a))u(π(b)ξ),

circle actions and partial automorphisms24which says that uπ(a) = π(θ(a))u on π(I)H. But both uπ(a) and π(θ(a))u vanish on theorthogonal complement of π(I)H hence the equality above holds on the whole of H. Souπ(a)u∗= π(θ(a))uu∗= π(θ(a)).Recall from (4.6) that λ((a1δ1)∗) = a∗1δ0 for a1 ∈D1. Given b−1 ∈D−1 = I, of theform b−1 = θ−1(c∗1d1) with c1, d1 in D1 note that(c1δ1)∗∗(d1δ1) = θ−1(c∗1d1)δ0 = b−1δ0.Soλ(b−1δ0) = λ((c1δ1)∗) ∗(d1δ1) = c∗1d1δ1 = θ(b−1)δ1.The above form of b−1 is sufficiently general to insure thatλ(b−1δ0) = θ(b−1)δ1for any b−1 ∈D−1.

This equality is the basis of the proof that π × u = σ. In fact, sincethe dual action is semi-saturated as proven in (4.7), it is enough to show that π × u andσ coincide on B0 ∪B1.

This is obvious for B0 so it remains to prove thatσ(b1δ1) = π(b1)ufor b1 ∈D1.Both sides of the equal sign above represent operators which vanish on the orthogonalcomplement of π(I)H so, in order to prove their equality, we may restrict ourselves toπ(I)H. Given a−1 ∈I and ξ ∈H we have for any b1 ∈D1π(b1)uπ(a−1)ξ = σ(b1δ0)σ(λ(a−1δ0))ξ = σ((b1δ0) ∗(θ(a−1)δ1))ξ =σ(θ(θ−1(b1)a−1)δ1)ξ = σ((b1δ1) ∗(a−1δ0))ξ = σ(b1δ1)π(a−1)ξwhich concludes our proof.⊓⊔

circle actions and partial automorphisms256. The Toeplitz Algebra of a Partial AutomorphismLet A be a C∗-algebra and Θ = (θ, I, J) be a partial automorphism of A.

We denote byB, the covariance algebra C∗(A, Θ) and regard A as a subalgebra of B in the obvious way.Fixing a faithful representation of B, allows us to assume that B ⊆B(H) for someHilbert space H. Using Theorem (5.6) we conclude that there is a partial isometry u inB(H) such that, for x ∈I and y ∈J, we have(i) uxu∗= θ(x)(ii) xu∗u = x(iii) yuu∗= y.Moreover, the set of finite sums of the form Pn∈Z anun with an ∈Dn = Dom(θ−n), isdense in B (see 5.4). The dual action of S1 on B is given, on the above mentioned denseset, byαz(Xanun) =Xznanunfor z ∈S1.So the nth spectral subspace for α is given by Bn = Dnun (see 3.11).In the following, N∗denotes the set of strictly positive integers, S denotes the uni-lateral shift on ℓ2(N∗) while P = 1 −SS∗and Q = SS∗.

Also let eij denote the standardmatrix units in B(ℓ2(N∗)) for i, j ≥1. Note that eij = SiS∗j −Si−1S∗j−1.6.1.Definition.

The Toeplitz algebra for the pair (A, Θ) is the sub-C∗-algebra E =E(A, Θ), of operators on H ⊗ℓ2(N∗), generated by the set{bn ⊗Sn : n ∈Z, bn ∈Bn}.Since the dual action α is semi-saturated, we have by (4.7) and (4.8), that Bn = (B1)nfor n ≥1. Our next result is an immediate consequence of this.6.2.

Proposition. E is generated by (B0 ⊗1) ∪(B1 ⊗S).Proof.

It is enough to note that, for n ≥1(Bn ⊗Sn) = ((B1)n ⊗Sn) = (B1 ⊗S)n.⊓⊔6.3. Definition.

Let Λ be the subset of B(H ⊗ℓ2(N∗)) given byΛ =Mi,j≥1(BiB∗j ⊗eij).From now on we will use the symbol L to denote the closure of the sum of any familyof independent subspaces of a Banach space. Also, following (2.2), we regard BiB∗j as theclosed linear span of the set of products xy where x ∈Bi and y ∈B∗j .Because, for any j, B∗j Bj ⊆B0 and also because BiB0 ⊂Bi, we have that Λ is asub-C∗-algebra of B(H ⊗ℓ2(N∗)).

circle actions and partial automorphisms266.4. Proposition.

Λ is an ideal in E.Proof. Since Bn = (B1)n, for n ≥1, the set of elements of the form bn−1b1b∗m, the indicesindicating the spectral subspace of B each factor lies in, is total (their linear span is dense)in BnB∗m.

So the identitybn−1b1b∗m ⊗enm = (bn−1 ⊗Sn−1)(bmb∗1 ⊗Sm−1)∗−(bn−1b1 ⊗Sn)(bm ⊗Sm)∗implies that Λ is contained in E.The verification that Λ is actually an ideal is nowstraightforward.⊓⊔As in [17], one should now show that the quotient E/Λ is isomorphic to B, providingthe short exact sequence0 →Λ →E →B →0Even though we could attempt to prove this right now, we shall deduce it from a muchstronger result (Lemma 6.7 below), which we will need for various other reasons.For each z ∈S1 let ∆z be the infinite matrix∆z =1zz2...regarded as a unitary operator on H ⊗ℓ2(N∗). Note that ∆z(bn ⊗Sn)∆−1z= zn(bn ⊗Sn)so that ∆zE∆−1z= E and so we can define an action γ of S1 on E byγz(x) = ∆zx∆−1zfor x ∈E, z ∈S1.Our next goal is to show that γ is a semi-saturated regular action, which will enableus to describe E as the covariance algebra for a certain partial automorphism, by Theorem(4.21).6.5.

Proposition. Let En denote the nth spectral subspace relative to γ.

Then(i) E0 = A ⊗1 + L∞n=1(Dn ⊗enn)(ii) E1 = B1 ⊗S + L∞n=1(Bn+1B∗n ⊗en+1,n)(iii) E∗1E1 = D−1 ⊗1 + L∞n=1((D−1 ∩Dn) ⊗enn)(iv) E1E∗1 = D1 ⊗Q + L∞n=2(Dn ⊗enn).Proof.Let Y0 and Y1 denote the right hand side of (i) and (ii), respectively. DefineYn = (Y1)n for n > 1, and Yn = (Y ∗1 )−n for n < 0.It can be shown without muchdifficulty, that Y = Ln∈Z Yn is a sub-C∗-algebra of E.Should the reader decide to verify this by himself, we suggest he starts by provingthat Y1Y0 and Y0Y1 are contained in Y1 and that Y ∗1 Y1 and Y1Y ∗1 are contained in Y0.We next note that B0 ⊗1 ⊆Y0 and that B1 ⊗S ⊆Y1 so, by (6.2) we get Y = E. Sinceit is obvious that Yn ⊆En we must have Yn = En for all n and in particular for n = 0 andn = 1, proving (i) and (ii).

The two remaining statements follow easily from (ii).⊓⊔

circle actions and partial automorphisms276.6. Proposition.

γ is a semi-saturated regular action of S1 on E. The roles of the mapsθ and λ mentioned in (4.4) are played, respectively, by θE and λE defined byθE: x ∈E∗1E1 7−→(u ⊗S)x(u ⊗S)∗∈E1E∗1λE: x∗∈E∗1 7−→(u ⊗S)x∗∈E1E∗1.Therefore E is isomorphic to the covariance algebra C∗(E0, (θE, E∗1E1, E1E∗1 )).Proof. The proof of regularity consists in verifying that the above, in fact, describes welldefined maps between the indicated sets and that they satisfy (4.4).

These are somewhatroutine tasks, which we leave for the reader. Finally, since it is obvious that B0 ⊗1 ⊆E0and that B1 ⊗S ⊆E1, γ is seen to be semi-saturated by (6.2).⊓⊔Our next Lemma is an important tool in obtaining representations of the Toeplitzalgebra (compare Lemma 7.1 in [1]).6.7.

Lemma. Let K be a Hilbert space and π be a representation of A on K. Supposethat V is a partial isometry operating on K such that for all x ∈I(i) V π(x) = π(θ(x))V(ii) V ∗V π(x) = π(x)Then there exists a representation πE of E on K such thatπE(anun ⊗Sn) = π(an)V nfor n ∈Z, an ∈Dn.If, in addition, for all y ∈J(iii) V V ∗π(y) = π(y)then there exists a representation πB of B on K such thatπB(anun) = π(an)V nfor n ∈Z, an ∈Dn.Proof.The second assertion, by far the easier, can be proved exactly as in the proofof (5.5).Now assume that V satisfies (i) and (ii).Our strategy will be to define arepresentation π0 of E0 on K, such that the pair (π0, V ) satisfies (i), (ii) and (iii) withrespect to the partial automorphism (θE, E∗1E1, E1E∗1) of E0.

Once that is done, we mayapply the part of the present Lemma which we have already taken care of. For n ∈N∗defineπn: an ∈Dn 7−→V n−1π(θ−(n−1)(an))V n−1∗−V nπ(θ−n(an))V n∗∈B(K).The reader may now verify that πn is a representation of Dn on K and thatπn(an)πm(am) = 0

circle actions and partial automorphisms28for n ̸= m, an ∈Dn and am ∈Dm. In doing so, the following identities are helpfulV n∗V nπ(a−n) = π(a−n)V n∗V n = π(a−n)forn ≥1, a−n ∈D−nπ(an)V nV n∗V n = π(an)V nforn ≥1, an ∈Dn.Next, we may define a representation ˜π of E0 by˜π(a ⊗1 +NXn=1an ⊗enn) = π(a) +NXn=1πn(an).We claim that (˜π, V ) satisfy (i), (ii) and (iii) with respect to (θE, E∗1E1, E1E∗1 ).

In fact letx = a−1 ⊗1 + PNn=1 hn ⊗enn ∈E∗1E1 where a−1 ∈D−1 and hn ∈D−1 ∩Dn. We have˜π(θE(x))V = ˜π((u ⊗S)(a−1 ⊗1 +NXn=1hn ⊗enn)(u ⊗S)∗)V =˜π(θ(a−1) ⊗Q +NXn=1θ(hn) ⊗en+1,n+1)V =˜π(θ(a−1) ⊗1 −θ(a−1) ⊗e11 +NXn=1θ(hn) ⊗en+1,n+1)V =(π(θ(a−1)) −π(θ(a−1)) + V π(a−1)V ∗++NXn=1V nπ(θ−n+1(hn))V n∗−V n+1π(θ−n(hn))V n+1∗)V =V π(a−1) + NXn=1V nV n−1∗π(hn)V ∗−V n+1V n∗π(hn)V ∗V =V π(a−1) +NXn=1V nV n−1∗π(hn) −V n+1V n∗π(hn) =V π(a−1) + VNXn=1V n−1π(θ−(n−1)(hn))V n−1∗−V nπ(θ−n(hn))V n∗=V ˜π(a−1 ⊗1 +NXn=1hn ⊗enn) = V ˜π(x).

circle actions and partial automorphisms29showing (i).To check (iii) let y = a1 ⊗Q + PNn=2 an ⊗enn be in E1E∗1 .Using thatV V ∗V = V we haveV V ∗˜π(y) = V V ∗˜π(a1 ⊗1 −a1 ⊗e11 +NXn=2an ⊗enn) =V V ∗(V π(θ−1(a1))V ∗+NXn=2V n−1π(θ−(n−1)(an))V n∗−V nπ(θ−n(an))V n∗) = π(y).The verification of (ii) is left to the reader. This implies, by our previous work, thatthere exists a representation πE of E on K such thatπE(ynwn) = ˜π(yn)V nfor yn ∈EnE∗n, where w is the partial isometry implementing the partial automorphismθE.

But, as it can be easily verified, w = u ⊗S. Therefore, if a1 ∈D1 we have thata1 ⊗Q ∈E1E∗1 and soπE(a1u ⊗S) = πE((a1 ⊗Q)(u ⊗S)) = ˜π(a1 ⊗Q)V =˜π(a1 ⊗1 −a1 ⊗e11)V = V π(θ−1(a1))V ∗V = π(a1)V.To conclude the proof one has, for an, bn ∈Dn and n > 1,anbnun ⊗Sn = (anun−1 ⊗Sn−1)(θ−(n−1)(bn)u ⊗S)so, by inductionπE(anbnun ⊗Sn) = π(an)V n−1π(θ−(n−1)(bn))V = π(anbn)V n.⊓⊔6.8.

Proposition. There exists a short exact sequence0 −→Λi−→Eφ−→B −→0where i is the inclusion and φ is given byφ(anun ⊗Sn) = anunfor n ∈Z, an ∈Dn.Proof.By (6.7) there indeed exists φ: E →B satisfying φ(anun ⊗Sn) = anun.

Theidentity used in our proof of (6.4) shows that φ vanishes on Λ so it defines a homomorphism˜φ: E/Λ →B which is obviously surjective.The action γ drops to E/Λ, since Λ is aninvariant ideal, and then ˜φ becomes a covariant homomorphism (with the dual action onB).For covariant homomorphisms under actions of S1, injectivity is equivalent to injec-tivity on the fixed point subalgebra by (2.9). The latter being true for ˜φ, we conclude that˜φ is an isomorphism and the proof is thus completed.⊓⊔

circle actions and partial automorphisms307. The Generalized Pimsner-Voiculescu Exact SequenceIn this section we shall obtain a generalization of the Pimsner-Voiculescu exact sequence[17] to our context of crossed products by partial automorphisms.An important tool in the sequel will be the fact that the inclusion of a full cornerof a C∗-algebra induces an isomorphism on K-theory, in the presence of strictly positiveelements ([2], see also [14], Proposition 2.1).

In fact a much stronger result holds: If B is afull corner in the C∗-algebra A and if A has a strictly positive element, then the inclusionι: B →A induces an invertible element in KK(B, A). This is a trivial consequence of ([2],Lemma 2.5).

In fact, if v is an isometry in M(A ⊗K) such that vv∗= p ⊗1, where pis the projection in M(A ⊗K) for which pAp = B, then let φ: A ⊗K →B ⊗K be theisomorphism given by φ(x) = vxv∗. The composition φ(ι ⊗idK), once composed with theinclusion of B ⊗K into M2(B ⊗K), can be easily shown to be homotopic to the lattermap.

Likewise, (ι⊗idK)φ when composed with inclusion of A ⊗K into M2(A ⊗K) is alsohomotopic to the latter map. In a word, φ provides an inverse for ι in KK(B, A)The model of KK-theory we shall adopt is the one introduced by Cuntz in [7].

Seealso the exposition in Chapter (5) of [10] which is where we borrow our notation from.As we already mentioned, we shall make intensive use of the result above and hence weneed to ensure that several algebras in our construction, including Λ, have strictly positiveelements. It is not hard to see that, in the case of Λ, this implies that Dn must have astrictly positive element for each n. However we cannot think of a reasonably gracefulset of hypothesis which could provide for this much, other than, of course, assuming ouralgebras to be separable (see 1.4.3 and 3.10.5 in [15]).As in the previous section, let A be a C∗-algebra with a fixed partial automorphismΘ = (θ, I, J) and let B = C∗(A, Θ).

In most of our results below, A will, therefore, beassumed separable.In the following, all occurrences of i refer to the inclusion homomorphism that shouldbe clear from the context. Also we shall denote by θ−1∗the map induced at the level of Kgroups by the compositionJθ−1→Ii→A.The following is our second main result.7.1.

Theorem. Let Θ = (θ, I, J) be a partial automorphism of the separable C∗-algebraA.

Then there exists an exact sequence of K-groupsK0(J)i∗−θ−1∗−→K0(A)i∗−→K0(C∗(A, Θ))↑↓K1(C∗(A, Θ))i∗←−K1(A)i∗−θ−1∗←−K1(J)Our proof will be, roughly speaking, based on [17] in the sense that we shall deriveour result from the usual exact sequence of K-Theory for the extension0 −→Λi−→Eφ−→C∗(A, Θ) −→0

circle actions and partial automorphisms31The crucial part in accomplishing this program is proving that K∗(A) and K∗(E) areisomorphic, at which point we can no longer follow the original method of Pimsner andVoiculescu but, instead, we must resort to techniques from KK-theory (see Proposition5.5 in [6] and Theorem 7.2 in [1]) which will, in fact, lead us to the stronger result that Aand E are KK-equivalent to each other.In the following we letd: A →Eandj: J →Λbe defined by d(a) = a ⊗1 for a in A and j(x) = x ⊗e11 for x ∈J.7.2. Proposition.

The diagramK∗(Λ)i∗−→K∗(E)j∗↑↑d∗K∗(J)i∗−θ−1∗−→K∗(A)is commutative. Moreover, if A is separable then j induces a KK-equivalence.Proof.

Let ˜J denote J with an added unit and let y = 1 + ab be an invertible element in˜J with a and b in J. The K1-class of y being denoted by [y]1, we haved∗θ−1∗[y]1 = [1 + u∗abu ⊗1]1 = [1 + (u∗a ⊗S∗)(bu ⊗S)]1.Recall from Lemma (1.1) in [14] that the K1 classes of 1 + rs and 1 + sr coincide,whenever 1 + rs is invertible.

So we haved∗θ−1∗[y]1 = [1 + ba ⊗Q]1 = [1 + (b ⊗1)(a ⊗Q)]1 =[1 + (a ⊗Q)(b ⊗1)]1 = [1 + ab ⊗Q].Following our diagram in the counterclockwise direction from K1(J) we then obtaind∗(i∗−θ−1∗)[y]1 = [1 + ab ⊗1]1 −[1 + ab ⊗Q]1 = [1 + ab ⊗e11]1 = i∗j∗[y]1.Clearly, any invertible element in ˜J is homotopic to an invertible element of the form1 +ab as above. Moreover, by tensoring everything with Mn(C) we get the above equalityfor any invertible y ∈Mn( ˜J).

The commutativity of our diagram is therefore proved forK1. The proof for K0 follows by taking suspensions.

circle actions and partial automorphisms32To conclude, we need to show that j induces a KK-equivalence. Note that j is anisomorphism from J onto J ⊗e11 = B1B∗1 ⊗e11.

The latter is obviously a corner of Λ andbecause, for i, j > 1(BiB∗1 ⊗ei1)(B1B∗1 ⊗e11)(B1B∗j ⊗e1j) =BiB∗1B1B∗1B1B∗j ⊗eij = BiB∗j ⊗eijone sees that J ⊗e11 is not contained in any proper ideal of Λ. In two words, J ⊗e11 isa full corner of Λ.

If we now invoke the separability of A, and hence also of Λ, the resultwill follow from our comments in the introduction to the present section.⊓⊔It follows that j∗is an isomorphism between the indicated K-groups and so all weneed, in order to prove (7.1), is to show that d∗is an isomorphism as well. Then we justhave to write down the K-theory exact sequence for0 −→Λi−→Eφ−→B −→0.and to replace the two segments K∗(Λ)i∗−→K∗(E) there, by K∗(J)i∗−θ−1∗−→K∗(A).7.3.

Definition. Let Λ0 be the subalgebra of B(H ⊗ℓ2(N)) given byΛ0 =Mi,j≥0BiB∗j ⊗eij.

(note that, as opposed to (6.3), the indices i and j here, start at zero). Also letj0: A −→Λ0be given by j0(a) = a ⊗e00.7.4.

Proposition. If A is separable then j0 is a KK-equivalence.Proof.

As in (7.2) it is clear that the image of j, namely A⊗e00 = B0B∗0 ⊗e00, is a cornerof Λ0. To show that it is also full note that for i, j ≥0(BiB∗0 ⊗ei0)(B0B∗0 ⊗e00)(B0B∗j ⊗e0j) = BiB∗j ⊗eij.The result now follows as in (7.2).⊓⊔Aiming at a proof that d induces an invertible element in KK(A, E), we shall nowintroduce an element of KK(E, Λ0) which will be shown to provide the required inverse,given the KK-equivalence between A and Λ0.In order to avoid confusion, we let T be the unilateral shift on ℓ2(N) (as opposed toS, which operates on ℓ2(N∗)).

Defineφ, φ: E −→B(H ⊗ℓ2(N))byφ(bn ⊗Sn) = bn ⊗T nandφ(bn ⊗Sn) = bn ⊗T nTT ∗.for n ∈Z and bn ∈Bn.

circle actions and partial automorphisms337.5. Proposition.

The images of φ and φ are contained in the multiplier algebra M(Λ0).Moreover, for every x ∈E we have φ(x) −φ(x) ∈Λ0.Proof. We have, for i, j ≥0, n ∈Z and bn ∈Bn(bn ⊗T n)(BiB∗j ⊗eij) = [n + i ≥0] bnBiB∗j ⊗en+i,jwhere [n + i ≥0] indicates the obvious boolean function assuming the real values 1 and 0.Also(BiB∗j ⊗eij)(bn ⊗T n) = [j −n ≥0] BiB∗j bn ⊗ei,j−nshowing that bn ⊗T n ∈M(Λ0).

Similarly the image of φ is also seen to be contained inM(Λ0).Note that for n ≥0, bn ∈Bn and x = bn ⊗Snφ(x) −φ(x) = bn ⊗en,0which belongs to Λ0. Since the set of x’s as above, generate E, it follows that φ(x) −φ(x)is in Λ0 for all x ∈E.⊓⊔At this point we start our main KK-theory computations.

We refer the reader tochapter (5) in [10] for the main facts about the KK-product as well as for notation.According to (5.1.1) in [10], q(φ, φ) defines a homomorphism from qE to Λ0 andtherefore its homotopy class gives an element, denoted [q(φ, φ)], in KK(E, Λ0).7.6. Proposition.

The product [q(d, 0)] · [q(φ, φ)], in KK(A, Λ0), equals [q(j0, 0)].Proof. Still using the notation from [10] and thus denoting by 1E the unit in the ringKK(E, E), we have d∗(1E) = [q(d, 0)].

So, by Theorem (5.1.15) in [10][q(d, 0)] · [q(φ, φ)] = d∗(1E) · [q(φ, φ)] = d∗(1E · [q(φ, φ)]) = d∗([q(φ, φ)]) =[q(φ, φ)q(d)] = [q(φd, φd)] = [q(φd + j0, φd)] = [q(j0, 0)].⊓⊔Since [q(j0, 0)] is invertible in KK(A, Λ0) by (7.4), if we assume A to be separable,we find that [q(d, 0)] is right invertible and we shall now concentrate our efforts in provingthat it is also left invertible.7.7. Definition.

Let Ω0 be the subset of B(H ⊗ℓ2(N∗) ⊗ℓ2(N)) given byΩ0 =Mi,j≥0((Bi ⊗1)E(B∗j ⊗1)) ⊗eijSimilarly to what we said with respect to both Λ and Λ0, we have that Ω0 is asubalgebra of operators on H ⊗ℓ2(N∗) ⊗ℓ2(N) while the mapk0: E −→Ω0

circle actions and partial automorphisms34defined by k0(x) = x ⊗e00, induces the invertible element [q(k0, 0)] in KK(E, Ω0), when Ais separable.Let d′ be the homomorphismd′: Λ0 −→Ω0defined byd′(bib∗j ⊗eij) = bib∗j ⊗1 ⊗eijfor i, j ≥0, bib∗j ∈BiB∗j .Note that the image of d′ in fact lies in Ω0 since, by (2.7)BiB∗j ⊗1 = BiB∗i BiB∗j BjB∗j ⊗1 = (Bi ⊗1)(B∗i BiB∗j Bj ⊗1)(B∗j ⊗1) ⊆⊆(Bi ⊗1)(B0 ⊗1)(B∗j ⊗1) ⊆(Bi ⊗1)E(B∗j ⊗1).We then have the commutative diagramΛ0d′−→Ω0j0 ↑↑k0Ad−→Ein which the vertical arrows are invertible in the corresponding KK-groups in the separablecase. It is then obvious that the existence of a left inverse to d can be verified by exhibitinga left inverse for d′.7.8.

Proposition. The product [q(φ, φ)] · [q(d′, 0)], in KK(E, Ω0), equals [q(k0, 0)].Proof.

Using (5.1.15) in [10], once more, as well as the fact that d′∗(1Λ0) = [q(d′, 0)], wehave[q(φ, φ)] · [q(d′, 0)] = [q(φ, φ)] · d′∗(1Λ0) = d′∗([q(φ, φ)] · 1Λ0) = d′∗([q(φ, φ)]) = [q(φ, φ)d′].Note that, in the last term above, we have written d′ where the definition (see [10]) ofd′∗calls for idK ⊗d′. We are justified in doing so because q(φ, φ) takes values in Λ0 whichshould be thought of as a subalgebra of K ⊗Λ0, as is customary in K-theory.Observing that d′ extends to a homomorphism from B(H⊗ℓ2(N)) to B(H⊗ℓ2(N∗)⊗ℓ2(N)) in an obvious way, one sees that q(φ, φ)d′ = q(ψ, ψ) where ψ and ψ are the mapsfrom E to the multiplier algebra M(Ω0) given byψ(bn ⊗Sn) = bn ⊗1 ⊗T nandψ(bn ⊗Sn) = bn ⊗1 ⊗T nTT ∗

circle actions and partial automorphisms35for n ∈Z and bn ∈Bn. Recalling that Q = SS∗and P = 1 −SS∗, consider the operatorW on H ⊗ℓ2(N∗) ⊗ℓ2(N) given byW = 1 ⊗P ⊗e00 + 1 ⊗S ⊗e01 + 1 ⊗S∗⊗e10 + 1 ⊗1 ⊗(1 −e00 −e11).It’s matrix with respect to the canonic basis of ℓ2(N) is given byW =1 ⊗P1 ⊗S0· · ·1 ⊗S∗00· · ·001· · ·1............Clearly, W is a self-adjoint unitary which, therefore can be connected to the identityoperator through the path of unitariesWt = (1 + W)/2 + eπit(1 −W)/2for 0 ≤t ≤1.If we let λ = λ(t) = (1 −eπit)/2 and µ = µ(t) = (1 + eπit)/2 we can write Wt asWt =1 ⊗P + µ ⊗(1 −P)λ ⊗S0· · ·λ ⊗S∗µ ⊗10· · ·001· · ·1............Let Vt = Wt(u ⊗1 ⊗T) or, in matrix form,Vt =λu ⊗S0· · ·µu ⊗10· · ·0u ⊗10· · ·0u ⊗1...............Observe that, because Wt commutes with A ⊗1 ⊗1, we have for all x ∈I(i) Vt(x ⊗1 ⊗1) = (θ(x) ⊗1 ⊗1)Vt(ii) V ∗t Vt(x ⊗1 ⊗1) = x ⊗1 ⊗1.These are precisely the assumptions in (6.7) so there exists a representation ψt of E onH ⊗ℓ2(N∗) ⊗ℓ2(N) such that for n ∈Z and an ∈Dnψt(anun ⊗Sn) = (an ⊗1 ⊗1)V nt

circle actions and partial automorphisms36Since V0 = u ⊗1 ⊗T, it is clear that ψ0 = ψ. On the other hand it is easy to see thatψ1(anun ⊗Sn) = anun ⊗Sn ⊗e00 + anun ⊗1 ⊗T nTT ∗for an ∈Dnwhich gives ψ1(x) = k0(x) + ψ(x) for all x ∈E.Therefore, the pair ψ1, ψ defines ahomomorphismq(ψ1, ψ): qE −→Ω0which obviously coincide with q(k0, 0).

To conclude, we need to show that ψt(x) ∈M(Ω0)and that ψt(x) −ψ(x) ∈Ω0 for all t and for all x ∈E. In fact, once this is done, we obtaina well defined family of homomorphismsq(ψt, ψ): qE →Ω0,providing a homotopy from q(ψ, ψ) to q(k0, 0) and showing that [q(ψ, ψ)] equals [q(k0, 0)]in KK(E, Ω0).The fact that ψt(x) −ψ(x) ∈Ω0 is obvious for x of the form x = a ⊗1 with a ∈A.

Ifx = a1u ⊗S, for a1 ∈D1, we haveψt(x) −ψ(x) = ψt(x) −ψ(x) + ψ(x) −ψ(x)whileψt(x) −ψ(x) = (a1 ⊗1 ⊗1)Wt(u ⊗1 ⊗T) −(a1u ⊗1 ⊗T) =(Wt −I)(a1u ⊗1 ⊗T) =λa1u ⊗S0· · ·−λa1u ⊗10· · ·00· · ·.........which clearly belongs to Ω0. This implies two important facts.

First, since we alreadyknow that ψ(E) ⊆M(Ω0), we conclude that ψt(x) ∈M(Ω0) for x ∈(B0 ⊗1) ∪(B1 ⊗S),which is a generating set. So ψt(x) ∈M(Ω0) for all x ∈E.

Secondly, because Ω0 is anideal in M(Ω0), our task of proving that ψt(x) −ψ(x) ∈Ω0 needs only be verified on agenerating set, which the above computation also yields.⊓⊔Proof of (7.1). Our last proposition shows that [q(d′, 0)] is left invertible since [q(k0, 0)]is invertible.

We conclude that [q(d, 0)] is invertible and hence that A and E are KK-equivalent to each other. In particular the mapd∗: K∗(A) →K∗(E)is an isomorphism.

Our result now follows from combining the K-theory exact sequencefor the Toeplitz extension0 −→Λi−→Eφ−→B −→0

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