c^*-envelopes and interpolation theoryusers.uoa.gr/~akatavol/newtexfil/paulsenetc/2206.read.pdf ·...

C∗-envelopes and Interpolation TheorySCOTT MCCULLOUGH & VERN PAULSEN

ABSTRACT. The C∗-envelope of an operator algebra A is theminimal C∗-algebra that can be generated by a completely iso-metric representation of A. In this paper we compute the C∗-envelopes of two operator algebras that arise in interpolation the-ory. The first example is the quotient of H∞(D2) by the idealof functions vanishing at 3 particular points. We prove, amongother results, that the C∗-envelope of this 3 dimensional oper-ator algebra is infinite dimensional and, moreover, possesses in-finite dimensional irreducible representations. The second ex-ample is the quotients of H∞(A), where A denotes an annulus,by the ideal of functions vanishing at n ≥ 3 points. We provethat the C∗-envelope of this n-dimensional operator algebra is*-isomorphic to Mn ⊗ C(T), where T denotes the unit circle.

0. INTRODUCTION

Fix n distinct points x1, . . . , xn in the unit disk D. The matrix valued versionof Pick’s Theorem says, given ` × ` matrices W1, . . . , Wn, there exists an analyticfunction F : D , M` such that ‖F(z)‖ ≤ 1 for all z ∈ D and F(xj) = Wj if andonly if the Pick matrix (

I −WjW∗m1− xjxm

)

is positive (semi-definite). Many proofs of Pick’s Theorem involve, either explic-itly or implicitly, the construction of a completely isometric representation of aquotient of H∞(D).

To understand this viewpoint, let E = {x1, . . . , xn}, let IE denote the idealof functions in H∞(D) which vanish at the points of E and let Q denote thequotient H∞(D)/IE . Not only does Q have a norm, but M`(Q), the ` × ` ma-trices with entries from Q, also naturally has a norm by identifying M`(Q) withM`(H∞(D))/M`(IE).

479

480 SCOTT MCCULLOUGH & VERN PAULSEN

Let P denote the n × n matrix (1/(1 − xjxm)) and, for f ∈ H∞(D), letDiag(f ) denote the n × n diagonal matrix with diagonal entries f(x1), . . . ,f(xn). The scalar version (` = 1) of Pick’s Theorem is equivalent to the factthat the homomorphism ρ : Q, Mn given by

(0.1) ρ(f + IE) = P−1/2Diag(f )P1/2

is an isometry; i.e., ‖f + IE‖ = ‖P−1/2 Diag(f )P1/2‖. If F ∈ M`(H∞(D)) andDiag(F) denotes the n×n block diagonal matrix with ` × ` matrix entries withdiagonal entries F(x1), . . . , F(xn), then the matrix version of Pick’s Theoremimplies that ‖F +M`(IE)‖ = ‖P−1/2Diag(F)P1/2‖, where now P1/2 is interpretedin the natural way as an n × n matrix with ` × ` matrix entries. Thus, afterthe canonical shuffle which identifies Mn(M`) with M`(Mn), the mapping ρ(`) :M`(Q) , Mn(M`), defined by applying ρ entry-wise, is also an isometry and soρ is a completely isometric unital representation of Q on the Hilbert space Cn.

The representation ρ is the minimal possible completely isometric unital rep-resentation of Q on Hilbert space, in a sense that we make precise below.

Let X be a compact Hausdorff space, let C(X) denote the Banach algebraof continuous complex-valued functions on X with the supremum norm, and letA ⊂ C(X) denote a uniform algebra on X. Let M` denote the ` × ` matricesover C and letM`(A) denote the `×` matrices with entries from A. An elementF = (fjm) ∈ M`(A) is naturally identified with the M` valued function, F(x) =(fjm(x)), and

‖F‖ = sup{‖F(x)‖M` | x ∈ X}defines a norm on M`(A).

Fix a finite subset E = {x1, . . . , xn} of X, let IE denote the ideal of functionswhich vanish on E, IE = {f ∈ A | f(xj) = 0, j = 1, . . . , n}, and let Q =A/IE denote the quotient algebra. The identification M`(Q) = M`(A)/M`(IE)naturally endowsM`(Q) with a matrix norm structure.

By the abstract characterization of operator algebras [7], the algebra Q is aunital operator algebra. That is, there exists a Hilbert space H and a unital com-pletely isometric homomorphism ρ : Q , B(H). Thus not only is ρ an isometry,but also for each positive integer ` the map ρ(`) : Mn(Q) , B(H(`)) defined byρ(`)(ajm) = (ρ(ajm)) is an isometry. Here H(`) = H ⊕ · · · ⊕H (` copies) andan `× ` matrix of operators (Tjm) is regarded as acting on H(`) by

(Tjm)

h1...hm

=∑m T1mh`

...∑m T`mhm

.Among all such completely isometric representations of an operator algebra

Q, Hamana [9] proved that there is a representation γ : Q , B(Hγ) which isminimal in the sense that it has the following universal property.

C∗-envelopes and Interpolation Theory 481

If ρ : Q , B(Hρ) is any completely isometric representation, then there existsa *-homomorphism σ : C∗(ρ(Q)) , C∗(γ(Q)) such that σ(ρ(a)) = γ(a), fora ∈ Q.

This universal property of γ determines C∗(γ(Q)) up to a *-isomorphismthat fixesQ. That is, if γ′ : Q , B(Hγ′) also has this universal property, then thereis a unique *-isomorphism σ : C∗(γ′(Q)) , C∗(γ(Q)) such that σ(γ′(a)) =γ(a) for all a ∈ Q. This essentially unique C∗-algebra is called the C∗-envelopeof Q and is denoted C∗e (Q). Thus, C∗e (Q) is the smallest, in the sense that it isthe universal quotient, C∗ algebra which determines Q completely isometrically.

Note that every irreducible representation of C∗e (Q) induces an irreduciblerepresentation of C∗(ρ(Q)) for any completely isometric representation ρ of Q.Thus, to find a minimal family of irreducible representations which determines Qup to complete isometry, it is essential to understand C∗e (Q).

Returning to the operator algebra Q = H∞(D)/IE , since the ρ of (0.1) is acomplete isometry, the universal property of γ and C∗e (Q) implies there existsan onto *-homomorphism σ : C∗(ρ(Q)) , C∗e (Q) such that σ(ρ(f + IE)) =γ(f + IE). Thus C∗e (Q) is a quotient of C∗(ρ(Q)). However, the fact that P hasno zero entries implies C∗(ρ(Q)) = Mn, and since the only quotients of Mn areMn itself and (0), it follows that C∗e (Q) = Mn and, up to canonical identification,ρ = γ.

We think of the problem of identifying C∗e (Q) as the C∗-approach to thegeneral matrix-valued interpolation problem for the set E, and in this paper wecompute two examples.

The first arises in the interpolation problem studied first by Solazzo [13]. LetQ denote the quotient of the analytic functions on the bidisk that are continu-ous on the boundary, A(D2), by the ideal of functions which vanish at the threepoints (0,0), (1/

√2,0), (0,1/

√2). Although it plays no essential role here, we

remark that this algebra is completely isometrically isomorphic to the algebra thatone obtains by considering the quotient of H∞(D2) by the ideal of functions van-ishing at the same points. We characterize the C∗-envelope of this 3-dimensionalquotient algebra as a C∗-subalgebra of M3 ⊗ C∗u(U, P), where C∗u(U, P) is theuniversal C∗-algebra for an arbitrary unitary and positive P satisfying 0 ≤ P ≤ I.The C∗-algebra C∗u(U, P) is characterized by the universal property that if U0 isany unitary and 0 ≤ P0 ≤ 1, then there exists a *-homomorphism from C∗u(U, P)onto C∗({U0, P0}) sending U to U0 and P to P0. We prove that if U0 is any uni-tary and 0 ≤ P0 ≤ δ < 1, then the induced quotient map from M3 ⊗ C∗u(U, P)to M3 ⊗ C∗({U0, P0}) sends C∗e (Q) onto M3 ⊗ C∗({U0, P0}). In particular, thisimplies that C∗e (Q) has irreducible representations onto M3k for every k ≥ 1 andirreducible representations of infinite dimension.

This characterization of C∗e (Q) implies that Q does not have a completelyisometric representation on a finite dimensional Hilbert space. Solazzo’s thesisshows that even the isometric theory of this quotient algebra is problematical. It isstill unknown whether Q has an isometric representation on a finite dimensional


Hilbert space, but the results in [13] suggest that it cannot.The second example concerns interpolation on an annulus. Fix 0 < q < 1

and let A denote the annulus {q < |z| < 1}. Let Q denote the quotient ofH∞(A)modulo the ideal of functions which vanish at n ≥ 3 points in A. Results of [10]show that Q has an isometric representation on a Hilbert space of dimension 2n.But results of the first author [11] show that Q can’t be represented completelyisometrically on a finite dimensional Hilbert space and hence C∗e (Q) is infinitedimensional. We show that in fact C∗e (Q) is *-isomorphic to C(T) ⊗Mn, whereT denotes the unit circle. Thus, although C∗e (Q) is infinite dimensional, all of itsirreducible representations are n-dimensional. Consequently this general matrix-valued interpolation is much more tractable than the bidisk problem discussedabove.

It is worthwhile to contrast the interpolation condition implicit in the com-putation of C∗e (Q) with Agler’s Pick Theorem for the bidisk [2]. Agler’s Theo-rem gives a numerically efficient method for determining whether a solution toa Pick problem on the bidisk has a solution, for a given set of data. Agler andMcCarthy [3] have even used the theorem to construct solutions to three pointinterpolation problems in the bidisk. By contrast, the completely isometric repre-sentation ρ : Q, C∗e (Q) provides, in some sense, a minimal way to decide if thecorresponding Pick problem has a solution that is independent of the particularmatrices or their size. It seems, C∗e (Q) can not be avoided in any attempt to simul-taneously solve all Pick problems for the tuple (0,0), (0,1/

√2), (1/

√2,0) using a

single representation of Q. It also gives a concrete indication of the complexity ofthe three dimensional operator algebra Q.

For the case of Q, it is interesting to note that it is expected that an Agler typecondition for interpolation would involve the existence of a one-parameter familyPs of n×n positive semidefinite matrices, solving an algebraic equation involvingthe interpolation data and the matrices ((Ps)jm(1−ϕs(xj)ϕs(xm))), whereϕsparameterizes, up to certain canonical equivalences, the analytic functions on theannulus which are unimodular on the boundary with exactly two zeros. Herethe parameter s can be chosen to vary over the unit circle. Similarly, the Pickcondition implicit in C∗e (Q) is the usual Pick condition for the annulus whichinvolves the positivity of a family of Pick kernels parameterized by the unit circle.

The remainder of the paper contains five sections. Our strategy for comput-ing both C∗-envelopes is to identify sufficiently many boundary representations.Definitions and our working result on boundary representations are in sectionone. Sections two through four treat the bidisk example. Basic preliminariesare in section two. In section three the existence of sufficiently many boundaryrepresentations is established. Section four contains the identification of C∗e (Q).Section five contains the annulus example.

1. BOUNDARY REPRESENTATIONS

In this section we gather a couple of general observations that are useful for com-


puting C∗-envelopes.Let A denote a unital operator algebra. Following Arveson [4], a com-

pletely isometric unital homomorphism ρ : A , B, where B is a C∗-algebraand C∗(ρ(A)) = B together with a *-homomorphism π : B , B(Hπ) iscalled a boundary representation for A provided that, if ϕ : B , B(Hπ) isa completely positive map such that ϕ(ρ(a)) = π(ρ(a)) for every a ∈ A,then ϕ = π . Now let γ denote the completely isometric representation of Ain the definition of the C∗-envelope and let σ : B , C∗e (A) denote the re-sulting canonical *-homomorphism from B onto the C∗-envelope of A withσ(ρ(a)) = γ(a). The mapping σ(ρ(a)) = γ(a) , π(ρ(a)) is completelycontractive from γ(A) ⊂ C∗e (A) into B(Hγ). Consequently, there exists a com-pletely positive map ψ : C∗e (A), B(Hπ) such that ψ(σ(ρ(a))) = π(ρ(a)) fora ∈ A. Since π is a boundary representation, ψ◦σ = π . Thus ker(σ) ⊂ ker(π)and therefore there exists a *-homomorphisms τ : C∗e (A) , B(Hπ) such thatτ ◦σ = π . Thus, boundary representations of A⊂ B induce *-homomorphismsof C∗e (A) that can easily be seen to be boundary representations forA⊂ C∗e (A).It is still unknown whether or not C∗e (A) is determined by its boundary repre-sentations. That is, if the representation of C∗e (A) given by the direct sum of allboundary representations is one-one. As for the examples in this paper, there areenough boundary representations to determine both C∗e (Q) and C∗e (Q).

We will use the following condition to produce boundary representations. IfA is a unital operator algebra and the unital subalgebra generated by the finitesubset a = {a1, . . . , an} is dense in A, the subset a is a generating set for A.

Definition 1.1. Let A denote a unital operator algebra with generating seta = {a1, . . . , an}. An n-tuple T = (T1, . . . , Tn) of operators on a Hilbert spaceH is a boundary tuple for A and a provided that:(1) there is a unital completely contractive homomorphism ρ : A , B(H) with

ρ(aj) = Tj , for j = 1, . . . , n;(2) if K is a Hilbert space, H is a subspace of K, and ν : A , B(K) is a unital

completely contractive homomorphism such thatH is invariant for ν(aj) andPHν(aj)|Hj = Tj , for j = 1, . . . , n, then H is reducing for ν(aj), for j = 1,. . . , n;

(3) if K is a Hilbert space, H is a subspace of K, and ν : A , B(K) is a unitalcompletely contractive homomorphism such that H⊥ is invariant for ν(aj)and PHν(aj)|Hj = Tj , for j = 1, . . . , n, then H is reducing for ν(aj), forj = 1, . . . , n.Theorem 1.2. Let A be a unital operator algebra with finite generating set a =

{a1, . . . , an}. The tuple T = (T1, . . . , Tn) is a boundary tuple for A and a if andonly if there exists a boundary representation π : C∗e (A) , B(Hπ) for A such thatπ(γ(aj)) = Tj .

Proof. First assume that T = (T1, . . . , Tn) is a boundary tuple. By (1) thereexists a unital completely contractive homomorphism ρ : A , B(H) such that


ρ(aj) = Tj , for j = 1, . . . , n. By the Arveson Extension Theorem [4], thereis a completely positive map ϕ : C∗e (A) , B(H) such that ϕ ◦ γ = ρ. ByStinespring’s Theorem [14], there exists a Hilbert space K containing H and a*-homomorphism τ : C∗e (A) , B(K) such that ϕ(b) = PHτ(b)|H for b ∈C∗e (A). Since ρ is a homomorphism, it follows that H is semi-invariant forτ(γ(A)) [12]. Hence, there exists a subspace K1 ⊂ K which is orthogonal toH and such that both K1 and K1 ⊕ H are invariant for τ(γ(A)). Let K2 =(K1 ⊕H)⊥, so that K = K1 ⊕H ⊕K2, and set τ1 and τ2 on γ(A) by τ1(γ(a)) =PK1⊕Hτ(γ(a))|K1⊕H and τ2(γ(a)) = PH⊕K2τ(γ(a))|H⊕K2 .

Since K1 is invariant for τ(γ(A)) and τ ◦ γ : A , B(K) is unital andcompletely contractive, both τ1 and τ2 are unital completely contractive homo-morphisms with H invariant for τ2(γ(aj)) and, relative to the K1 ⊕H, H⊥ = K1invariant for τ1(γ(aj)). The hypothesis that T is a boundary tuple now impliesthat H is reducing for τ`(γ(aj)), which in turn implies that H is reducing forτ(γ(aj)).

Since C∗(γ(A)) = C∗e (A), the set {γ(a1), . . . , γ(an), γ(a1)∗, . . . , γ(an)∗}generates the C∗-algebra C∗e (A). Since H is reducing for τ(γ(aj)), it is reducingfor τ(C∗e (A)) and hence ϕ is a *-homomorphism of C∗e (A).

Hence, not only does there exist a unique *-homomorphism ϕ of C∗e (A)into B(H) sending γ(aj) to Tj , but any completely positive extension of the mapsending γ(aj) to Tj must be ϕ so that ϕ is a boundary representation.

Conversely, assume that there exists a *-homomorphism π : C∗e (A) , B(H)with π(γ(aj)) = Tj which is a boundary representation forA, and let ρ = π ◦γ.To prove that item (2) in the definition of boundary tuple holds, let H ⊂ K = H⊕K1 and a completely contractive unital homomorphism ν : A , B(K) such thatH is invariant for ν(aj) and PHν(aj)|H = Tj be given. To prove that H reducesν(aj) it is enough to show that PHν(aj)ν(aj)∗|H = TjT∗j . To this end, observethat, as γ(a) , ν(a) is completely contractive, there exists a completely positiveunital mapϕ : C∗e (A), B(K) so thatϕ◦γ = ν. Sinceπ is a boundary represen-tation, b , PHϕ(b)|H is completely positive, and PHϕ(γ(a))|H = π(γ(a)), itfollows that π(b) = PHϕ(b)|H , for b ∈ C∗e (A). Since ϕ is completely positive,it satisfies the Schwarz inequality, ϕ(bb∗) ≥ ϕ(b)ϕ(b)∗. Thus,

TjT∗j = π(γ(aj)γ(aj)∗)= PHϕ(γ(aj)γ(aj)∗)

∣∣H

≥ PHϕ(γ(aj))ϕ(γ(aj))∗∣∣H

≥ PHϕ(γ(aj))PHϕ(γ(aj))∗∣∣H

= TjT∗j .Thus TjT∗j = PHϕ(γ(aj))ϕ(γ(aj))∗|H = PHν(aj)ν(aj)∗|H as claimed.

The proof of item (3) in the definition of boundary tuple is similar to theproof of item (2). ❐


Remark 1.3. It is fairly easy to see that T is a boundary tuple if and only ifwhenever H ⊂ K, ν : A , B(K) is a unital completely contractive homomor-phism with PHν(aj)|H = Tj for j = 1, . . . , n and H is semi-invariant, then H isreducing.

2. THE BIDISK EXAMPLE

Let A(D2) denote the uniform algebra of analytic functions on the bidisk thatare continuous on the boundary of the bidisk, let x0 = (0,0), x1 = (1/

√2,0),

x2 = (0,1/√

2), let E = {x0, x1, x2}, and let IE denote the ideal of functions inA(D2) which vanish on E. In this section we begin the study of C∗e (Q), where Qis the unital operator algebra A(D2)/IE.

Since A(D2) is generated by the coordinate functions z1, z2, the operatoralgebra Q is generated by

t1 = z1 + IE and t2 = z2 + IE.

Our approach is to find boundary tuples T = (T1, T2) for t = {t1, t2}.There is another natural generating set for Q. Choose function fj , j = 0,

1, 2, such that fj(xm) = δjm, where δjm is the Kronecker delta. The cosetsej = fj + IE are three commuting idempotents whose sum is the identity andsuch that ejem = δjmem. Finally, note that

ej =√

2tj, for j = 1, 2, ande0 = 1−

√2(t1 + t2).

Proposition 2.1. Let T1 and T2 be operators on H. The map ρ : Q , B(H)with ρ(tj) = Tj , j = 1, 2 gives a well-defined unital completely contractive ho-momorphism if and only if ‖Tj‖ ≤ 1, j = 1, 2, and Ej =

√2Tj are commuting

idempotents satisfying EjEm = δjmEm.

Proof. The algebraic conditions guarantee that there is a well defined ho-momorphism ρ : Q , B(H) with ρ(tj) = Tj , j = 1, 2. Since the ma-trix norms on Q are quotient norms, to check that ρ is completely contrac-tive it is enough to check that the composition α : A(D2) , B(H), given byα(f) = ρ(f +IE) is completely contractive. But, since α(zj) = Tj and ‖Tj‖ ≤ 1,either Ando’s Theorem or the Commutant Lifting Theorem imply α is completelycontractive. ❐

Proposition 2.2. If Ej , j = 0, 1, 2, are non-zero commuting idempotents onH satisfying EjEm = δjmEm and E0 + E1 + E2 = I, then there is a decompositionH = H0⊕H1⊕H2, with each Hj non-zero, and operators A, B, C so that, as matrices


relative to this decomposition,

E0 =

I A B

0 0 0

0 0 0

, E1 =

0 −A −AC0 I C

0 0 0

,

E2 =

0 0 AC − B0 0 −C0 0 I

.Proof. Let H0 = range(E0), H1 = range(E0+E1)∩H⊥0 , and H2 = H⊥0 ∩H⊥1 .

Since E0 is idempotent and H0 = range(E0), it follows that E0 has the desiredform. Next, write E = E0 + E1 = (Mjm)2jm=0 as a matrix with respect to thedecomposition of H. Since range(E0 + E1) = H0 ⊕ H1, M2m = 0 for each m.Since EE0 = E0, M00 = I and M10 = 0. Since E2 = E, it follows that M01M11 = 0and M01M12 = 0. But, as range(E) = H0 ⊕ H1, the sum of the ranges of M11and M12 is all of H1. Thus, M01 = 0. Reasoning similarly from the identitiesM211 = M11 and M11M12 = M12, it follows that the range of M11 is dense in H1,but since M11 is also idempotent, M11 = I. Now, the identities E21 = E1 andE1 = E − E0 imply that E1 has the desired form. Finally, by virtue of the relationE2 = I − E0 − E1, E2 also has the desired form. ❐

Note that the construction makes sense in the case one of the Ej = 0, in whichcase the corresponding Hj = 0.

Theorem 2.3. Let T1, T2 be operators on H. There exists a one-one completelycontractive unital homomorphism ρ : Q , B(H) with ρ(tj) = Tj , j = 1, 2, if andonly if there is an orthogonal decomposition of H as H = H0 ⊕ H1 ⊕ H2 such thatrelative to this decomposition

T1 = 1√2

0 A −AZ0 I −Z0 0 0

and T2 = 1√2

0 0 Y

0 0 Z

0 0 I

,with

I +A∗A ≤ 2(I + ZZ∗)−1,(2.1)

and

Y∗Y + Z∗Z ≤ I.(2.2)


Proof. Suppose that T1 and T2 have the indicated forms. By computingT∗2 T2 = 12(I + Y∗Y + Z∗Z), it follows that ‖T2‖ ≤ 1 if and only if condition(2.2) holds. Next verify that T1T∗1 = XX∗, where

(2.3) X = 1√2

AR

R

0

and R = (I+ZZ∗)1/2. Thus ‖T1‖ ≤ 1 if and only if X∗X ≤ I. This last inequalityis equivalent to condition (2.1).

If there is a completely contractive unital homomorphism ρ : Q , B(H)with ρ(tj) = Tj , then Propositions 2.1 and 2.2 together imply that Tj have theindicated form with, in the notation of Proposition 2.2, Z = −C and Y = B−AC.Since both ‖Tj‖ ≤ 1 by Proposition 2.1, the conditions (2.1) and (2.2) hold.

Conversely, if Tj have the indicated forms and ‖Tj‖ ≤ 1, then Ej =√

2Tj ,j = 1, 2, are commuting idempotents satisfying EjEm = δjmEm. Moreover, ifboth (2.1) and (2.2) hold, then ‖Tj‖ ≤ 1 so that Proposition 2.1 implies theexistence of ρ. ❐

We shall refer to a pair of operators T1, T2 of the form given in Theorem 2.3as operators in standard form and call ρ the representation induced by (A, Y , Z).

3. BOUNDARY REPRESENTATIONS FOR THE BIDISK EXAMPLE

Proposition 3.1. If (A, Y , Z) satisfy the conditions (2.1) and (2.2) of Theorem2.3, then there exists (Ã, Ỹ , Z̃) satisfying

Ã∗Ã+ I = 2(I + Z̃Z̃∗)−1,(2.1′)Ỹ∗Ỹ + Z̃∗Z̃ = I,(2.2′)‖Z̃‖ = ‖Z‖.(3.1)

and such that ρ, the representation induced by (A, Y , Z), dilates to ρ̃, the representa-tion induced by (Ã, Ỹ , Z̃).

If ‖Zh2‖ < ‖h2‖ for all h2 ≠ 0 and ‖Z∗h1‖ < ‖h1‖ for all h1 ≠ 0, then Ã, Ỹmay be chosen so that

(3.2) Ã, Ã∗, Ỹ , Ỹ∗ are one-one.

Finally, if both (2.1) and (2.2) are strict inequalities and ‖Z‖ < 1, then (Ã, Ỹ , Z̃)can be chosen so that (2.1′), (2.2′), and (3.2) hold,

(3.1′) ‖Z̃‖ < 1,and Z̃∗ has a polar decomposition as UP where the partial isometry U is unitary.


Proof. Fix a Hilbert space L and choose operators A1 : H1 , L and Y1 :H2 , L so that A∗A+ I +A∗1A1 = 2(I + Z∗Z)−1 and Y∗Y + Z∗Z + Y∗1 Y1 = I.

Let Kj = Hj⊕L⊕L⊕· · · , j = 0, 1, 2, and define Â : K1 , K0, Ŷ : K2 , K0and Ẑ : K2 , K1 by Â(h⊕f1⊕f2⊕· · · ) = Ah⊕A1h⊕f1⊕f2 · · · , Ŷ (h⊕f1⊕f2⊕· · · ) = Yh⊕Y1h⊕f1⊕f2 · · · , and Ẑ(h⊕f1⊕f2⊕· · · ) = Zh⊕0⊕0⊕· · · .Then Â, Ŷ , Ẑ satisfy (2.1′), (2.2′), and (3.1), and the representation induced by(Â, Ŷ , Ẑ) is a dilation of the representation induced by (A, Y , Z).

To prove the second part, note that from the construction of Â and Ŷ theirkernels, if any, must be a subset of {hj ⊕ 0 ⊕ 0 ⊕ · · · | hj ∈ Hj}, j = 1, 2,respectively. Identifying hj ∈ Hj with hj ⊕ 0 ⊕ 0 ⊕ · · · and using (2.1′), checkthat Âh1 = 0 if and only if ‖Ẑ∗h1‖ = ‖h1‖. Similarly, by (2.2′), Ŷh2 = 0 if andonly if ‖Ẑh2‖ = ‖h2‖. But these happen if and only if hj = 0, as can be seenfrom the form of Ẑ and the hypothesis ‖Zh2‖ < ‖h2‖ and ‖Z∗h1‖ < ‖h1‖ forall hj ≠ 0. Thus, Â and Ẑ are both one-one.

Let UP and VQ denote the polar decompositions of Â and Ŷ respectively.Since both Â and Ŷ are one-one, U and V are isometries and P and Q have nokernel. Let

Ã =U I − UU∗

0 U∗

P 00 P

,Ỹ =

V I − VV∗0 V∗

Q 00 Q

,and

Z̃ =Ẑ 0

0 Ẑ

.It is easily checked that Ã, Ỹ , Z̃ satisfy (2.1′), (2.2′), (3.1), and (3.2), and thatthe representation induced by (Ã, Ỹ , Z̃) dilates the representation induced by(A, Y , Z).

To prove the last assertion, suppose that both (2.1) and (2.2) are strict in-equalities and ‖Z‖ < 1. Let K denote the kernel of Z and let ∆ denote a positivemultiple of the inclusion K ⊂ H3 so that

Y∗Y + Z∗Z +∆∆∗ ≤ δI,(3.3)and

Z∗Z +∆∆∗ ≤ δI,(3.4)for some δ < 1. Let Kj = Hj ⊕ K ⊕ K ⊕ ·· · and define Ǎ(h ⊕ f1 ⊕ f2 ⊕· · · ) = Ah ⊕ 0 ⊕ 0 ⊕ · · · , Y̌ (h ⊕ f1 ⊕ f2 ⊕ · · · ) = Y(h ⊕ 0 ⊕ 0 ⊕ · · · ), and


Ž(h⊕ f1 ⊕ f2 ⊕ · · · ) = Zh⊕∆∗h⊕ 1/√2f1 ⊕ 1/√2f2 ⊕ · · · . With respect tothe decompositions Hj ⊕K⊕ (

⊕L), Ǎ∗Ǎ = A∗A⊕0⊕0, Y̌∗Y̌ = Y∗Y ⊕0⊕0,Ž∗Ž = (Z∗Z + ∆∆∗) ⊕ 12 I ⊕ 12 I, and ŽŽ∗ = ZZ∗ ⊕ ∆∗∆ ⊕ 12 I, where Z∆ = 0was used in the last computation.

The inequalities (3.3) and (3.4) imply that Y̌ , Ž satisfy (2.2) with a strictinequality. Since ∆∗∆ ≤ δI, it follows that 2(I +∆∗∆)−1 ≥ (2/(1+δ)I and thusǍ, Ž satisfy (2.1) with a strict inequality. The construction guarantees that Ž hasno kernel, ‖Ž‖ < 1, both inequalities (2.1) and (2.2) are strict for (Ǎ, Y̌ , Ž), andthat the representation induced by (A, Y , Z) dilates to the representation inducedby (Ǎ, Y̌ , Ž). Accordingly, to finish the proof, we may assume that (A, Y , Z)satisfy (2.1) and (2.2) with strict inequalities, Z has no kernel, and ‖Z‖ < 1.

The inequality (2.2) is equivalent to the inequality

(3.5) I + ZZ∗ ≤(A∗A+ I

2

)−1and if the inequality (2.2) is strict, then so is the inequality (3.5). Thus, with K∗equal to the kernel of Z∗, there is a positive multiple ∆∗ of the inclusionK∗ ⊂ H2such that

I + ZZ∗ +∆∗∆∗∗ ≤ (A∗A+ I2)−1

(3.6)

andZZ∗ +∆∗∆∗∗ ≤ δI,(3.7)

for some δ < 1.Let Kj∗ = Hj ⊕K∗ ⊕K∗ ⊕ · · · , j = 0, 1, 2 and define Ā∗(h1 ⊕ f1 ⊕ f2 ⊕

· · · ) = Ah1 ⊕ 0⊕ 0⊕ · · · , Ȳ∗(h2 ⊕ f1 ⊕ f2 ⊕ · · · ) = Y(h1⊕ 0⊕ 0⊕ · · · ), andZ̄∗(h2 ⊕ f1 ⊕ f2 ⊕ · · · ) = (Zh2 + ∆∗f1 ⊕ 1/√2f2 ⊕ 1/√2f3 ⊕ · · · ). Since Zhas no kernel and the range of ∆∗ is the kernel of Z∗, Z̄ is one-one. Since ∆∗ isone-one on the kernel of Z∗, Z̄∗ is also one-one.

With respect to the decompositionsHj⊕K∗⊕(⊕K∗), Ā∗Ā = A∗A⊕0⊕0,

Ȳ∗Ȳ = Y∗Y ⊕0⊕0, Z̄Z̄∗ = (ZZ∗+∆∗∆∗∗)⊕ 12 I, and Z̄∗Z̄ = Z∗Z⊕∆∗∗∆∗⊕ 12 I,where Z∗∆∗ = 0 was used in the last computation. In view of (3.6), Ā, Z̄ satisfy(2.1). Since ∆∗∆∗∗ ≤ I, the pair Ȳ , Z̄ satisfies (2.2). The inequality (3.7) implies‖Z̄‖ < 1. The operators Ā, Ȳ , Z̄ were constructed so that the representationinduced by (A, Y , Z) dilates to the representation induced by (Ā, Ȳ , Z̄). Since thislater tuple satisfies (2.1), (2.2), and ‖Z̄‖ < 1, by what has already been proved,it then dilates to a tuple (Ã, Ỹ , Z̃) which satisfies (2.1′), (2.2′), (3.2), and ‖Z̃‖ =‖Z̄‖ < 1. Thus, to finish the proof, we may assume that (A, Y , Z) satisfies (2.1)and (2.2), ‖Z‖ < 1, and Z and Z∗ have no kernel.

If (A, Y , Z) satisfies (2.1) and (2.2) and ‖Z‖ < 1, then by the first part of theTheorem, which has already been proved, there exists (Ã, Ỹ , Z̃) satisfying (2.1′),


(2.2′), (3.1), and (3.2). Moreover, since Z and Z∗ have no kernel, in the polardecomposition Z∗ = UP , the partial isometry U is unitary. From the constructionof Z̃ from Z, up to unitary equivalence Z̃ = Z ⊕ Z ⊕ 0. Thus, Z̃∗ has polardecomposition Ũ∗P̃ with Ũ = U ⊕ U ⊕ I. ❐

Proposition 3.2. Let ρ denote the representation induced by (A, Y , Z). If (2.1′)and (2.2′) hold, then T1 = ρ(t1) and T2 = ρ(t2) are partial isometries.

Proof. Since

T2T∗2 =0 0 00 0 0

0 0 I

,T2 is a partial isometry.

Let X denote the operator of (2.3), where R = (I + ZZ∗)1/2, and recall thatT1T∗1 = XX∗. Substituting I+ZZ∗ = 2(I+A∗A)−1, and usingA(I+A∗A)−1/2 =(I + AA∗)−1/2A, it follows that X∗X = I so that X, and thus T1, is a partialisometry. ❐

Thus, every one-one completely contractive unital representation ρ of Q di-lates to a one-one completely contractive unital representation ρ̃ of Q for whichboth T̃1 = ρ̃(t1) and T̃2 = ρ̃(t2) are partial isometries.

Proposition 3.3. If ρ : Q , B(Hρ) is a completely contractive unital repre-sentation, then there exists a completely contractive unital representation α induced by(R,U(I−P2)1/2, P), where U is unitary, 0 ≤ P ≤ I, and R = (I−P2)1/2(I+P2)−1/2such that for all ` and F ∈M`(Q), ‖ρ(`)(F)‖ ≤ ‖α(`)(F)‖.

The proof of Proposition 3.3 relies on the following three lemmas.

Lemma 3.4. If (A, Y , Z) satisfies (2.1) and (2.2) and if 0 < δ < 1, then(δA,δY , δZ) satisfies (2.1) and (2.2) with strict inequalities.

Proof. Only the strict inequality in (2.1) requires an explanation. In additionto the inequality (2.1), we have δ2A∗A ≤ A∗A and

2(I + ZZ∗)−1 ≤ 2(I + δ2ZZ∗)−1.

Hence, 2(I+δ2ZZ∗)−1−(I+δA∗A) can be written as a sum of positive operators,

2(I + δ2ZZ∗)−1 − (I + δ2A∗A) = (2(I + δ2ZZ∗)−1 − 2(I + ZZ∗)−1)+ (2(I + ZZ∗)−1 − (I +A∗A))+ ((I +A∗A)− (I + δ2A∗A)).


Thus, if strict inequality does not hold in (2.1) for δA, δZ, then there exists asequence of unit vectors xn such that all of the sequences

(2(I + δ2ZZ∗)−1 − 2(I + ZZ∗)−1)xn,(2(I + ZZ∗)−1 − (I +A∗A))xn,

(1− δ2)A∗Axn,

tend to zero. From the last of these sequences, it follows that Axn tends to zero.Thus, from the second, (2(I + ZZ∗)−1 − I)xn also tends to zero. Thus (I +ZZ∗)−1(I − ZZ∗)xn, and therefore (I − ZZ∗)xn, tend to zero. Thus, ‖Z∗xn‖tends to one. On the other hand, from the first limit it follows that Z∗xn tendsto zero. ❐

Lemma 3.5. Suppose (A, Y , Z) satisfies (2.1) and (2.2). Let δn = 1 − 1/nand (An, Yn, Zn) = (δnA,δnY , δnZ). Let ρ denote the representation induced by(A, Y , Z) and ρn the representation induced by (An, Yn, Zn). If F ∈ M`(Q), then‖ρ(`)n (F)‖ converges to ‖ρ(`)(F)‖.

Proof. Let T(n)j and Tj , j = 1, 2, denote the operators in standard formcorresponding to (An, Yn, Zn) and (A, Y , Z) respectively as in Proposition 2.2.Note that T(n)j converges to Tj , j = 1, 2. There exists matrices Fm ∈M` such thatF = ∑2m=0 Fmem, where e0, e1, e2 is the generating set for Q given immediatelybefore Proposition 2.1. Thus,

ρ(`)n (F) = F0 ⊗ (1−√

2(T (n)1 + T(n)2 ))+ F1 ⊗√

2T(n)1 + F2 ⊗√

2T(n)2 ,

which then converges to ρ(`)(F). ❐Lemma 3.6. If 0 ≤ P ≤ I, then the triple (R,W(I − P2)1/2, P) satisfies (2.1′)

and (2.2′) for any unitary W , where R = (I − P2)1/2(I + P2)−1/2.Conversely, if (A, Y , Z) satisfies (2.1′) and (2.2′), if A and Y have polar decom-

positions in which the partial isometry is a unitary, and if there is a polar decompo-sition Z∗ = U∗P with U unitary, then there is a unitary W so that the represen-tation induced by (A, Y , Z) is unitarily equivalent to the representation induced by(R,W(I − P2)1/2, P).

Proof. Straightforward computation proves the first part.To prove the second part, note that A∗A = 2(I+ZZ∗)−1− I = 2(I+P2)−1−

I = R2, and Y∗Y = U∗(I − P2)U , so that A and Y have polar decompositionsUAR and UYU∗(I − P2)1/2U where UA and UY are unitary. Let ρ denote therepresentation induced by (A, Y , Z), let Tj = ρ(tj), and represent T1 and T2 in


standard form as in Theorem 2.3. Let

V =

UA 0 0

0 I 0

0 0 U

.

To finish the proof note that V is unitary and

V∗T1V = 1√2

0 R −RP0 I −P0 0 0

and

V∗T2V = 1√2

0 0 W(I − P2)1/20 0 P

0 0 I

,

where W = U∗AUYU∗ ❐

Proof of Proposition 3.3. There exists (A, Y , Z) satisfying (2.1) and (2.2) suchthat ρ is the representation induced by (A, Y , Z). Let (An, Yn, Zn) and ρn be asin Lemma 3.5. By Lemma 3.4, (An, Yn, Zn) satisfy (2.1) and (2.2) with strict in-equalities and ‖Zn‖ < 1. Hence, by Proposition 3.1, ρn dilates to a representationρ̃n induced by (Ãn, Ỹn, Z̃n), where each of Ãn, Ỹn, Z̃∗n has a polar decompositionsin which the partial isometry is unitary and where (2.1′) and (2.2′) hold. Thus,by Lemma 3.6, there exists 0 ≤ Pn ≤ I and a unitary Wn such that ρ̃n is unitarilyequivalent to the representation αn induced by the triple (Rn,Wn(I−P2n)1/2, Pn),where Rn = (I − P2n)1/2(I + P2n)−1/2. The representation α =

⊕αn is the rep-

resentation induced by (⊕Rn,

⊕Wn(I − P2n)1/2,

⊕Pn) = (R,W(I − P2)1/2, P).

Thus, by Lemma 3.5,

‖α(`)(F)‖ = sup{‖α(`)n (F)‖ : n}= sup{‖ρ̃(`)n (F)‖ : n}≥ sup{‖ρ(`)n (F)‖ : n}≥ ‖ρ(`)(F)‖. ❐

Proposition 3.7. If A, Y , Z satisfy (2.1′), (2.2′), and (3.2), and if ‖Z‖ < 1,then the representation induced by (A, Y , Z) is a boundary representation of Q.


Proof. Let Tj = ρ(tj). In view of Proposition 2.1, it is enough to prove thatif

ν(tj) =Tj ∆j

0 Dj

and µ(tj) =Tj 0Γj Gj

induce completely contractive representations of Q, then Γj = ∆j = 0.

First consider ν. Set Jj = ν(tj), represent Tj in standard form onH0⊕H1⊕H2as in Theorem 2.3, and view Jj as a 4× 4 operator matrix on H0⊕H1⊕H2⊕H3.With respect to this decomposition, write

∆j =∆j0∆j1∆j2

.In view of Proposition 2.1, ν is completely contractive if and only if ‖Jj‖ ≤ 1,

J2j = 12Jj , and JjJm = 0 if j ≠ m. Further, since (A, Y , Z) satisfies both (2.1′)and (2.2′), by Proposition 3.2 both T1 and T2 are partial isometries so that

(3.8) TjT∗j Tj = Tj,

j = 1, 2.The inequality ‖Jj‖ ≤ 1 implies TjT∗j + ∆j∆∗j ≤ I. Since TjT∗j is the pro-

jection onto the closure of the range of Tj , it follows that the range of ∆j isorthogonal to the range of Tj so that

(3.9) T∗j ∆j = 0.Since J1J2 = 0, T1∆2 + ∆1D2 = 0. Multiplying on the left by T1T∗1 and using(3.8) and (3.9) gives

(3.10) T1∆2 = 0.Similarly,

(3.11) T2∆1 = 0.The identity J2j = 1/

√2Jj implies Tj∆j + ∆jDj = 1/√2∆j . Multiplying on

the left by TjT∗j and using (3.8) and (3.9) obtains

(3.12) Tj∆j = 0for j = 1, 2.


Since

T2∆1 =Y∆12Z∆12∆12

= 0,by (3.11), ∆12 = 0. By (3.12),

T1∆1 =A∆11∆11

0

= 0.Thus, ∆11 = 0. Finally, by (3.9),

T∗1 ∆1 =

0

A∗∆10−Z∗A∗∆10

= 0,so that A∗∆10 = 0. Since A∗ is one-one, ∆10 = 0. Thus ∆1 = 0.

To see that ∆2 = 0, observe that T2∆2 = 0 implies ∆22 = 0. Since T1∆2 = 0,it follows that ∆21 = 0. Finally, as T∗2 ∆2 = 0 and Y∗ has no kernel, ∆20 = 0.

Note that only the hypothesis that A∗ and Y∗ are one-one was used to showH reduces ν.

Now let Jj = µ(tj) and view Jj as a 4× 4 block operator matrix as before sothat

J∗j =T∗j Γ∗j

0 D∗j

.Since J1 and J2 satisfy the hypothesis of Proposition 2.1 if and only if J∗1 , J

∗2 do,

the mapping µ∗ defined by µ∗(tj) = J∗j is completely contractive.Since T∗j is a partial isometry,

(3.8∗) T∗j TjT∗j = T∗j .

Arguing as above,

TjΓ∗j = 0,(3.9∗)T∗1 Γ∗2 = 0,(3.10∗)T∗2 Γ∗1 = 0,(3.11∗)

andT∗j Γ∗j = 0.(3.12∗)


Write

Γ∗j =Γ∗j0Γ∗j1Γ∗j1

.Since T2Γ∗2 = 0 by (3.9∗), Γ∗22 = 0. By (3.10∗) and (3.12∗), T∗j Γ∗2 = 0, for j = 1,2. Thus,

(3.13)A∗Γ∗20 + Γ∗21 = 0,

Y∗Γ∗20 + Z∗Γ∗21 = 0.Solving (3.13) gives (Y∗ − Z∗A∗)Γ20 = 0.

The equalities (2.1′) and (2.2′) and the condition (3.2) yield unitaries UA andUY so that A = UAPA and Y = UYPY , where PA = (I − ZZ∗)1/2(I + ZZ∗)−1/2and PY = (I − Z∗Z)1/2. Hence

Y∗ − Z∗A∗ = (I − Z∗Z)1/2U∗Y − Z∗(I − ZZ∗)1/2(I + ZZ∗)−1/2U∗A= (I − Z∗Z)1/2U∗Y − (I − Z∗Z)1/2(I + Z∗Z)−1/2Z∗U∗A= (I − Z∗Z)1/2U∗Y (I −UY(I + Z∗Z)−1/2Z∗U∗A ).

The last part of the last term is invertible since

‖UY (I + Z∗Z)−1/2Z∗U∗A‖ ≤ ‖Z∗‖ < 1.

Thus Y∗ − Z∗A∗ is invertible and so Γ∗20 = 0. It now follows from (3.13) thatΓ∗21 = 0, and thus Γ2 = 0.Since T1Γ∗1 = 0 and T∗j Γ∗1 = 0 for j = 1, 2,

(3.14)

∣∣∣∣∣∣∣∣Γ∗11 − ZΓ∗12 = 0,A∗Γ∗10 + Γ∗11 = 0,

Y∗Γ∗10 + Z∗Γ∗11 + Γ∗12 = 0.Solving for Γ∗11 in the first equation and substituting into the third and secondequations gives the block matrix equation

(3.15)

(I + Z∗Z) Y∗Z A∗

Γ∗12Γ∗10 = 0.

Since the (1,1) entry of the above operator matrix is invertible, the matrix itself isinvertible if and only if the Schur complement A∗−Z(I+Z∗Z)−1Y∗ is invertible.


Again, the equalities (2.1′) and (2.2′) and the condition (3.2) yield unitariesUA and UY so that A = UAPA and Y = UYPY , where PA = (I − ZZ∗)1/2(I +ZZ∗)−1/2 and PY = (I − Z∗Z)1/2. HenceA∗ − Z(I + Z∗Z)−1Y∗

= (I − ZZ∗)1/2(I + ZZ∗)−1/2U∗A − Z(I + Z∗Z)−1(I − Z∗Z)1/2U∗Y= (I − ZZ∗)1/2(I + ZZ∗)−1/2U∗A − (I + ZZ∗)−1(I − ZZ∗)1/2Z∗U∗Y= (I − ZZ∗)1/2(I + ZZ∗)−1/2U∗A (I −UA(I + ZZ∗)−1/2Z∗U∗Y ).

As before, the last part of the last term is invertible since

‖UA(I + ZZ∗)−1/2Z∗U∗Y ‖ ≤ ‖Z∗‖ < 1,and thus the last term is invertible.

Since the matrix in (3.15) is invertible, Γ∗10 = Γ∗12 = 0. From (3.14), it nowfollows that Γ∗11 = 0.

Note that only UY and AY are isometries, which is implied by the hypothesisA and Y are one-one, was need to show H reduces µ. ❐

4. THE C∗-ENVELOPE FOR THE BIDISK EXAMPLE

We are now in a position to describe C∗e (Q) up to a *-isomorphism. To thisend, let C∗u(U, P) be the universal C∗-algebra generated by a unitary U and apositive P satisfying 0 ≤ P ≤ I. This universal C∗-algebra is determined by theproperty that, given any unitary U0 and positive P0 such that 0 ≤ P0 ≤ I on someHilbert space H, there exists a *-homomorphism σ0 : C∗u(U, P) , C∗({U0, P0})with σ0(U) = U0 and σ0(P) = P0. To form this C∗-algebra, start with the freeunital *-algebra on symbols U and P and assign a C∗-(semi)norm to the algebraby taking the supremum over σ0. (Note that it may be assumed that all the U0and P0 act on a fixed separable Hilbert space.)

Theorem 4.1. Define elements Tµ1 and Tµ2 of M3 ⊗ C∗u(U, P) by

Tµ1 =1√2

0 R −RP0 I −P0 0 0

(4.1)and

Tµ2 =1√2

0 0 U(I − P2)1/20 0 P

0 0 I

,(4.2)where R = (I + P2)1/2(I − P2)−1/2. Then there exists a *-isomorphismσ : C∗({Tµ1 , Tµ2 }), C∗e (Q), with σ(Tµj ) = tj , j = 1, 2.


Proof. By Theorem 2.3, there is a completely contractive unital homomor-phism α : Q, C∗({Tµ1 , Tµ2 }) with α(tj) = Tµj , j = 1, 2. To see that α is a com-plete isometry, represent C∗e (Q) ⊂ B(K) so that the representation γ : Q , B(K)is a complete isometry. By Proposition 3.3, there exists a unitary U0 and a positiveP0, 0 ≤ P0 ≤ I, such that the completely contractive unital homomorphism ρ0induced by (R0, U0(I − P20 )1/2, P0), where R0 = (I − P20 )1/2(I + P20 )−1/2, has theproperty ‖γ(`)(F)‖ ≤ ‖ρ(`)0 (F)‖, for all ` and F ∈ M`(Q). Thus ρ0 is completelyisometric, and

ρ0(t1) = T 01 =1√2

0 R0 −R0P00 I −P00 0 0

and

ρ0(t2) = T 02 =1√2

0 0 U0(I − P20 )1/20 0 P0

0 0 I

.By the universal property of C∗u(U, P), there exists a *-homomorphism σ0 :

C∗u(U, P) , C∗({U0, P0}) such that σ0(U) = U0 and σ0(P) = P0. Hence 13 ⊗σ0 = σ(3)0 maps Tµj to T 0j . Since σ(3)0 ◦α = ρ0, the map α is a complete isometry.

By the universal property of C∗e (Q), there exists an onto *-homomorphismτ : C∗({Tµ1 , Tµ2 }) , C∗e (Q), with τ(Tµj ) = γ(tj). To finish the proof it isenough to show that τ is one-one.

Let U , P denote the universal U and P of C∗u(U, P). Given 0 < δ < 1, letρδ denote the representation of Q induced by (Rδ,U(I − (δP)2)1/2, δP), whereRδ = (I − (δP)2)1/2(I + (δP)2)−1/2 and let Tδj = ρδ(tj) denote the corre-sponding operators in standard form. By Lemma 3.6 and Proposition 3.7, thetuple (Tδ1 , T

δ2 ) is a boundary tuple. Hence, by Theorem 1.2, there exists πδ :

C∗e (Q) , C∗({Tδ1 , Tδ2 }), a *-homomorphism, such that Tδj = πδ(γ(tj)). If pis any *-polynomial in two non-commuting variables, then πδ(τ(p(T

µ1 , T

µ2 ))) =

p(Tδ1 , Tδ2 ) and, as δ tends to 1, p(T

δ1 , T

δ2 ) tends to p(T

µ1 , T

µ2 ). Thus τ is one-one.

This completes the proof. ❐

Remark 4.2. We don’t know whether the mapping

π : C∗e (Q), C∗({Tµ1 , Tµ2 }) with π(γ(tj)) = Tµj , j = 1,2,

is a boundary representation.

Corollary 4.3. Let x0, x1, x2 be the three points in the unit bidiskD2, as above.Given `×` matrices W0, W1, W2, there exists an analytic function F : D2 , M` such


that ‖F(z)‖ ≤ 1 for all z ∈ D2 and F(xj) = Wj , j = 0, 1, 2, if and only ifW0 ⊗ I (W1−W0)⊗ R (W0−W1)⊗ RP+(W2−W0)⊗ U(I−P2)1/2

0 W1 ⊗ I (W2−W1)⊗ P0 0 W2 ⊗ I

has norm at most one.

Proof. The existence of such a function F and ‖W0⊗E0+W1⊗E1+W2⊗E2‖ ≤1 are equivalent. Applying Proposition 2.1 and Theorem 4.1, we obtain an explicitformula for the idempotents E0, E1, and E3, which, when substituted for theseidempotents, yields the desired result. ❐

Lemma 4.4. Let U be a unitary, P positive and invertible with ‖P‖ < 1, andset R = (I + P2)1/2(I − P2)−1/2. The unital C∗-subalgebra of M3 ⊗ C∗({U,P}),generated by

T1 = 1√2

0 R −RP0 I −P0 0 0

and

T1 = 1√2

0 0 U(I − P2)1/20 0 P

0 0 I

is all of M3 ⊗ C∗({U,P}).

Proof. Let H denote the space that U , P act on so that the matrix representa-tions of Tj are with respect to an orthogonal decompositionH ⊕H ⊕H. Let Qjmdenote the block matrix unit equal to the identity operator in the (j,m) position,where the indexing starts with 0, so that 0 ≤ j,m ≤ 2. By considering the identityI, T∗2 T2, and (I−

√2(T1+T2))(I−

√2(T1+T2))∗, eachQmm is in C∗({T1, T2}).

By consideringQ11T2T∗2 Q11, T∗2 Q11T2, andQ00T1Q11T

∗1 Q00, it can be seen that

C∗({T1, T2}) contains∑X`Qjj for Xj ∈ C∗({P}). From this fact it follows that

Q01, Q02, and UQ02 are all in C∗({T1, T2}). Hence C∗({T1, T2}) contains all ofQjm and UQjm. ❐

Theorem 4.5. If U is unitary and P is positive and invertible with ‖P‖ < 1,then M3⊗C∗({U,P}) is a quotient of C∗e (Q). In particular, C∗e (Q) has irreduciblerepresentations on Hilbert spaces of dimension 3k for k = 1, 2, 3, . . . and on separableinfinite dimensional Hilbert space.


Proof. For any such pair (U, P) the representation induced by (R,U(I −P2)1/2, P), where R = (I − P2)−1/2(I + P2)1/2, is a boundary representationby Proposition 3.7 and hence there is an onto *-homomorphism π : C∗e (Q) ,C∗({T1, T2}), where Tj = π(γ(tj)) and Tj are as in Proposition 3.2. An applica-tion of Corollary 4.3 shows that M3 ⊗ C∗({U,P}) is a quotient of C∗e (Q).

Given k, let P denote the diagonal matrix with a 1 in the (1,1) entry andzeros elsewhere, and let U denote the matrix which circularly permutes the basisso that U has a 1 in the (m,m + 1), 1 ≤ m < k, and (k,1) entries and is 0elsewhere. Since C∗({U,P}) = Mk, this pair yields an irreducible representationof C∗e (Q) on a Hilbert space of dimension 3k.

For the infinite dimensional irreducible representation, consider the Hilbertspace `2(Z) and let U be the bilateral shift and P the matrix with a 1 in the (0,0)entry and 0 elsewhere. Then C∗({U,P}) contains all the matrix units and is henceirreducible. ❐

Remark 4.6. For any k, one can choose U , P , with 0 < P < I, such thatC∗({U,P}) = Mk⊗C∗(Fk), where C∗(Fk) is the full C∗-algebra of the free groupon k generators. Thus M3k ⊗ C∗(Fk) is also a quotient of C∗e (Q). To see how toconstruct such a pair U , P , let U1, . . . , Uk be unitaries that generate C∗(Fk) andlet (ai,j) be a scalar k×k unitary with all non-zero entries. If we let U = (Uiai,j)and let P be a diagonal matrix whose diagonal entries are of the form diI, where0 < d1 < · · · < dk < 1, then it is easily checked that C∗(U, P) = Mk ⊗ C∗(Fk)as desired.

5. THE ANNULUS EXAMPLE

In this section we compute C∗e (Q), where Q is the quotient of the algebra ofbounded analytic functions on an annulus, modulo the ideal of functions whichvanish at a finite number of points, x1, . . . , xn. The case of two points is easily an-alyzed and distinct from the case of three or more points. Accordingly, throughoutthis section n ≥ 3.

To establish the notation, fix q, 0 < q < 1, let A = {z : q < |z| < 1}, letE = {x1, . . . , xn} be a set of distinct points in A, let H∞(A) denote the algebra ofbounded analytic functions on A, and let IE denote the ideal of functions whichvanish on E. With these notations, Q = H∞(A)/IE .

Abrahamse’s [1] generalization of Pick’s Theorem to multiply connected do-mains involves a family of kernels. In the doubly connected case of A, up tocanonical equivalences, the family is given by

(5.1) k(z,w; t) =∞∑−∞

(zw̄)n

1+ q2n+1q−2t .


According to Abrahamse, given w1, . . . , wn in D, there exists f ∈ H∞(A) suchthat ‖f‖ ≤ 1 and f(xj) = wj if and only if, for every t, the n×n Pick matrix

M(t) = ((1−wjw`)k(wj,w`; t))

is positive semi-definite. Note that k(z,w; t + 1) = zk(z,w; t)w̄, so that itsuffices to consider 0 ≤ t < 1.

Ball [5] extended Pick interpolation to matrix-valued functions on multiplyconnected domains. In general, `×`matrix-valued interpolation requires a familyof `×` matrix-valued kernels. However, in the doubly connected case, these `×`matrix-valued kernels reduce to a direct sum of the scalar kernels (5.1).

Theorem 5.1 (Abrahamse and Ball). Given ` × ` matrices W1, . . . , Wn, thereexists F ∈ M`(H∞(A)) such that ‖F‖ ≤ 1 and F(xj) = Wj if and only if for every0 ≤ t < 1 the n×n block Pick matrix with ` × ` matrix entries

M(t) = ((1−WjW∗m)k(wj,wm; t))

is positive semi-definite.

Choose h1, . . . , hn such that ehj = xj and let P(t) denote the n×n matrix

P(t) = (pjm(t)) = (exp(thj)k(xj, xm; t) exp(thm)).

With respect to a measure µt on the boundary of A which is a constant multipleof arclength measure on each boundary component (say arclength measure onthe outer boundary and a multiple, which varies with t, of arclength measure onthe inner boundary), integrating against k(·, z; t) reproduces functions analyticin a neighborhood of the closure of A, from which it follows that P(t) is strictlypositive definite.

If D is a diagonal matrix, then B(t) = P(t)−1/2DP(t)1/2 is a continuousmatrix-valued function with B(t + 1) = B(t) and thus can be regarded as a con-tinuous function on the unit circle T.

Given f ∈ H∞(A), we obtain an element Bf (t) ∈ C(T)⊗Mn by setting

Bf (t) = P(t)−1/2 Diag(f (x1), . . . , f (xn))P(t)1/2,

where Diag(d1, . . . , dn) is the diagonal matrix with diagonal entries d1, . . . , dn.Since Bf depends only upon the values of f on E, we obtain a mapping γ0 : Q,C(T)⊗Mn, which is easily seen to be a homomorphism.

Theorem 5.2. With notations as above,(a) γ0 is a complete isometry;(b) C∗(γ0(Q)) =Mn ⊗ C(T); and(c) the induced homomorphism from C∗(γ0(Q)) onto C∗e (Q) is a *-isomorphism.


Thus, the C∗-envelope of Q is Mn ⊗ C(T).We begin by collecting a couple of lemmas.

Lemma 5.3. The matrix valued function P(t) is continuous and P(t + 1) =P(t). For each t and j there exists at most one m such that pjm(t) = 0. For eachf ∈ H∞(A), Bf (t) is continuous and Bf (t) = Bf (t + 1).

Proof. It is well known that the kernels vary continuously with t [1]. Theexponential factors exp(thj)were chosen exactly to kill the period k(z,w; t+1) =zk(z,w; t)w̄ so that P(t+1) = P(t). From the theta function representation forthe kernels k(z,w; t) [6], [11], for fixedw and t, k(z,w; t) has at most one zeroin A. This proves the statements regarding P(t).

For each t, P(t) is strictly positive definite. Thus, as P(t) is continuous,P(t)1/2 and P(t)−1/2 are continuous. The remainder of the lemma follows. ❐

Lemma 5.4 ([11, Theorem 2.4, n = 3]). Choose the negative real axis for abranch cut of the logarithm. Let L denote the lattice Z + (−iπ/ log(q))Z and e∗denote the half period: e∗ = 12[1+−iπ/(2 log(q))]. Given tuples of distinct pointsz1, z2, z3 andw1,w2,w3 from A, and t real, if the matrix k(zj,wm; t) is singular,then

t − 12 log(q)

(∑log(zj)+

∑log(wm)

)= e∗ mod L.

Remark 5.5. As a consequence of Lemma 5.4, if z1, z2, z3 and w1, w2 bothare distinct sets of points, then the matrix (k(zj,wm; t)) has rank two.

To see why this is true, suppose the matrix were only rank one, then bothof the matrices k(zj,wm; t)j≠3 and k(zj,wm; t)j≠2 would be singular so that,modulo L,

t − 12 log(q)

( ∑j≠r

log(zj)+2∑1

log(wm))= e∗

for r = 3, 2. Taking the difference would then imply (1/(2 log(q))(log(z3) −log(z2)) is zero modulo L, but that implies z2 = z3.

Lemma 5.6. Suppose M is a positive k × k matrix with rank two and v is avector. There is at most one r such that(

M vv∗ r

)

is positive and of rank two.

Proof. If both M and the displayed matrix are positive, then there exists aunique w in the range of M so that v = M1/2w. If both M and the displayedmatrix also have rank two, then r = w∗w. ❐


Lemma 5.7. If there exists a diagonal matrix D such that

(k(xm,xj ; t)) = D(k(xm,xj ; s))D∗,

then s = t modulo Z.

Fix w ∈ A. Let E denote the functions ϕ which are analytic in A, haveexactly two zeros (counting with multiplicity), one at w and the other at aϕ, andwhich are unimodular on the boundary of A. These functions can be explicitlyconstructed in terms of theta functions. Moreover, modulo multiplication by aunimodular constant, E can be parameterized by the unit circle in such a way thatthe second zero aϕ varies continuously with the parameter.

Proof of Lemma 5.7. Fix a point w ∈ A distinct from E and let xn+1 = w.Let E denote the collection of functions above.

The operator St of multiplication by ζ(z) = z on the Hilbert space Ht offunctions analytic in A with reproducing kernel k(z,w; t) is a rank one Bundleshift over A. Thus St is a pure subnormal operator whose minimal normal ex-tension has spectrum in A, if ϕ ∈ E; then ϕ(St) is a pure isometry, and St isunitarily equivalent to Ss if and only if s − t ∈ Z. Since ϕ has two zeros, w andaϕ, the shift ϕ(St) has multiplicity two and I −ϕ(St)ϕ(St)∗ is the projectiononto the kernel ofϕ(St)∗. Sinceϕ(St)∗k(·,w; t) =ϕ(St)∗k(·, aϕ; t) = 0, thiskernel is spanned by {k(·,w; t), k(·, aϕ; t)}. Restricting to the points x ∈ E, itfollows that the n×n matrix

Nt(ϕ) = ((1−ϕ(xj)ϕ(xm))k(xj, xm; t))= (〈(I −ϕ(St)ϕ(St)∗)k(·, xm; t), k(·, xj ; t)〉)

is positive semidefinite with rank at most two and that its range is spanned by{(k(xj,w; t))nj=1, (k(xj, aϕ; t))nj=1}. On the other hand, by (5.5.1), these vec-tors are linearly independent so that Nt(ϕ) has rank exactly two.

Note that (k(xj,w; t))nj=1 is in the range of Nt(ϕ) indepedent of ϕ. More-over, ⋂

ϕrange(Nt(ϕ)) = span{(k(xj,w; t))nj=1}.

To verify this last assertion, choose ψ ∈ E with aψ ≠ aϕ. If Nt(ϕ) ∩ Nt(ψ)contains anything but a multiply of (k(xj,w; t))nj=1, then there exists constantscϕ, dϕ and cψ, dϕ with neither dϕ nor dψ equal to zero and

cϕk(·,w; t)+ dϕk(·, aϕ; t) = cψk(·,w; t)+ dψk(·, aψ; t).


Hence, amongst {k(·,w; t), k(·, aϕ; t), k(·, aψ; t)} there is a nontrivial linearrelation so that the matrix

k(x1,w; t) k(x1, aϕ; t) k(x1, aψ; t)

k(x2,w; t) k(x2, aϕ; t) k(x2, aψ; t)

k(x3,w; t) k(x3, aϕ; t) k(x3, aψ; t)

is singular. By Lemma 5.4, modulo L,

t − 12 log(q)

(∑log(xj)+ log(w)+ log(aϕ)+ log(aψ)

)= e∗.

Varying aψ continuously obtains a contradiction.Since δ(xj)k(xj, xm; s)δ(xm) = k(xj, xm; t), we have

Nt(ϕ) = ((1−ϕ(xj)ϕ(xm))δ(xj)k(xj, xm; s)δ(xm))= D((1−ϕ(xj)ϕ(xm))k(xj, xm; s))D∗ = DNs(ϕ)D∗.

Consequently, the range of Nt(ϕ) is also the range of DNs(ϕ), which is spannedby {(δ(xj)k(xj,w; s))nj=1, (δ(xj)k(xj, aϕ; s))nj=1}. Thus, (δ(xj)k(xj,w; s))nj=1is in the range of Nt(ϕ), independent of ϕ. Thus, by what was proved above,there is a constant c ≠ 0 such that

(5.2) δ(xj)k(xj,w; s) = k(xj,w; t)c,

j = 1, . . . , n. Let δ(w) = 1/c.Now let Nat (ϕ) denote the same matrix as Nt(ϕ), except with the index

running to n + 1 instead of n and with xn+1 = w. Similarly, let Da denote thediagonal matrix which is the same as D, but with an extra row and column andδ(w) in the (n + 1, n + 1) entry. From (5.2) and the fact that ϕ(w) = 0, thematrices Nat (ϕ) and DaNas (ϕ)D∗a are the same, with the possible exception ofthe entry in the (n+1)× (n+1) position, are both positive and of rank two, andthe upper n×n block is positive and rank two. Hence by Lemma 5.6 and the factthat ϕ(w) = 0, k(w,w; t) = δ(w)k(w,w; s)δ(w).

Choose xn+1, xn+2, . . . so that {xj}∞j=1 is a determining set for A. Arguingby induction, there exists δ(xj) so that

(5.3) k(xj, xm; t) = δ(xj)k(xj, xm; s)δ(xm),for all j, m. Define U : Ht , Hs densely on finite linear combinations ofk(·, xm; t) by Uk(·, xm; t) = δ(xm)k(·, xm; t). By (5.3) U is a unitary andthus extends to a unitary, still denoted by U , from Ht onto Hs . Moreover, Uintertwines S∗t and S∗s . Hence St and Ss are unitarily equivalent and thus t − s ∈Z. ❐


Proof of Theorem 5.2. To prove (a), let F ∈ M`(H∞(A)) be given and letWj = F(xj). By Theorem 5.1, ‖F +M`(IE)‖ ≤ 1 if and only if(5.4) ((I −WjW∗m)k(xj, xm; t)) ≥ 0for all t ∈ R. Let D denote the n × n block diagonal matrix with the ` × `matrices W1, . . . , Wn as diagonal entries. Let H(t) denote the diagonal matrixwith diagonal entries exp(thj). Thus, (5.1) holds if and only if

0 ≤ H(t)∗((I −WjW∗m)k(xj, xm; t))H(t)= ((I −WjW∗m)pjm(t))= P(t)−DP(t)D∗.

Conjugating by P(t)−1/2, it follows that ‖F + M`(IE)‖ ≤ 1 is equivalent to‖P(t)−1/2DP(t)1/2‖ ≤ 1 for all t. Thus, ‖F + M`(IE)‖ ≤ 1 if and only if‖γ(`)0 (F)‖ ≤ 1 and therefore γ0 is a complete isometry.

To prove (b), it will be useful to choose functions fj ∈ H∞(A) such thatfj(x`) = δj`, and let Ej(t) = γ0(fj) = P(t)−1/2EjjP(t)1/2, since C∗(γ0(Q)) =C∗({E1(t), . . . , En(t)}).

We will prove that B = C∗(γ0(Q)) is a rich C∗-subalgebra of A = Mn⊗C(T)in the sense of [8, 11.1.1]. It will then follow by [8, 11.1.4] that B = A.

Fix t and let C ⊂ Mn denote the C∗-algebra generated by {E1(t), . . . , En(t)}.Since every irreducible representation of A is equivalent to evaluation at a point,and since C is the image of B under the *-homomorphism of point evaluation, toverify [8, 11.1.1(i)], it suffices to show that C = Mn.

To this end, note that by considering Ej(t)Ej(t)∗ and using the fact thatpjj(t) > 0, C contains P(t)−1/2EjjP(t)−1/2. Summing over j, it follows thatC contains P(t)−1 and therefore P(t)−1/2. Consequently, as Ej(t)Em(t)∗ =P(t)−1/2pjm(t)P(t)−1/2 is also in C, C contains Ejm whenever pjm(t) ≠ 0.Suppose pjm = 0. Then, by Lemma 5.3, if we choose ν different from both jandm, then pνm(t) ≠ 0 and pjν(t) ≠ 0 and thus C contains both Eνm and Ejν .Since EjνEνm = Ejm, it follows that C contains all the matrix units Ejm and istherefore all of Mn.

Next we prove that the restrictions to B of inequivalent irreducible repre-sentations of A are inequivalent on B. To achieve this we must show that, for0 ≤ s, t < 1, and s ≠ t, there is no unitary U such that U∗Ej(t)U = Ej(s)for all j. For if U∗P(t)−1/2EjjP(t)1/2U = P(s)−1/2EjjP(s)1/2 for each j, thenD = P(s)1/2U∗P(t)−1/2 commutes with each Ejj and is therefore diagonal. Thisimplies that P(s) = D∗P(t)D, so that s and t satisfy the hypothesis of Lemma5.7. Thus t − s ∈ Z and so s = t.

This last calculation shows that B satisfies [8, 11.1.1(ii)], and so B is a richC∗-subalgebra of A as was to be shown.

Since γ0 is a complete isometry, there exists an onto *-homomorphism τ :C∗(γ0(Q)) , C∗e (Q) such that τ(γ0(f )) = γ(f). However, by [11, Theorem


0.10], γ0 is a boundary representation and so there exists a *-homomorphism fromC∗e (Q) to C∗(γ0(Q)). Since the composition of this map with γ0 is the identityon Q, this latter map must be an inverse for τ and hence τ is a *-isomorphism.This proves (c) and completes the proof of the theorem. ❐

REFERENCES

[1] M.B. ABRAHAMSE, The Pick interpolation theorem for finitely connected domains, MichiganMath. J. 26 (1979), 195-203.

[2] JIM AGLER, Interpolation, unpublished manuscript.[3] JIM AGLER & JOHN E. MCCARTHY, The three point Pick problem on the bidisk, New York J.

Math. 6 (2000), 227-236.[4] WILLIAM ARVESON, Subalgebras of C*-algebras, Acta Math. 123 (1969), 141-224.[5] JOSEPH A. BALL, A lifting theorem for operator models of finite rank on multiply-connected do-

mains, Journal of Operator Theory 1 (1979), 35-57.[6] JOSEPH A. BALL & KEVIN F. CLANCEY, Reproducing kernels for Hardy spaces on multiply

connected domains, Integral Equations Operator Theory 25 (1996), 35-57.[7] DAVID P. BLECHER, ZHONG-JIN RUAN & ALLAN M. SINCLAIR, A characterization of

operator algebras, J. Funct. Anal. 89 (1990) 188-201.[8] JACQUES DIXMIER, C∗-algebras, North-Holland, New York, 1977.[9] MASAMICHI HAMANA, Injective envelopes of operator systems, Publ. Res. Inst. Math. Sci. 15

(1979), (773-785).[10] S.I. FEDOROV & V.L. VINNIKOV, On the Nevanlinna-Pick interpolation in multiply connected

domains, (Russian) Dept. of Mathematics University of Auckland Report Series 325, October1995; Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI) 254 (1998); Anal.Teor. Chisel. i Teor. Funkts. 15, 5-27.

[11] S. MCCULLOUGH, Isometric representations of some quotients ofH∞ of an annulus Integral Equa-tions and Operator Theory 39 (2001), 335-362.

[12] D. SARASON, On spectral sets having connected complement, Acta. Sci. Math. 26 (1965), 289-299.[13] JAMES P. SOLAZZO, Interpolation and computability, Ph. D. Thesis, University of Houston,

2000.[14] W. FORREST STINESPRING, Positive functions on C∗-algebras, Proc. Amer. Math. Soc. 6

(1955), 211-216.

SCOTT MCCULLOUGH:Department of MathematicsUniversity of Florida490 Little HallGainesville, FL 32611-8105, U. S. A.E-MAIL: [email protected]

VERN PAULSEN:Department of MathematicsUniversity of HoustonHouston, TX 77204-3476, U. S. A.E-MAIL: [email protected]

KEY WORDS AND PHRASES: Hamana boundary, bidisk, interpolation.Received : July 9th, 2001.

IntroductionBoundary representationsThe bidisk exampleBoundary representations for the bidisk exampleThe C^*-envelope for the bidisk exampleThe annulus exampleReferences

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