Extension Theorems for

Contraction Operators on Kreın Spaces

Michael A. Dritschel and James Rovnyak∗

To the memory of Mark Grigor’evic Kreın

Abstract notions of Julia and defect operators are used as a foundation for atheory of matrix extension and commutant lifting problems for contraction operators onKreın spaces. The account includes a self-contained treatment of key propositions fromthe theory of Potapov, Ginsburg, Kreın, and Shmul’yan on the behavior of a contractionoperator on negative subspaces. This theory is extended by an analysis of the behaviorof the adjoint of a contraction operator on negative subspaces. Together, these resultsprovide the technical input for the main extension theorems.

Contents

Introduction

Chapter 1: Operator Theory on Kreın Spaces

1.1 Definitions and preliminaries1.2 Defect operators and Julia operators1.3 Contraction and bicontraction operators1.4 Additional results on contractions and bicontractions

Notes

Chapter 2: Matrix Extensions of Contraction Operators

2.1 The adjoint of a contraction2.2 Column extensions2.3 Row extensions2.4 Two-by-two matrix completions

Notes

∗ The second author was supported by the National Science Foundation.

1

Chapter 3: Commutant Lifting of Contraction Operators

3.1 Dilation theory3.2 Commutant lifting3.3 Characterization of extensions3.4 Abstract Leech theorem

Notes

Appendix A: Complementation Theory

Appendix B: More on Julia Operators

Bibliography

Introduction

Kreın spaces are indefinite generalizations of Hilbert spaces which areimportant in both abstract operator theory and its applications. We are concernedwith everywhere defined and continuous linear operators on a Kreın space H toa Kreın space K. The set of all such operators is denoted B(H,K). An operatorT ∈ B(H,K) is a contraction if

〈Tf, Tf〉K ≤ 〈f, f〉H

for all vectors f in H. If both T and T ∗ are contractions, then T is said to be abicontraction.

Our purpose is to show the possibility of proving extension theorems forcontraction operators on Kreın spaces which are strikingly similar to those of theHilbert space case. By an extension of an operator T ∈ B(H,K) we mean, forexample, a row extension

( T E ) ∈ B(H⊕ E ,K),

column extension(

TF

)

∈ B(H,K ⊕F),

or a two-by-two matrix extension

(

T EF G

)

∈ B(H⊕ E ,K ⊕F),

2

where E and F are Kreın spaces. Many theorems have the hypothesis that F isa Hilbert space. In applications, this hypothesis is often assured whenever T isa contraction, and so the hypothesis is not as restrictive as might appear. Ap-plications include commutant lifting theorems, in which a commutator relation isextended to minimal isometric and minimal unitary dilations.

We assume familiarity with operator theory on Hilbert spaces, but we donot presume that the reader is necessarily at ease in the indefinite environment ofKreın spaces. Our aim has been to give a self-contained treatment for such readers,and experts may view parts of the paper as expository. Background material issummarized in §1.1. While it is essential to master concepts of regular subspaces,projections, negative subspaces, isometries, and direct sums, if a few things aretaken for granted or gleaned from standard monographs, not much more is neededas preparation. Kreın space extensions of the Hilbert space notions of defect andJulia operators are introduced in §1.2. Julia operators are a kind of unitary two-by-two matrix extension of a given operator and are needed to formulate certainof the extension theorems in Chapters 2 and 3.

The backbone of the study of contraction operators on Kreın spaces isa theory, due to Ginsburg, Kreın, and Shmul’yan, which analyzes how a contrac-tion operator acts with respect to negative subspaces. One should also mentionPotapov, whose treatment of the finite dimensional case helped to motivate de-velopments. The discussion in §1.3 includes the Potapov-Ginsburg transform andits interpretation as a scattering operator, characterizations of bicontractions, andmappings of operator spheres.

The main new results are those of Chapters 2 and 3. These characterizecontractive and bicontractive extensions of given contractive and bicontractive op-erators. Theorems on row, column, and two-by-two matrix extensions, as well ascommutant lifting theorems, are proved for Kreın spaces in forms essentially iden-tical to the Hilbert space case. Similar results were previously obtained by Alpay,de Branges, Constantinescu and Gheondea, and Dritschel, using other hypothesesand somewhat different methods. Our approach is based on an analysis of theadjoint of a contraction relative to negative subspaces in §2.1.

The principal methods and main results of Chapters 2 and 3 appear in thefirst author’s doctoral dissertation (University of Virginia, May, 1989). The presentaccount is an expanded version of the dissertation which includes simplificationsand additional results.

3

Chapter 1: Operator Theory on Kreın spaces

1.1 Definitions and Preliminaries

Kreın spaces are generalizations of Hilbert spaces and are the fundamentalunderlying objects in our study. Our philosophy on notation is to reserve thesimplest notation for Kreın spaces and operators which act on them. This isgenerally standard Hilbert space notation.

A. Kreın Spaces and Continuous Operators

A Kreın space H is a scalar product space which is isomorphic to thedirect sum of a Hilbert space and the anti-space of a Hilbert space. By a scalar

product space we mean a complex vector space H together with a scalar product〈·, ·〉 which obeys the same axioms of linearity and symmetry as for Hilbert spacesand is nondegenerate in the sense that the only vector f in H such that 〈f, g〉 = 0for all g in H is f = 0. The anti-space of a scalar product space (H, 〈·, ·〉) is(H,−〈·, ·〉). Notions of isomorphism, subspace, orthogonality, direct sum, andlinear operator are defined as in linear algebra. Orthogonality is indicated by ⊥,direct sum by +, and orthogonal direct sum by ⊕. Occasionally, a subscript H isused to show dependence on the underlying space.

A fundamental decomposition of a Kreın space H is a direct sum repre-sentation H = H+⊕H− of H, where H+ and H− are subspaces of H such that H+

is a Hilbert space and H− the anti-space of a Hilbert space in the scalar productof H. In general, fundamental decompositions are not unique. The choice of afundamental decomposition H = H+ ⊕ H− induces a Hilbert space inner prod-uct, norm, and strong topology on H. Namely, the Hilbert space inner product off+ + f− and g+ + g− (f+, g+ ∈ H+ and f−, g− ∈ H−) is 〈f+, f+〉 − 〈f−, f−〉. Thestrong topology of this Hilbert space is independent of the choice of fundamentaldecomposition and is also called the Mackey topology of H. It is used to define con-vergence and continuity in the usual way. The norm of the Hilbert space dependson the choice of fundamental decomposition, but two such norms are equivalent.

If H is a Kreın space, the dimensions of H+ and H− in any fundamentaldecompositionH = H+⊕H− are independent of the choice of fundamental decom-position. These dimensions are called the positive and negative indices and of H.A Pontryagin space is a Kreın space with finite negative index. Two Kreın spacesare isomorphic if and only if they have the same positive and negative indices.

Given a Kreın space H with fundamental decomposition H = H+ ⊕H−,we define operators J , P+, P− on H by

Jf = f1 − f2, P+f = f1, P−f = f2,

4

wheneverf = f1 + f2, f1 ∈ H+, f2 ∈ H−.

We call J the signature operator or fundamental symmetry , and P+ and P− theassociated projections for the given fundamental decomposition. The signatureoperator J serves to identify the fundamental decomposition. We write HJ for Hviewed as a Hilbert space relative to the given fundamental decomposition. Thus,

〈f, g〉HJ= 〈Jf, g〉H

for any vectors f and g in H. The absolute value of H−, |H−|, is a Hilbert spacedefined as the anti-space of H−. In this notation, HJ = H+ ⊕ |H−|.

If H is a Kreın space and J1 and J2 are signature operators for twofundamental decompositions of H, then J2 = U−1J1U where U is an isomorphismof H onto itself.

If H and K are Kreın spaces, B(H) and B(H,K) denote the sets of ev-erywhere defined continuous operators on H to itself and on H to K, respectively.Every A ∈ B(H,K) has a unique adjoint A∗ ∈ B(K,H) satisfying

〈Af, g〉K = 〈f,A∗g〉H, f ∈ H, g ∈ K.

The identity operator is written 1. By viewing H and K as Hilbert spaces relativeto some fundamental decompositions, we may induce norm, weak operator, andstrong operator topologies on B(H,K) which are independent of the choice offundamental decompositions. Any two operator norms obtained in this way areequivalent.

Kreın space adjoints and Hilbert space adjoints must be distinguished.Let H and K be Kreın spaces with fundamental symmetries JH and JK. TheKreın space adjoint of an operator A ∈ B(H,K) is an operator A∗ ∈ B(K,H).The Hilbert space adjoint of A ∈ B(HJH

,KJK) is an operator A× ∈ B(KJK

,HJH)

related to A∗ byA∗ = JHA×JK.

In general, we reserve ∗ for Kreın space adjoints and × for Hilbert space adjoints.Let H and K be Kreın spaces. As in the Hilbert space case, we say that

(i) A ∈ B(H) is selfadjoint if A∗ = A,(ii) A ∈ B(H) is a projection if A is selfadjoint and A2 = A,(iii) A ∈ B(H,K) is isometric if A∗A = 1, and(iv) A ∈ B(H,K) is unitary if both A and A∗ are isometric.

A partial ordering of selfadjoint operators is defined in the usual way: ifA,B ∈ B(H) are selfadjoint, A ≥ 0 means that 〈Af, f〉H ≥ 0 for all f in H, andA ≥ B means that A−B ≥ 0.

Note that the associated projections P± for a fundamental decompositionH = H+ ⊕H− are projections in the sense of the preceding definition.

5

B. Subspaces and Projections

A subspace of a Kreın space H is a nonempty linear set M in H (notnecessarily closed). The Kreın space orthogonal complementM⊥ of M coincideswith the Hilbert space orthogonal complement of JM in HJ for any signatureoperator J for H. For any subspacesM and N of H,

M⊥⊥ =M and (M+N )⊥ =M⊥ ∩ N⊥,

whereM is the closure ofM. If M and N are closed, then also

(M∩N )⊥ = (M⊥ +N⊥).

A subspaceM of H is dense in H if and only ifM⊥ = {0}.In contrast with the Hilbert space case, the relationM+M⊥ = H may

fail for a closed subspaceM of a Kreın space H. Moreover, a closed subspaceM ofa Kreın space H need not itself be a Kreın space in the scalar product of H. Thesepathologies are excluded in an important class of subspaces. By a regular subspace

of a Kreın space H we mean a closed subspaceM of H which is a Kreın space inthe scalar product of H. An analogue of the projection theorem for Hilbert spacesholds for regular subspaces of a Kreın space.

Theorem 1.1.1. If M is a closed subspace of a Kreın space H, thefollowing assertions are equivalent:

(i) M is regular;

(ii) H =M⊕M⊥;

(iii) M is the range of a projection operator P .

In this case, ifM is viewed as a Kreın space in the scalar product of H, the strongtopology of M coincides with the restriction of the strong topology of H to M,and the inclusion ofM in H is continuous.

We occasionally write PrHM or simply PrM for the projection operator ona Kreın space H whose range is the regular subspace M. If P is a projection onH with range M, then 1 − P is a projection with range M⊥. Therefore M⊥ isregular wheneverM is regular. The class of regular subspaces of a Kreın space His not in general closed under intersection and union.

A subspaceM of a Kreın space H is

(i) negative if 〈f, f〉H ≤ 0 for all f in M,

(ii) maximal negative if M is negative and not a proper subset of anothernegative subspace,

6

(iii) uniformly negative if for some (and hence any) fundamental symmetry Jon H, there is a δJ > 0 such that

〈f, f〉H ≤ −δJ‖f‖2HJ

for all f inM, and

(iv) maximal uniformly negative ifM is uniformly negative and not a propersubset of another uniformly negative subspace.

An equivalent form of (iv) is thatM is

(iv′) maximal uniformly negative ifM is maximal negative and uniformly neg-ative.

By reversing the sense of inequalities of scalar products, we obtain paralleldefinitions for a subspace to be positive, maximal positive, uniformly positive, and

maximal uniformly positive. A subspace which is either positive or negative is saidto be definite. Maximal positive and maximal negative subspaces are closed.

Properties of definite subspaces are derived from a graph representation.LetM be a negative subspace of a Kreın space H. Fix a fundamental decomposi-tion H = H+ ⊕H− of H. If h = g + f , g ∈ H+, f ∈ H−, we also write

h =

(

g

f

)

.

In this representation, no nonzero element ofM has the form(

g0

)

, and soM is thegraph

G(K) =

{(

Kf

f

)

: f ∈ domK

}

of a Hilbert space contraction operator K, with domain domK ⊂ |H−| and rangeranK ⊂ H+, which is called the angle operator for M. Every Hilbert spacecontraction operator K with domK ⊂ |H−| and ranK ⊂ H+ is the angle operatorof some negative subspace M of H. The following properties are more or lessimmediate:

(i) M is closed if and only if domK is closed in |H−|;(ii) M is maximal negative if and only if domK = |H−|;(iii) M is uniformly negative if and only if ‖K‖ < 1;

(iv) M is maximal uniformly negative if and only if both domK = |H−| and‖K‖ < 1.

Positive subspaces have a similar graph representation, and parallel results hold.

7

Theorem 1.1.2. Let H be a Kreın space.

(i) A closed subspace M of H is maximal negative if and only if M⊥ ismaximal positive in H.

(ii) A closed subspace M of H is maximal uniformly negative if and only ifM⊥ is maximal uniformly positive in H.

(iii) Every negative subspace ofH is contained in a maximal negative subspaceof H.

(iv) Every uniformly negative subspace of H is contained in a maximal uni-formly negative subspace of H.

Moreover, each of these statements remains true if the words “positive” and “neg-ative” are interchanged.

The condition for a negative subspaceM of a Kreın space H to be maxi-mal negative is frequently used in this form: For some and hence any fundamentaldecomposition H = H+ ⊕H− of H, PrH−

M = H−.A closed positive subspace of a Kreın spaceH is a Hilbert space if and only

if it is uniformly positive. An example is H+ in any fundamental decompositionH = H+⊕H− ofH. This subspace is also maximal uniformly positive. Conversely,if H+ is a maximal uniformly positive subspace of a Kreın space H and H− = H⊥

+,then H = H+ ⊕H− is a fundamental decomposition of H.

C. Isometries and Partial Isometries

The definition of a partial isometry in the Kreın space setting is similarto that for Hilbert spaces. Let A ∈ B(H,K), where H and K are Kreın spaces.We call A a partial isometry if there exist regular subspaces M of H and N ofK such that A maps M isometrically onto N and kerA = M⊥. We call M theinitial space and N the final space of A in this situation. The following propertiesare immediate.

(i) For any f and g in H, the identity

〈Af,Ag〉K = 〈f, g〉H

holds if either f or g is inM.(ii) The adjoint A∗ of the operator A is a partial isometry with initial spaceN and final spaceM.

(iii) If P is the projection of H onM and Q is the projection of K on N , thenA∗A = P , kerA = kerP , and AA∗ = Q, kerA∗ = kerQ.

It is not evident from the definition of a partial isometry that an isometryis a partial isometry. This is true and a consequence of the following nonspatialcharacterization of partial isometries.

8

Theorem 1.1.3. Let A ∈ B(H,K), where H and K are Kreın spaces.Then A is a partial isometry if and only if AA∗A = A.

Proof. Suppose that A is a partial isometry with initial space M andfinal space N . If f is in M⊥ = kerA, then

AA∗Af = 0 = Af.

If f is inM, then A∗Af = f by what was noted above, and hence again AA∗Af =Af . By linearity, AA∗A = A.

Conversely, assume that AA∗A = A, and set P = A∗A. Then P isselfadjoint and idempotent:

P 2 − P = A∗(AA∗A−A) = 0.

Hence P is a projection operator on H. In a similar way, A∗AA∗ = A∗ andQ = AA∗ is a projection operator on K. By Theorem 1.1.1, M = PH andN = QK are regular subspaces of H and K. Now

AM = AA∗AM⊂ AA∗K = QK = N

andA∗N = A∗AA∗N ⊂ A∗AH = PH =M.

Hence AM = N and A|M maps M isometrically onto N . It remains to showthat kerA =M⊥. Let f ∈ H. If Af = 0, then Pf = A∗Af = 0, and so f ∈ M⊥.Conversely, if f ∈M⊥, then A∗Af = Pf = 0 and Af = AA∗Af = 0. Thus A is apartial isometry.

Corollary 1.1.4. Let H and K be Kreın spaces. An isometry A ∈B(H,K) is a partial isometry with initial space H. In particular, the range of anisometry A ∈ B(H,K) is a regular subspace of K.

Proof. If A ∈ B(H,K) is an isometry, then A∗A = 1 and so AA∗A = A.Therefore A is a partial isometry by Theorem 1.1.3. The initial space of A is(A∗A)H = H.

Theorem 1.1.5. Let H and K be Kreın spaces, and let A ∈ B(H,K) bean isometry.

(i) IfM is a closed uniformly positive or closed uniformly negative subspaceof H, AM is of the same type in K.

(ii) If H = H+ ⊕ H− is a fundamental decomposition of H, then AH =AH+ ⊕AH− is a fundamental decomposition of AH.

9

Theorem 1.1.6. Let A ∈ B(H,K), where H and K are Kreın spaces.The following assertions are equivalent:

(i) A is a partial isometry;

(ii) A∗A is a projection operator and kerA∗A = kerA;

(iii) AA∗ is a projection operator and kerAA∗ = kerA∗.

Proofs of these results are not difficult and omitted. An example clarifiesthe condition on kernels in parts (ii) and (iii) of Theorem 1.1.6. Let H and K eachbe the space of pairs

(

ab

)

of complex numbers with

⟨(

ab

)

,

(

ab

)⟩

= |a|2 − |b|2.

If A =(

11

11

)

, then A∗ =(

1−1

−11

)

, and A∗A = 0 is trivially a projection. But A isnot a partial isometry because kerA∗A and kerA do not coincide.

In contrast with the Hilbert space case, densely defined isometries, that is,linear mappings which preserve scalar products, do not necessarily have continuousextensions to everywhere defined isometries. There even exist everywhere definedisometries on Kreın spaces which are not continuous (Bognar [12], p. 125). Asimple condition for continuity, given in the next result, is sometimes useful.

Theorem 1.1.7. Let C be a densely defined isometry from a Hilbertspace H to a Kreın space K. Then C has an extension to an operator C ∈ B(H,K)if and only if ranC is uniformly positive.

Proof. Necessity follows from Theorem 1.1.5. Conversely, if ranC isuniformly positive, then ranC is a Hilbert space in the scalar product of K. Wecan extend C by continuity to a Hilbert space isometry C from H to ranC viewedas a Hilbert space in the scalar product of K. Since the inclusion of ranC in K iscontinuous by Theorem 1.1.1, C is continuous as an operator on H to K.

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D. Orthogonal Direct Sums

Orthogonal direct sums of Kreın spaces are convenient for matrix repre-sentations of operators. If a Kreın space H is represented as an orthogonal directsum H = H1⊕H2, then for any f in H1 and g in H2, the expressions h =

(

fg

)

andh = f + g are used interchangeably. If a second Kreın space K is represented asK = K1⊕K2, then in the usual way any operator A ∈ B(H,K) has a representationin matrix form

A =

(

A11 A12

A21 A22

)

.

This representation extends to orthogonal direct sums with any finite number ofsummands, and the usual rules of matrix calculus are valid. In fact, we can extendthe matrix calculus to orthogonal direct sums with countably many summands. Weshall only indicate the appropriate definitions of external and internal orthogonaldirect sums of sequences, leaving the rest to the reader.

Let H1, H2, . . . be given Kreın spaces with fundamental decompositionsH+

1 ⊕H−1 , H+

2 ⊕H−2 , . . .. Define

H = H1 ⊕H2 ⊕ · · ·to be the space of sequences f = {f1, f2, . . .} such that fn ∈ Hn for all n and

∞∑

n=1

‖fn‖2 <∞,

where norms are computed relative to the given fundamental decompositions. Iff = {f1, f2, . . .} and g = {g1, g2, . . .} are in H, set

〈f, g〉H =

∞∑

n=1

〈fn, gn〉Hn.

Then H is a Kreın space. We call H an external orthogonal direct sum or simplydirect sum of H1, H2, . . .. For each n, Hn has a natural embedding in H as aregular subspace. If P±

n is the projection of H onto H±n for every n, then

supn

∥

∥

∥

∥

∥

n∑

k=1

P+k

∥

∥

∥

∥

∥

<∞ and supn

∥

∥

∥

∥

∥

n∑

k=1

P−k

∥

∥

∥

∥

∥

<∞,

where norms are computed relative to any fundamental decomposition of H. Itis sufficient to prove this for a definite choice of fundamental decomposition of H.Choosing H = H+ ⊕H−, where H± = H±

1 ⊕H±2 ⊕ · · ·, we have, in fact,

∥

∥

∥

∥

∥

n∑

k=1

P±k

∥

∥

∥

∥

∥

≤ 1

11

for all n. It should be noted that the definition of H depends on the choice offundamental decompositions for the summands. However, any two spaces obtainedin this way with different choices of fundamental decompositions for the summandsare naturally isomorphic.

There are several possibilities for the definition of an internal orthogo-nal direct sum. We use a strong hypothesis which results in a notion which isisomorphic with the previous definition of an external orthogonal direct sum.

Theorem 1.1.8. Let H be a Kreın space, and let M1, M2, . . . begiven pairwise orthogonal regular subspaces of H with fundamental decomposi-tions M+

1 ⊕M−1 , M+

2 ⊕M−2 , . . .. Assume that the projections P±

1 , P±2 , . . . on

the subspacesM±1 , M±

2 , . . . satisfy

supn

∥

∥

∥

∥

∥

n∑

k=1

P+k

∥

∥

∥

∥

∥

<∞ and supn

∥

∥

∥

∥

∥

n∑

k=1

P−k

∥

∥

∥

∥

∥

<∞,

where norms are computed relative to any fundamental decomposition of H. Then:

(i) The closed spanM ofM1,M2, . . . is a regular subspace ofH. The closedspanM± of M±

1 , M±2 , . . . is uniformly positive with the plus signs and

uniformly negative with the minus signs, andM =M+ ⊕M−.(ii) Every f ∈ M has a norm convergent representation f = f1 + f2 + · · ·,

where fn ∈ Mn for all n and∑∞

n=1 ‖fn‖2 < ∞ with norms computedrelative to any fundamental decomposition of H. The representation isunique. Conversely, every square summable sequence of elements ofM1,M2, . . . determines an element ofM in this way.

(iii) If P , P±, P±1 , P±

2 , . . . are the projections of H on the subspacesM,M±,M±

1 ,M±2 , . . ., then P =

∑∞n=1 Pn and P± =

∑∞n=1 P±

n with convergencein the strong operator topology.

In this situation, we writeM =M1 ⊕M2 ⊕ · · · and callM the internal

orthogonal direct sum or direct sum ofM1,M2, . . .. An internal orthogonal directsum is defined whenever there exist projections satisfying the hypotheses of thetheorem, but the direct sum itself is independent of the choice of projections. Therole of the hypothesis on projections is to insure that the difference between internaland external orthogonal direct sums is essentially a notational one. Every internalorthogonal direct sum is naturally isomorphic to an external orthogonal directsum, and conversely. In practice, the distinction between internal and externalorthogonal direct sums is typically ignored.

12

Lemma 1.1.9. Let N1, N2, . . . be regular subspaces of a Kreın space Hsuch that N1 ⊃ N2 ⊃ · · ·. Assume that the projections Q1, Q2, . . . of H onto thesubspaces N1, N2, . . . satisfy supn ‖Qn‖ <∞, where norms are computed relativeto any fundamental decomposition of H. Then

N =∞⋂

1

Nn and M =∞∨

1

N⊥n

are regular subspaces of H with N = M⊥. If Q is the projection of H onto N ,then Q = limn→∞ Qn with convergence in the strong operator topology.

Proof of Lemma 1.1.9. For any f ∈ H, Q1f , Q2f , . . . is a boundedsequence, and hence g = limk→∞ Qnk

f exists weakly for some subsequence. El-ementary Hilbert space considerations show that g ∈ N . Write h = f − g, sothat

h = limk→∞

(1−Qnk)f

weakly. Since(1−Qnk

)f ∈ N⊥nk⊂ N⊥

for every k, we have h ∈ N⊥. Thus f = g + h with g ∈ N and h ∈ N⊥. ThereforeN +N⊥ = H. This implies that N ∩ N⊥ = {0}, because any vector in N ∩ N⊥

is orthogonal to N +N⊥ = H. By Theorem 1.1.1, N is a regular subspace of H.In the preceding construction, suppose that f ∈ N⊥. Then g = 0 because

H = N ⊕N⊥. Therefore

f = h = limk→∞

(1−Qnk)f

belongs to M = ∨∞1 N⊥n , and so N⊥ ⊂ M. The reverse inclusion is obvious, and

so N⊥ =M. In particular,M is a regular subspace of H. We clearly have

limn→∞

Qnf = Qf

if f is in the span of N , N⊥1 , N⊥

2 , . . .. Since this span is dense in H and theprojections Q1, Q2, . . . are bounded in norm, the relation holds for all f ∈ H by aroutine approximation.

Proof of Theorem 1.1.8. For each n ≥ 1, let Nn be the orthogonalcomplement of M1 ⊕ · · · ⊕ Mn in H. The projection Qn of H onto Nn isQn = 1− P1 − · · · − Pn. By Lemma 1.1.9,

N =∞⋂

1

Nn and M =∞∨

1

N⊥n

13

are regular subspaces of H with N = M⊥. If Q is the projection of H onto N ,then Q = limn→∞ Qn in the strong operator topology. By the definition of thesubspaces N1, N2, . . .,M is the closed span ofM1,M2, . . .. Moreover, if P is theprojection of H ontoM, then

P = 1−Q = limn→∞

(1−Qn) = limn→∞

n∑

k=1

Pk

with convergence in the strong operator topology.We repeat the construction withM1,M2, . . . replaced byM±

1 ,M±2 , . . ..

Thus the closed span ofM±1 ,M±

2 , . . . is a regular subspaceM± of H. SinceM±

is positive/negative and regular, M± is uniformly positive/uniformly negative. IfP± is the projection of H ontoM±, then

P± = limn→∞

n∑

k=1

P±k

with convergence in the strong operator topology. Clearly P = P+ + P− andP+P− = P−P+ = 0. ThusM =M+ ⊕M−. The assertions (i) and (iii) are nowproved.

The assertion (ii) is independent of the choice of fundamental decomposi-tion of H. It is convenient to choose a fundamental decomposition H = H+ ⊕H−

of H such that H± ⊃M±. Then the subspacesM+1 , M+

2 , . . . andM−1 , M−

2 , . . .are pairwise orthogonal in both the Kreın space sense and the Hilbert space sense.In this situation, (ii) is clear.

1.2 Defect Operators and Julia Operators

As preparation for the study of operators on Kreın spaces, we considertwo related problems. One is to represent a given selfadjoint operator H in theform H = AA∗, where A is an operator with zero kernel. The other is to embedany given operator T in a unitary matrix U =

(

T∗

∗∗

)

. These constructions areneeded to circumvent certain Hilbert space notions which do not have adequatecounterparts in the indefinite case.

Let H be a Kreın space, and let H ∈ B(H) be a selfadjoint operator.By h+(H) we mean the supremum of all r = 1, 2, . . . such that there exists anonnegative and invertible matrix of the form

[

〈Hfj , fk〉H]r

j,k=1, f1, . . ., fr ∈ H.

Set h+(H) = 0 if no such r exists, and h−(H) = h+(−H). We call h±(H) thepositive and negative hermitian indices of H . Notice that H ≥ 0 if and only ifh−(H) = 0, and H ≤ 0 if and only if h+(H) = 0.

14

Theorem 1.2.1. Let A and H be Kreın spaces, A separable, and letA ∈ B(A,H). If A has zero kernel, then the positive and negative indices of Acoincide with h+(AA∗) and h−(AA∗), respectively.

Proof. Let A = A+⊕A− be a fundamental decomposition. It is sufficientto show that h+(AA∗) = dimA+, since we then obtain h−(AA∗) = dimA− bysuitably reversing signs.

The dimension of A+ is the supremum of all r = 1, 2, . . . such that thereexists a nonnegative and invertible matrix of the form

[

〈gj , gk〉A]r

j,k=1, g1, . . ., gr ∈ A,

and zero if no such r exists. If f1, . . ., fr ∈ H,

[

〈AA∗fj , fk〉H]r

j,k=1=

[

〈A∗fj , A∗fk〉A

]r

j,k=1,

and so h+(AA∗) ≤ dimA+.To see that equality holds, consider vectors g1, . . ., gr in A such that

[

〈gj , gk〉A]r

j,k=1

is nonnegative and invertible. Since kerA = {0}, the range of A∗ is dense in A,and there exist vectors f1n, . . ., frn in H such that

limn→∞

A∗fjn = gj , j = 1, . . ., r.

Hence for all sufficiently large n, the matrix

[

〈A∗fjn, A∗fkn〉A]r

j,k=1=

[

〈AA∗fjn, fkn〉H]r

j,k=1

is nonnegative and invertible, and so h+(AA∗) ≥ dimA+. Thus equality holds.

Theorem 1.2.2. Let H be a Kreın space, and let H ∈ B(H) be a selfad-joint operator. Then there is a Kreın space A and an operator A ∈ B(A,H) withzero kernel such that H = AA∗.

Proof. Let JH be the signature operator for some fundamental decom-position H = H+ ⊕ H−. Then HJH is a selfadjoint operator on the Hilbertspace H+ ⊕ |H−|. Let its spectral decomposition be HJH =

∫

λdE(λ), and set

R =∫

|λ|1/2dE(λ), M+ = E((0,∞)), andM− = E((−∞, 0)).

Let A be a Kreın space with fundamental decomposition A = A+ ⊕A−

and signature operator JA such that dimA± = dimM±. Choose an isometry

15

W on A+ ⊕ |A−| to H+ ⊕ |H−| such that WA± = M±. Define A ∈ B(A,H)by A = RW . Then kerA = {0}, and AA∗ = RWW ∗R∗ = (RWJAW×R)JH =(HJH)JH = H .

Definition 1.2.3. Let T ∈ B(H,K), H and K Kreın spaces.

(i) By a defect operator for T we mean an operator D ∈ B(D,H), whereD a Kreın space, such that D has zero kernel and 1− T ∗T = DD∗.

(ii) By a Julia operator for T we mean a unitary operator U having theform

U =

(

T D

D∗ L

)

∈ B(H ⊕D,K ⊕ D),

where D and D are Kreın spaces, D ∈ B(D,K) and D ∈ B(D,H) havezero kernels, and L ∈ B(D, D).

In (i), the positive and negative indices of D are determined by T andgiven by h±(1−T ∗T ) when D is separable by Theorem 1.2.1. In (ii), D is a defectoperator for T and D is a defect operator for T ∗. Hence in (ii), the positive andnegative indices of D and D are determined by T and coincide when these spacesare separable with h±(1 − TT ∗) and h±(1− T ∗T ), respectively. It is also easy tosee that if U is a Julia operator for T , then U∗ is a Julia operator for T ∗.

For an example, let H and K be Kreın spaces, and let T ∈ B(H,K) bea partial isometry with initial space M and final space N . Choose D = N⊥ andD = M⊥ in the scalar products of K and H, respectively. Define D ∈ B(D,K)and D ∈ B(D,H) to be the inclusion mappings, and let L ∈ B(D, D) be the zerooperator. Then

(

TD∗

DL

)

∈ B(H⊕D,K ⊕ D) is a Julia operator for T .Theorem 1.2.2 insures the existence of a defect operator for any given

operator T . As we show next, it also implies the existence of a Julia operator forany given operator T .

Theorem 1.2.4. Let H and K be Kreın spaces, and let T ∈ B(H,K). IfD ∈ B(D,H) is a defect operator for T , there exists a Julia operator of the form

U =

(

T D

D∗ L

)

∈ B(H⊕D,K ⊕ D).

Proof. We seek a Kreın space D and operators D ∈ B(D,K) and L ∈B(D, D) such that D has zero kernel and U =

(

TD∗

DL

)

is unitary, that is,

T ∗T + DD∗ = 1,

T ∗D + DL = 0,

D∗D + L∗L = 1,

(1.2.1a, b, c)

16

and

TT ∗ + DD∗ = 1,

D∗T ∗ + LD∗ = 0,

D∗D + LL∗ = 1.

(1.2.2a, b, c)

The relation (1.2.1a) holds by the assumption that D ∈ B(D,H) is a defect oper-ator for T . It implies that

V =

(

T

D∗

)

∈ B(H,K ⊕ D)

is an isometry. Then 1−V V ∗ is the projection onto kerV ∗. Factor 1−V V ∗ = BB∗,where B ∈ B(D,K ⊕ D) for some Kreın space D and kerB = {0}. Since BB∗ isa projection and kerB∗ = kerBB∗, B is a partial isometry by Theorem 1.1.6. Infact, it is an isometry with range kerV ∗. Thus V ∗B = 0 and B∗B = 1.

If we write elements of K ⊕ D in column form, then

B =

(

D

L

)

,

where D ∈ B(D,K) and L ∈ B(D, D). The relations (1.2.1b,c) follow from V ∗B =0 and B∗B = 1. The relations (1.2.2a,b,c) follow from 1−V V ∗ = BB∗. It remainsto verify that D has zero kernel. If Df = 0 for some vector f in D, then Lf = 0by (1.2.1b) because D has zero kernel by assumption. Hence Bf = 0 and f = 0.

17

1.3 Contraction and Bicontraction Operators

Let H and K be Kreın spaces. An operator T ∈ B(H,K) is said to be acontraction if

〈Tf, Tf〉K ≤ 〈f, f〉H, f ∈ H,

and a bicontraction if both T and T ∗ are contractions. An example of a contractionwhich is not a bicontraction is the embedding of a Hilbert space into its directsum with the anti-space of a nonzero Hilbert space. The structural properties ofcontractions depend on the way in which such operators map negative subspaces.

Theorem 1.3.1. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction. Choose fundamental decompositions for H and K, and let norms becomputed with respect to the associated Hilbert spaces. Set

δ =

{

‖T‖+[

1 + ‖T‖2]1/2

}−1

.

(i) For any f ∈ H with 〈f, f〉H ≤ 0, ‖Tf‖ ≥ δ‖f‖.(ii) The kernel of T is a closed uniformly positive subspace of H.

Proof. Let J be the chosen fundamental symmetry on H. Since T is acontraction, 1 − T ∗T ≥ 0 in the partial ordering of selfadjoint operators on H.Therefore C = J(1− T ∗T ) is nonnegative as an operator on HJ . For any f ∈ H,

‖f‖ − ‖T‖‖Tf‖ ≤ ‖Cf‖ ≤ ‖C‖1/2〈Cf, f〉1/2HJ≤

[

1 + ‖T‖2]1/2

〈Cf, f〉1/2HJ

.

If 〈f, f〉H ≤ 0, then also

〈Cf, f〉HJ= 〈f, f〉H − 〈Tf, Tf〉K ≤ −〈Tf, Tf〉K ≤ ‖Tf‖2.

Combining these inequalities, we obtain

‖f‖ − ‖T‖‖Tf‖ ≤[

1 + ‖T‖2]1/2

‖Tf‖,

which proves (i). If g ∈ kerT , then Cg = J(1− T ∗T )g = Jg and

‖g‖2 = ‖Cg‖2 ≤ ‖C‖〈Cg, g〉HJ= ‖C‖〈g, g〉H.

Thus (ii) follows.

18

Corollary 1.3.2. Let H and K be Kreın spaces, and let T ∈ B(H,K)be a contraction. Then

(i) T maps any closed negative subspace of H in a one-to-one way onto aclosed negative subspace of K, and

(ii) T maps any closed uniformly negative subspace of H in a one-to-one wayonto a closed uniformly negative subspace of K.

Proof. Part (i) follows from Theorem 1.3.1. To prove (ii), we must showthat ifM is a uniformly negative subspace of H, then TM is a uniformly negativesubspace of K.

Assume that fundamental symmetries are chosen for H and K and normsare computed with respect to the associated Hilbert spaces. If M is uniformlynegative in H, there is an η > 0 such that

〈f, f〉H ≤ −η‖f‖2, f ∈M.

Hence for f ∈M,

〈Tf, Tf〉K ≤ 〈f, f〉H ≤ −η‖f‖2 ≤ −η‖T‖−2‖Tf‖2,

and TM is uniformly negative in K. This proves (ii).

The main results on contraction operators are derived using a scatteringformalism. The formalism uses two Kreın spaces H and K and fundamental de-compositions H = H+ ⊕H− and K = K+ ⊕ K−. In addition, let L be the directsum of the anti-space of H together with K. Thus L is the space of pairs

(

fg

)

withf ∈ H, g ∈ K, and

⟨(

fg

)

,

(

fg

)⟩

L

= −〈f, f〉H + 〈g, g〉K.

We use the fundamental decomposition of L given by L = L+ ⊕ L−,

L+ = K+ ⊕H− and L− = H+ ⊕K−.

The elements of L are also represented as pairs(

vu

)

with v in L+ and u in L−.If T ∈ B(H,K) is a contraction, its graph

G(T ) =

{(

f

Tf

)

: f ∈ H}

is a closed negative subspace of L. It is negative because for any f ∈ H,⟨(

f

Tf

)

,

(

f

Tf

)⟩

L

= −〈f, f〉H + 〈Tf, Tf〉K ≤ 0.

19

It is closed because the domain of T is closed. By the scattering operator orPotapov-Ginsburg transform of T we mean the angle operator S for G(T ) relativeto the fundamental decomposition L = L+ ⊕ L−. Thus

G(T ) =

{(

Suu

)

: u ∈ domS

}

⊂ L+ ⊕ L−.

By construction, the scattering operator S is a Hilbert space contraction withclosed domain domS ⊂ |L−| and ranS ⊂ L+.

Write T in matrix form

T =

(

T11 T12

T21 T22

)

∈ B(H+ ⊕H−,K+ ⊕K−). (1.3.1)

Denote the signature operators and projections for the given fundamental decom-positions of H and K by JH, JK and

P± : H → H±, Q± : K → K±. (1.3.2)

Define operators

Q+T + P− =

(

T11 T12

0 1

)

∈ B(H+ ⊕H−,K+ ⊕H−), (1.3.3)

and

P+ + Q−T =

(

1 0

T21 T22

)

∈ B(H+ ⊕H−,H+ ⊕K−). (1.3.4)

The preceding notation is assumed in Theorems 1.3.3–1.3.5.

Theorem 1.3.3. Let T ∈ B(H,K) be a contraction with scattering op-erator S. The domain of S is equal to ran (P+ + Q−T ), and

S = (Q+T + P−)(P+ + Q−T )−1|domS.

Proof. We show that P+ +Q−T is one-to-one and has closed range. Firstnote that T22 is contraction on H− to K−. For if f ∈ H−, then

〈T22f, T22f〉K−= 〈Q−Tf,Q−Tf〉K ≤ 〈Tf, Tf〉K ≤ 〈f, f〉H = 〈f, f〉H−

.

By Theorem 1.3.1, T22 is one-to-one and has closed range. By (1.3.4), P+ + Q−Tis one-to-one and has closed range.

20

For any f ∈ H,

PrL−

(

f

Tf

)

=

(

P+f

Q−T (P+ + P−)f

)

=

(

1 0

T21 T22

)(

P+f

P−f

)

= (P+ + Q−T )f

and

PrL+

(

f

Tf

)

=

(

Q+T (P+ + P−)f

P−f

)

=

(

T11 T12

0 1

) (

P+f

P−f

)

= (Q+T + P−)f.

The computation of S is immediate from these identities and the definition of theangle operator.

The scattering operator S models a system in which a left input f1 ispartially transmitted in f2, partially reflected in g1, and partially absorbed. Aright input g2 is partially transmitted in g1, partially reflected in f2, and partiallyabsorbed.

f1−−−−−−−−−−−−−−−→ −−−−−−−−−−−−−−−→f2

g1←−−−−−−−−−−−−−−− ←−−−−−−−−−−−−−−−g2

Here f1, f2, g1, g2 belong to H+, K+, |H−|, |K−|, respectively. The matrix entriesin

S =

(

Tℓ Rr

Rℓ Tr

)

,

represent left and right transmission and reflection coefficients. The scatteringoperator

S :

(

f1

g2

)

→(

f2

g1

)

associates net output to net input. The original operator

T :

(

f1

g1

)

→(

f2

g2

)

is a cascade scattering operator for the system. Passivity of the system means thatenergy is not increased, or, what is the same thing, S is contractive in the Hilbertspace sense:

‖f2‖2 + ‖g1‖2 ≤ ‖f1‖2 + ‖g2‖2.

21

This is equivalent to the inequality

‖f2‖2 − ‖g2‖2 ≤ ‖f1‖2 − ‖g1‖2,

which holds because T is a contraction. When domS 6= |L−|, only certain inputsand outputs occur in the system.

The scattering formalism is particularly simple when domS = |L−|. Inthis case, from Theorem 1.3.3 and the identities (1.3.3) and (1.3.4), we see thatT22 is invertible and

S =

(

T11 − T12T−122 T21 T12T

−122

−T−122 T21 T−1

22

)

∈ B(H+ ⊕ |K−|,K+ ⊕ |H−|). (1.3.5)

Identities involving the scattering operator can therefore be reduced to straight-forward matrix calculations.

Theorem 1.3.4. Let T ∈ B(H,K) be a contraction with scattering op-erator S. The following assertions are equivalent:

(i) T is a bicontraction;(ii) the entry T22 in (1.3.1) is invertible;(iii) domS = |L−|.

If T is a bicontraction, then

T = (Q+S + Q−)(P+ + P−S)−1, (1.3.6)

and the scattering operator for T ∗ is S×. Moreover, in this case,

T ∗ = (P+S× + P−)(Q+ + Q−S×)−1 (1.3.7)

and

1− T ∗T = JH(P+ + Q−T )×(1− S×S)(P+ + Q−T ), (1.3.8)

1− TT ∗ = JK(Q+ + P−T ∗)×(1− SS×)(Q+ + P−T ∗). (1.3.9)

Recall that ∗ indicates Kreın space adjoint and × Hilbert space adjoint.Notation in (1.3.6)–(1.3.9) is similar to that used in (1.3.3) and (1.3.4). Explicitly,if

S =

(

S11 S12

S21 S22

)

,

22

then

Q+S + Q− =

(

S11 S12

0 1

)

∈ B(H+ ⊕K−,K+ ⊕K−),

P+ + P−S =

(

1 0

S21 S22

)

∈ B(H+ ⊕K−,H+ ⊕H−),

and,

P+S× + P− =

(

S×11 S×

21

0 1

)

∈ B(K+ ⊕H−,H+ ⊕H−),

Q+ + Q−S× =

(

1 0

S×12 S×

22

)

∈ B(K+ ⊕H−,K+ ⊕K−).

Proof of Theorem 1.3.4. (i) ⇒ (ii) Assume that T is a bicontraction. Asin the proof of Theorem 1.3.3, T22 = Q−TP−|H− is one-to-one and has closedrange as an operator on H− to K−. Suppose that f is in K− and f⊥Q−TP−H−.Then f⊥TH− and T ∗f⊥H−. So T ∗f ∈ H+, and since T is bicontractive,

〈T ∗f, T ∗f〉H ≥ 0 ≥ 〈f, f〉K ≥ 〈T ∗f, T ∗f〉H.

Therefore equality holds throughout, and since f ∈ K− we obtain f = 0. HenceranT22 = K−, and T22 is invertible.

(ii)⇒ (iii) If T22 is invertible, so is the operator (1.3.4). Then by Theorem1.3.3, domS = ran (P+ + Q−T ) = |L−|.

(iii) ⇒ (i). If domS = |L−|, the scattering operator S is given by (1.3.5).Multiplication of operator matrices shows that

S× = (P+T ∗ + Q−)(Q+ + P−T ∗)−1, (1.3.10)

where

P+T ∗ + Q− =

(

T×11 −T×

21

0 1

)

∈ B(K+ ⊕ |K−|,H+ ⊕ |K−|),

Q+ + P−T ∗ =

(

1 0

−T×12 T×

22

)

∈ B(K+ ⊕ |K−|,K+ ⊕ |H−|).

Since S× is a contraction, for all g ∈ K,

‖(P+T ∗ + Q−)g‖2H+⊕|K−| ≤ ‖(Q+ + P−T ∗)g‖2K+⊕|H−|.

Therefore‖P+T ∗g‖2 + ‖Q+g‖2 ≤ ‖Q+g‖2 + ‖P−T ∗g‖2

and‖P+T ∗g‖2 − ‖P−T ∗g‖2 ≤ ‖Q+g‖2 − ‖Q−g‖2,

23

where the norms are computed in H+, |H−|, K+, |K−|. It follows that T ∗ iscontractive, and hence T is bicontractive.

Now assume that (i) - (iii) hold. Then (1.3.10) says that the scatteringoperator for T ∗ is S×, and (1.3.6) and (1.3.7) follow by multiplying matrices.

The identities (1.3.8) and (1.3.9) are equivalent because the scatteringoperator for T ∗ is S×. To prove (1.3.8), calculate using the identity for S inTheorem 1.3.3:

(P++Q−T )×(1− S×S)(P+ + Q−T )

= (P+ + Q−T )×(P+ + Q−T )− (Q+T + P−)×(Q+T + P−)

=

(

1 T×21

0 T×22

) (

1 0

T21 T22

)

−(

T×11 0

T×12 1

) (

T11 T12

0 1

)

.

By multiplication of matrices, this is the same as

(

1 00 −1

)

−(

T×11 T×

21

T×12 T×

22

)(

T11 T12

−T21 −T22

)

= JH(1− T ∗T ),

which yields the result.

The next result characterizes the Hilbert space contractions S which occuras scattering operators of bicontractions.

Theorem 1.3.5. Let H and K be Kreın spaces with fundamental de-compositions H = H+⊕H− and K = K+⊕K−. Let S ∈ B(H+⊕|K−|,K+⊕|H−|)be a contraction with matrix

S =

(

S11 S12

S21 S22

)

.

Then S is the scattering operator for a bicontraction T ∈ B(H,K) if and only ifS22 is invertible.

Proof. Necessity follows from Theorem 1.3.4 and (1.3.5). Conversely,suppose that S22 is invertible. Define T ∈ B(H,K) by

T =

(

S11 − S12S−122 S21 S12S

−122

−S−122 S21 S−1

22

)

.

Multiplication of matrices verifies the identity

S = (Q+T + P−)(P+ + Q−T )−1.

24

Since S is a contraction, for any f ∈ H,

‖(Q+T + P−)f‖2K+⊕|H−| ≤ ‖(P+ + Q−T )f‖2H+⊕|K−|,

equivalently,‖Q+Tf‖2 + ‖P−f‖2 ≤ ‖P+f‖2 + ‖Q−Tf‖2,‖Q+Tf‖2 − ‖Q−Tf‖2 ≤ ‖P+f‖2 − ‖P−f‖2,

which means that T ∈ B(H,K) is a contraction. Since T22 = S−122 is invertible, T

is a bicontraction by Theorem 1.3.4. By construction, S is the scattering operatorfor T .

Theorem 1.3.1 describes how a contraction maps negative subspaces. Bi-contractions are characterized by how they map maximal negative subspaces.

Theorem 1.3.6. Let H and K be Kreın spaces. If T ∈ B(H,K) is acontraction, the following assertions are equivalent:

(i) T is a bicontraction;(ii) αT ∗ is a contraction for some positive number α;(iii) T maps some maximal negative subspace of H onto a maximal negative

subspace of K;(iv) T maps every maximal negative subspace of H onto a maximal negative

subspace of K.

In this case, T maps any maximal uniformly negative subspace ofH in a one-to-oneway onto a maximal uniformly negative subspace of K.

Proof. Let H = H+ ⊕H− and K = K+ ⊕K− be fundamental decompo-sitions of the spaces H and K.

(i) ⇒ (ii) If (i) holds, (ii) holds with α = 1.(ii) ⇒ (iv) Assume (ii), and let M be a maximal negative subspace of

H. By Corollary 1.3.2, TM is a closed negative subspace of K. By the graphrepresentation of negative subspaces, TM is maximal negative if and only if thereis no nonzero vector in K− which is orthogonal to TM. If f is in K− and orthogonalto TM, then for g in M,

〈αT ∗f, g〉H = 〈f, αTg〉K = 0.

It follows that αT ∗f is in M⊥, which is maximal positive since M is maximalnegative. Therefore,

〈αT ∗f, αT ∗f〉H ≥ 0 ≥ 〈f, f〉K.

Since αT ∗ is contractive by assumption and f is in K−, 〈f, f〉K = 0 and f = 0.Hence TM is maximal negative.

25

(iv) ⇒ (iii) This is trivial.(iii) ⇒ (i) Assume that there is a maximal negative subspace M of H

such that TM is maximal negative in K. We show that the entry T22 in (1.3.1) isinvertible, and hence T is a bicontraction by Theorem 1.3.4.

RepresentM as the graph G(K) of a contraction operator K on |H−| toH+. Then sK is a contraction and its graph G(sK) is a maximal negative subspaceof H for 0 ≤ s ≤ 1. Since T is contractive, so is Q−T where Q− is the projectionof K onto K−. Choose δ > 0 for Q−T as in Theorem 1.3.1. For all f in H−,

‖(T21sK + T22)f‖2|K−| = ‖Q−T (sKf + f)‖2|K| ≥ δ‖sKf + f‖2|H| ≥ δ‖f‖2|H−|

uniformly for 0 ≤ s ≤ 1. The range of T21K + T22 is all of K− because T mapsM = G(K) onto a maximal negative subspace of K by assumption. ThereforeT21K + T22 is an invertible operator on K−. By a Neumann series argument,T21sK + T22 is invertible for 0 ≤ s ≤ 1. In particular, T22 is invertible.

We have proved the equivalence of the statements (i)–(iv). The last as-sertion of the theorem follows from Corollary 1.3.2.

Theorem 1.3.7. If H is a Pontryagin space, every contraction operatorT from H into itself is bicontractive.

Proof. In any fundamental decomposition H = H+ ⊕H−, H− has finitedimension. If P− is the projection onto H−, then P−T is a contraction because Tis a contraction. By Corollary 1.3.2, T22 = P−TP−|H− is a one-to-one mapping ofH− into itself. Hence T22 is invertible, and T is bicontractive by Theorem 1.3.6.

Examples 1.3.8. (i) Let H and K be Kreın spaces. An isometry V ∈B(H,K) is a bicontraction if and only if kerV ∗ is a uniformly positive subspace ofK. Necessity of the condition follows from Theorem 1.3.1. For the other direction,note that V is a partial isometry with initial space H and final space VH byCorollary 1.1.4. Since K = VH⊕kerV ∗ with kerV ∗ uniformly positive, sufficiencyof the condition follows from elementary properties of partial isometries.

(ii) Theorem 1.3.7 fails for Kreın spaces. Consider the Kreın space H ofsquare summable sequences a = (a0, a1, a2, . . .) with

〈a, a〉H = |a0|2 − |a1|2 − |a2|2 − · · ·.

The operator T ∈ B(H) defined by

T : (a0, a1, a2, . . .)→ (0, a0, a1, . . .)

is contractive, but its adjoint

T ∗ : (a0, a1, a2, . . .)→ (−a1, a2, a3, . . .)

26

is not contractive. The range of T is a maximal negative subspace of H which ismapped by T onto a proper subspace of itself and hence onto a negative subspacewhich is not maximal.

(iii) In connection with Theorem 1.3.6, we note a counter-example to arelated statement in Ando ([4], p. 31, Corollary 3.3.2). In our language, the state-ment is that an operator T ∈ B(H) is a scalar multiple of a bicontraction if it mapsmaximal negative subspaces onto maximal negative subspaces. A counterexampleis T = V ∗ ∈ B(H), where H is the anti-space of a Hilbert space and V is anisometry whose range is not all of H. The only maximal negative subspace of H isH, and this mapped by T onto itself. No constant multiple of T is a contraction,since the kernel of T includes a vector f with 〈f, f〉H < 0.

Bicontractions are also characterized in terms of mappings of operatorspheres.

Theorem 1.3.9. Let H and K be Kreın spaces, and let T ∈ B(H,K)be a contraction with matrix (1.3.1) relative to some fundamental decompositionsH = H+⊕H− and K = K+⊕K−. Then T is a bicontraction if and only if wheneverX ∈ B(|H−|,H+) is a contraction, then T21X + T22 is an invertible operator on|H−| to |K−| and

Y = (T11X + T12)(T21X + T22)−1 (1.3.11)

is a contraction in B(|K−|,K+).

Proof. If T is a bicontraction and X ∈ B(|H−|,H+) is a contraction,then

G(X) =

{(

Xf

f

)

: f ∈ H−

}

is a maximal negative subspace of H and

TG(X) =

{(

T11Xf + T12f

T21Xf + T22f

)

: f ∈ H−

}

is a maximal negative subspace of K by Theorem 1.3.6. Let Q− be the projectionof K onto K−. Then Q−T is contractive, and T21X+T22 is one-to-one by Corollary1.3.2. Since Q−TG(X) = K−, T21X + T22 is invertible. Therefore (1.3.11) definesan operator Y ∈ B(|K−|,K+). Since TG(X) is a negative subspace of K,

‖(T11X + T12)f‖2K+≤ ‖(T21X + T22)f‖2|K−|, f ∈ H−,

and Y is a contraction.In the other direction, if the condition holds for X = 0, then T22 is

invertible and T is a bicontraction by Theorem 1.3.4.

27

In the next result, the domain and range spaces H and K coincide.

Theorem 1.3.10. Let H be a Kreın space with fundamental decompo-sition H = H+ ⊕H−, and let T ∈ B(H) be a bicontraction. Define a mapping Φof the set C of all contractions X ∈ B(|H−|,H+) into itself by

Φ(X) = (T11X + T12)(T21X + T22)−1, X ∈ C.

Then for X ∈ C, we have Φ(X) = X if and only if the graph

M =

{(

Xf

f

)

: f ∈ H−

}

of X is invariant under T .

Proof. By its form, M is a maximal negative subspace of H. If M isinvariant under T , then TM =M by Theorem 1.3.6. Since

TM =

{(

T11Xf + T12f

T21Xf + T22f

)

: f ∈ H−

}

,

we have T11X + T12 = X(T21X + T22), that is, Φ(X) = X. The other direction isobtained by reversing these steps.

Existence theorems for definite invariant subspaces follow with the aidof Theorem 1.3.10. A full treatment is beyond the scope of our discussion. Buthaving come this far, we show the connection by giving one of the early results ofthe subject which is due to Kreın (Theorem 1.3.11). See the notes at the end of thechapter for literature references. Note that the set C in Theorem 1.3.10 is convexand compact in the weak operator topology of B(|H−|,H+). We also recall theSchauder-Tychonoff Theorem: Every continuous mapping of a convex and

compact subset of a locally convex linear topological vector space has a fixed point.The Schauder-Tychonoff theorem is proved, for example, in Dunford and Schwarz([31], p. 456). It is applicable in the situation of Theorem 1.3.10 whenever Φ iscontinuous. This is automatic in the case of a Pontryagin space.

Theorem 1.3.11. Let H be a Pontryagin space. If T ∈ B(H) is acontraction (and hence a bicontraction), there exists a maximal negative subspaceM of H which is mapped by T onto itself.

Proof. It is sufficient to show that the mapping Φ in Theorem 1.3.10 iscontinuous.

28

Let {Xα}α∈D be a generalized sequence in C which converges in the weakoperator topology to X ∈ C. Then

limα

(T11Xα + T12) = T11X + T12

in the weak operator topology of B(|H−|,K+).Let P− be the projection of H onto H−. Choose δ > 0 for the contraction

P−T ∈ B(H) as in Theorem 1.3.1. Then

‖T21Xαf + T22f‖ ≥ δ‖f‖, α ∈ D, f ∈ H−.

Therefore∥

∥(T21Xα + T22)−1

∥

∥ ≤ 1/δ for all α ∈ D. Now

limα

(T21Xα + T22) = T21X + T22

in the weak operator topology of B(|H−|) and hence also in the norm topologybecause H− is finite dimensional. By Theorem 1.3.9, T21X +T22 is invertible, and

limα

(T21Xα + T22)−1 = (T21X + T22)

−1

in the operator norm of B(|H−|). Thus

limα

Φ(Xα) = Φ(X)

in the weak operator topology of B(|H−|,H+), and so Φ is continuous.

1.4 Additional Results on Contractions and Bicontractions

The results here catalog properties of defect operators and Julia operatorsof contractions. We also give a useful theorem on the existence of a bicontractiveextension of a densely defined contraction.

Theorem 1.4.1. Let T ∈ B(H,K), H and K Kreın spaces, and let

U =

(

T D

D∗ L

)

∈ B(H ⊕D,K ⊕ D)

be a Julia operator for T . The adjoint of the operator (T D) ∈ B(H ⊕D,K) isan isometry. The following assertions are equivalent:

(i) T is a contraction;(ii) D is a Hilbert space;(iii) (T D) is a contraction;(iv) (T D) is a bicontraction;(v) the kernel of (T D) is uniformly positive in H⊕D.

29

Proof. In the proof, we write R = (T D) and make repeated use ofthe relations (1.2.1) and (1.2.2). The operator R∗ is an isometry because RR∗ =TT ∗ + DD∗ = 1 by the definition of a Julia operator.

(i) ⇒ (ii) If T is a contraction, then DD∗ = 1 − T ∗T ≥ 0. Since D haszero kernel, the range of D∗ is dense in D. The inequality DD∗ ≥ 0 then impliesthat 〈g, g〉D ≥ 0 for all g ∈ D, and therefore D is a Hilbert space.

(ii) ⇒ (iii) If D is a Hilbert space, then

1−R∗R =

(

1− T ∗T −T ∗D

−D∗T 1−D∗D

)

=

(

DD∗ DL

L∗D∗ L∗L

)

=

(

DL∗

)

( D∗ L ) ≥ 0,

because (D∗ L) ∈ B(H ⊕D, D). Thus R is a contraction.(iii)⇒ (iv) Since R∗ is an isometry, if R is a contraction it is automatically

a bicontraction.(iv) ⇒ (v) This follows from Theorem 1.3.1.(v) ⇒ (i) Since R∗ is isometric, this follows from Example 1.3.8 (i).

Corollary 1.4.2. Let T ∈ B(H,K), H and K Kreın spaces, and letD ∈ B(D,K) be a defect operator for T ∗. Then the operator (T D) has anisometric adjoint, and the following assertions are equivalent:

(i) T is a contraction;(ii) (T D) is a contraction;(iii) (T D) is a bicontraction;(iv) the kernel of (T D) is uniformly positive in H⊕D.

If these equivalent conditions are satisfied, then for any defect operator D ∈B(D,H) of T , D is a Hilbert space.

Proof. By Theorem 1.2.4, this is a consequence of Theorem 1.4.1.

Corollary 1.4.3. Let T ∈ B(H,K), H and K Kreın spaces, and let

U =

(

T D

D∗ L

)

∈ B(H ⊕D,K ⊕ D)

be a Julia operator for T . If T is a bicontraction, then D and D are Hilbert spaces,and both

( T D ) ∈ B(H⊕D,K) and

(

TD∗

)

∈ B(H,K ⊕ D)

are bicontractions with isometric adjoints.

Proof. Since T ∗ has Julia operator

U∗ =

(

T ∗ D

D∗ L∗

)

∈ B(K ⊕ D,H⊕D),

30

the result follows by applying Theorem 3.1.1 to both T and T ∗.

In Hilbert spaces, densely defined contractions extend automatically bycontinuity to everywhere defined contractions. This is not true in Kreın spaces,but with an extra condition the existence of a continuous bicontractive extensionis assured. The result can be used, for example, to construct continuous isometricoperators from densely defined isometries.

Theorem 1.4.4. Let H and K be Kreın spaces, and let T0 be a denselydefined linear mapping from H to K. Assume that

〈T0f, T0f〉K ≤ 〈f, f〉H, f ∈ domT0.

Assume also that domT0 contains a maximal uniformly negative subspace M ofH and that T0M is a maximal uniformly negative subspace of K. Then T0 has anextension by continuity to a bicontractive operator T ∈ B(H,K).

Lemma 1.4.5. Let H be a Kreın space with fundamental decompositionH = H+ ⊕H−. Set α =

√2− 1. If f ∈ H−, ‖f‖ ≥ 1, and g ∈ H, ‖g‖ < α, then

〈f + g, f + g〉 < 0.

The norms in the lemma are computed with respect to the given funda-mental decomposition of H.

Proof of Lemma 1.4.5. Fix f ∈ H−, ‖f‖ ≥ 1, and g ∈ H, ‖g‖ < α. Then

〈f + g, f + g〉 = −‖f‖2 + 2Re〈f, g〉 + 〈g, g〉< −‖f‖2 + 2α‖f‖+ α2.

The function φ(x) = −x2 +2αx+α2 attains its maximum at x = α, and thereforeφ(x) ≤ φ(1) for x ≥ 1. Thus 〈f + g, f + g〉 < −1 + 2α + α2 = 0.

Proof of Theorem 1.4.4. Choose fundamental decompositions H = H+ ⊕H− and K = K+ ⊕K− such that

H+ =M⊥, H− =M,

andK+ = (T0M)⊥, K− = T0M.

The conclusion is immediate if H is a Hilbert space, and we exclude this case inwhat follows. Let S be the restriction of T0 to H−, viewed as a linear mapping ofH− onto K−. For f ∈ H−,

‖Sf‖2 = −〈T0f, T0f〉K ≥ −〈f, f〉H = ‖f‖2.

31

Therefore S has a continuous inverse, and so S ∈ B(H−,K−) by the open mappingtheorem.

Set α =√

2− 1. We show that for g ∈ domT0, ‖g‖ < α,

∥

∥PrK−T0g

∥

∥ < ‖S‖.

Argue by contradiction, assuming that the inequality is not true. Choose h ∈ H−

such that Sh = PrK−T0g. Then ‖S‖ ≤

∥

∥PrK−T0g

∥

∥ = ‖Sh‖ ≤ ‖S‖‖h‖, and so‖h‖ ≥ 1. By Lemma 1.4.5,

0 > 〈h− g, h− g〉H ≥ 〈T0(h− g), T0(h− g)〉K,

a contradiction, because T0(h− g) has zero projection in K− and hence is in K+.We show that for g ∈ domT0, ‖g‖ < α,

∥

∥PrK+T0g

∥

∥ < 2‖S‖.

Since H is not a Hilbert space, we can choose h ∈ H− with ‖h‖ = 1. By Lemma1.4.5,

0 > 〈h− g, h− g〉H ≥ 〈T0(h− g), T0(h− g)〉K=

∥

∥PrK+T0(h− g)

∥

∥

2 −∥

∥PrK−T0(h− g)

∥

∥

2

=∥

∥PrK+T0g

∥

∥

2 −∥

∥PrK−T0(h− g)

∥

∥

2.

Therefore

∥

∥PrK+T0g

∥

∥

2<

∥

∥PrK−T0(h− g)

∥

∥

2 ≤[

∥

∥PrK−T0h

∥

∥ +∥

∥PrK−T0g

∥

∥

]2

< 4‖S‖2,

which proves the assertion.By what we have shown, for any g ∈ domT0, ‖g‖ < α,

‖T0g‖2 =∥

∥PrK+T0g

∥

∥

2+

∥

∥PrK−T0g

∥

∥

2< 5‖S‖2.

It follows that T0 has an extension to an operator T ∈ B(H,K) with

‖T‖ <

√5√

2− 1‖S‖.

Clearly T is a contraction. Since T maps a maximal negative subspace of H ontoa maximal negative subspace of K, T is a bicontraction.

32

Notes on Chapter 1

Accounts of Kreın spaces and operators on them are given in Ando [4], Azizovand Iokhvidov [10], Bognar [12], and Kreın [43]. The book by Iokhvidov, Kreın, andLanger [40] is another useful source: it emphasizes Pontryagin spaces, but the results andmethods often apply to general Kreın spaces. Kreın [43] is an authoritative source forthe early history of the subject and applications that motivated its development. Azizovand Iokhvidov [9] survey literature in the years 1953–1978. For the finite dimensionalcase, see Gohberg, Lancaster, and Rodman [38] and Potapov [55].

Additional information on isometries and direct sums may be found in Bognar[12], Gheondea [36], and McEnnis [52,53]. In particular, Theorem 1.1.8 on internalorthogonal direct sums is not as strong as possible. Gheondea [36] gives necessary andsufficient conditions for the closed span of pairwise orthogonal regular subspaces to beregular. This condition is weaker than the uniform boundedness condition we give onthe associated projections, which he shows to be equivalent to both convergence in thestrong and the weak operator topologies of the sums given in 1.1.8(iii). McEnnis [53]gives necessary and sufficient conditions for the square summability of the representationof elements of an internal direct sum given in 1.1.8(ii).

Factorizations of selfadjoint operators in the form H = AA∗, discussed in §1.2,appear in many places, including Bognar [12]. The existence of a Julia operator in theKreın space setting was first shown by Arsene, Constantinescu, and Gheondea [8] by adifferent method (see Appendix B). This is an important result and plays a key role inwhat follows. The abstract definition of a Julia operator and existence proof in Theorem1.2.4 seem to be new; uniqueness is discussed in Chapter 2 and Appendix B. For theHilbert space case as well as references to the original papers of Julia and Halmos, seeSz.-Nagy and Foias [66].

The main results of §1.3 are due to Ginsburg [37] and Kreın and Shmul’yan[45,47]. In the finite dimensional case, some ideas go back to Potapov [55]. Additionalresults are given in Kreın and Shmul’yan [46].

Theorem 1.3.11 is the simplest of its kind. The method is due to Kreın [42],but the problem was known previously. The history of the problem is given in the surveyof Azizov and Iokhvidov [9]. See their monograph [10] for a recent account. Additionalinformation may be found in Iokhvidov, Kreın, and Langer [40]. Other sources are Kreın[43], Ando [4], and Bognar [12]. Best possible conditions for the existence of definiteinvariant subspaces are not known.

Literature citations for the Potapov-Ginsburg transform are given in Azizovand Iokhvidov [9]. Our treatment is adapted from Iokhvidov, Kreın, and Langer [40].The transform is used in other situations, such as Arov [5], Dym [32], and Kreın andLanger [44]. The scattering interpretation appears in Alpay [3] and Dym [32].

Theorem 1.4.4 is due to Shmul’yan [64]. A similar result is proved by Yan [67].

33

Chapter 2: Matrix Extensionsof Contraction Operators

2.1 The Adjoint of a Contraction

The theory of Ginsburg, Kreın, and Shmul’yan discussed in Chapter 1 isbased on an analysis of how a contraction operator maps negative subspaces. In thecase of bicontractions, the adjoint of a given contraction is again a contraction andsubject to the same analysis. Contraction operators which are not bicontractionsare also of interest. The main point of this section is that one can decomposethe domain space of the adjoint of any Kreın space contraction in such a way asto discard a uniformly negative subspace, and then the restriction of the adjointto what remains is a bicontraction. In this way we are able to apply the strongresults on bicontractions to arbitrary contractions. In particular, we obtain a newcharacterization of bicontractions in terms of adjoint operators (Theorem 2.1.5).

To begin, we assume that bicontractive restrictions of the adjoint of a con-traction operator exist, and we determine some consequences of this fact. Chiefly,there exist maximal uniformly negative subspaces of the range space with specialproperties. Then we shall reverse the process.

Theorem 2.1.1. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction operator. Let K be a regular subspace of K such that K⊥ is uniformlynegative and T ∗|K is a bicontraction on K to H. Let L− be a uniformly negativesubspace of K which is maximal in K, and let K− = L− ⊕ K⊥. Define

L+ = {v : v ∈ K− and T ∗v⊥T ∗L−} .

Then K− is a maximal uniformly negative subspace of K, and K− = L−+L+.

Recall that + is the symbol for a direct sum which is not necessarilyorthogonal.

Proof. Since K⊥ is uniformly negative, K− = L− ⊕ K⊥ is closed anduniformly negative. Since L− is a maximal uniformly negative subspace in K, thesubspace defined by K+ = K ⊖ L− is closed and uniformly positive. Now

K = K ⊕ K⊥ = (K+ ⊕ L−)⊕ K⊥ = K+ ⊕K−,

and so K− is maximal uniformly negative in K.Since T ∗|K is a bicontraction and L− is maximal uniformly negative in K,

T ∗L− is a maximal uniformly negative subspace of H by Theorem 1.3.6. Choose

34

a fundamental decomposition H = H+ ⊕ H− by setting H− = T ∗L− and H+ =(T ∗L−)⊥.

Clearly K− ⊃ L− + L+. Let k ∈ K−. Write T ∗k = h− + h+, h± ∈ H±.Then h− = T ∗u for some u ∈ L−, and so h+ = T ∗v, where v = k − u ∈ K− andT ∗v⊥T ∗L−. Thus k = u+v where u ∈ L− and v ∈ L+. Therefore K− = L− +L+.

If f ∈ L−∩L+, then T ∗f ∈ T ∗L−∩T ∗L+ ⊂ H−∩H+ and T ∗f = 0. SinceT ∗|L− is one-to-one by Corollary 1.3.2, f = 0. Therefore the sum is direct.

Notice that in the situation of Theorem 2.1.1, T ∗|L− is one-to-one, itsrange is a maximal uniformly negative subspace of H, and T ∗L+⊥T ∗L−. We showthat for any contraction T ∈ B(H,K), H and K Kreın spaces, there exist subspacesof K having these properties.

Theorem 2.1.2. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction. Let D ∈ B(D,K) be a defect operator for T ∗, and assume that K−,D− are maximal uniformly negative subspaces of K, D. Then there exist closedsubspaces L− and L+ of K− such that

K− = L−+L+,

and T ∗ maps L− in a one-to-one way onto a maximal uniformly negative subspaceof H, T ∗L+⊥T ∗L−, and D∗L−⊥D−.

Lemma 2.1.3. Let T ∈ B(H,K) be a bicontraction, H and K Kreınspaces. Let H be a regular subspace of H such that H⊥ is uniformly positive.Then T = T |H is a bicontraction on H to K.

Proof of Lemma 2.1.3. Since T is a contraction and T is a restriction ofT , T is a contraction. For any g ∈ K,

⟨

T ∗g, T ∗g⟩

H=

⟨

PrH T ∗g,PrH T ∗g⟩

H

≤⟨

PrH T ∗g,PrH T ∗g⟩

H+

⟨

PrH⊖H T ∗g,PrH⊖H T ∗g⟩

H

= 〈T ∗g, T ∗g〉H≤ 〈g, g〉K

because H⊖ H = H⊥ is uniformly positive.

Proof of Theorem 2.1.2. Define D− = D|D− ∈ B(D−,K). The operator

R = (T D) ∈ B(H⊕D,K)

35

is a bicontraction by Corollary 1.4.2, and

R− = (T D− ) ∈ B(H ⊕D−,K)

is a bicontraction by Lemma 2.1.3. Hence R∗−K− is a maximal uniformly negative

subspace of H⊕D− by Theorem 1.3.6.Let H = H+⊕H− be a fundamental decomposition, and write the graph

representation of R∗−K− as

R∗−K− =

{(

Kff

)

: f ∈ H− ⊕D−

}

,

where K ∈ B(|H− ⊕D−|,H+) and ‖K‖ < 1. Then

H− =

{(

Kff

)

: f ∈ H−

}

is a maximal uniformly negative subspace of H. Set H+ = H⊖H−. Since R∗−|K−

is one-to-one and H− ⊂ R∗−K−, there is a subspace L− of K− which is mapped by

R∗− in a one-to-one way onto H−.

We show that L− is closed. Let f1, f2, . . . ∈ L− and let fn → f for somef ∈ K−. Then R∗

−fn → R∗−f , and R∗

−f ∈ H− because R∗−fn ∈ H− for all n. By

the definition of L−, we can write R∗−f = R∗

−g where g ∈ L−. Since R∗−|K− is

one-to-one, f = g ∈ L−, and L− is closed.Next note that T ∗|L− = R∗

−|L−, because

T ∗f = PrH R∗−f = R∗

−f

for any f ∈ L− by the definition of L−. Since R∗−|L− is one-to-one and maps L−

onto H−, T ∗|L− is one-to-one and maps L− onto H−. The preceding identity alsoshows that D∗

−f = 0 for any f ∈ L−, which implies that D∗L−⊥D−.

Define L+ = {v : v ∈ K− and T ∗v⊥T ∗L−} . The subspace T ∗L− = H−

is maximal uniformly negative, and exactly as in the proof of Theorem 2.1.1 wecan show that K− = L−+L+. By construction, T ∗L+⊥T ∗L−.

The existence of bicontractive restrictions of the adjoint of any contractionoperator follows.

36

Theorem 2.1.4. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction. There exists a regular subspace K of K such that K⊥ is uniformlynegative and T ∗|K is a bicontraction. If K′ is a second such subspace with K′ ⊃ K,then K′ = K.

Proof. Choose a defect operator D ∈ B(D,K) for T ∗. Let D− be amaximal uniformly negative subspace of D, and define D− = D|D−. By Corollary1.4.2, (T D) is a bicontraction. In particular, D is a contraction and D−D− =DD− is a closed uniformly negative subspace of K. Choose a maximal uniformlynegative subspace K− of K which contains D−D−.

Choose closed subspaces L− and L+ of K− having the properties specifiedin Theorem 2.1.2. Define K = L− ⊕ K+, where K+ = K⊥

−. We show that K has

the required properties. By construction, K⊥ is uniformly negative.For f ∈ K,

〈f, f〉K − 〈T ∗f, T ∗f〉H = 〈(1− TT ∗)f, f〉K= 〈D∗f,D∗f〉D≥ 0.

The inequality is proved by contradiction in this way. If 〈D∗f,D∗f〉K < 0, then

D∗−f = PrD−

D∗f 6= 0.

But D∗−K = D∗

−(L− ⊕ K+) = {0} because D∗L−⊥D− by the choice of L− andD−D− ⊂ K− by the choice of K−. In view of this contradiction, we have shownthat T ∗|K is a contraction from K into H.

The subspace L− is maximal negative in K, and it is mapped by T ∗|Konto a maximal negative subspace of H by the choice of L−. Therefore T ∗|Kis a bicontraction by Theorem 1.3.6. The last assertion is clear, since otherwiseTheorem 1.3.6 would be violated.

The methods of this section yield a new characterization of bicontractionoperators in the spirit of Theorem 1.3.6, but now with a condition on how adjointsmap negative subspaces.

Theorem 2.1.5. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction. Then T is a bicontraction if and only if T ∗ maps some maximalnegative subspace of K in a one-to-one way into a negative subspace of H.

Proof. Necessity follows from Theorem 1.3.6.Conversely, assume that T ∗ maps the maximal negative subspace N of K

into a negative subspace of H. Let D ∈ B(D,K) be a defect operator for T ∗. Let

37

D = D+⊕D−, H = H+⊕H−, and K = K+⊕K− be fundamental decompositions.The fundamental decomposition of K can be chosen so that DD− ⊂ K−, becauseDD− is uniformly negative by an argument in the proof of Theorem 2.1.4. SetD− = D|D−. Since the range of D− is contained in K−, D∗

−|K+ = 0.As in the proof of Theorem 2.1.2, the operator

R∗− = (T D− ) ∈ B(H ⊕D−,K)

is a bicontraction. By Theorem 1.3.6, R∗− maps N in a one-to-one way onto a

maximal negative subspace of H⊕D−. Therefore

R∗−N =

{(

Kff

)

: f ∈ H− ⊕D−

}

for some contraction K ∈ B(|H− ⊕D−|,H+). Define

M =

{(

Kff

)

: f ∈ H−

}

.

By its form,M is a maximal negative subspace of H.Since the range of R∗

−|N includes M, there is a subspace L of N whichis mapped by R∗

− ontoM. For f ∈ L,

T ∗f = PrH R∗−f = R∗

−f,

because R∗−f ∈ M ⊂ H. In other words, T ∗|L = R∗

−|L. Now by assumption,T ∗ maps N in a one-to-one way into a negative subspace of H. Since L ⊂ Nand T ∗L =M is maximal negative, we conclude that L = N . We therefore haveD∗

−|N = 0, because for f ∈ N ,

T ∗f = R∗−f = T ∗f + D∗

−f

as elements of H⊕D−.But D∗

−|K+ = 0, and so D∗− annihilates K+ + N . This is all of K. In

fact, N is maximal negative and has a representation as the set of vectors Nu+u,u ∈ K−, where N ∈ B(|K−|,K+) is a contraction. Therefore any u in K− hasa representation u = −Nu + (Nu + u) with −Nu ∈ K+ and Nu + u ∈ N . SoK− ⊂ K+ +N , and hence K+ +N = K. We have shown that D∗

− = 0.Since PrD−

D∗ = D∗− = 0, the range of D∗ is contained in D+. Therefore

〈(1− TT ∗)f, f〉K = 〈D∗f,D∗f〉D ≥ 0

for every f in K, and T ∗ is a contraction. Since T is a contraction by assumption,T is a bicontraction.

38

2.2 Column Extensions

Let H and K be Kreın spaces, and let T ∈ B(H,K). A column extension

of T is an operator of the form

C =

(

TE∗

)

∈ B(H,K ⊕ E),

where E is a Kreın space and E ∈ B(E ,H). In this section we assume that E is aHilbert space. We characterize the contractive column extensions of T when T is acontraction. We also characterize the bicontractive column extensions of T whenT is a bicontraction.

The motivating example is obtained from a defect operator D ∈ B(D,H)for T . If T is a contraction,D is a Hilbert space by Corollary 1.4.2. The operator

(

T

D∗

)

∈ B(H,K ⊕ D)

is an isometry and hence a contractive column extension of T . If T is a bicontrac-tion, then

(

TD∗

)

is a bicontraction by Corollary 1.4.3.

Theorem 2.2.1. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea given operator with defect operator D ∈ B(D,H). Let

C =

(

TE∗

)

∈ B(H,K ⊕ E)

be a column extension of T with E a Hilbert space.

(i) If T is a contraction, then C is a contraction if and only if E = DG whereG ∈ B(E , D) is a contraction.

(ii) If T is a bicontraction, then C is a bicontraction if and only if E = DGwhere G ∈ B(E , D) is a contraction.

Proof. (i) Assume that T is a contraction. Then D is a Hilbert space byCorollary 1.4.2. If C is a contraction, then

0 ≤ 1− C∗C = 1− T ∗T − EE∗ = DD∗ − EE∗.

Hence‖E∗f‖2E ≤ ‖D∗f‖2D, f ∈ H.

So E∗ = G∗D∗ for a contraction G∗ ∈ B(D, E). Thus the condition is necessary.Sufficiency is verified by reversing these steps.

39

(ii) Assume that T is a bicontraction. Then T is a contraction, and thenecessity of the condition follows from (i).

Conversely, let E = DG where G ∈ B(E , D) is a contraction. By (i), Cis a contraction. We prove that C is a bicontraction by showing that it maps anymaximal negative subspace N of H onto a maximal negative subspace of K ⊕ E .Since E is a Hilbert space, any maximal uniformly negative subspace K− of K is amaximal uniformly negative subspace of K ⊕ E . We have

CN =

{(

Tf

E∗f

)

: f ∈ N}

,

and so by Theorem 1.3.6,

PrK−CN = PrK−

TN = K−

because T is a bicontraction. Hence CN is maximal negative in K⊕E , and anotherapplication of Theorem 1.3.6 shows that C is a bicontraction.

Corollary 2.2.2. Let H and K be Kreın spaces, and let T ∈ B(H,K)be a bicontraction. Let

C =

(

TE∗

)

∈ B(H,K ⊕ E)

be a column extension of T with E a Hilbert space. If C is a contraction, then Cis a bicontraction.

Proof. Choose a defect operator for T as in Theorem 2.2.1. If C is acontraction, then by part (i) of the theorem, E = DG where G ∈ B(E , D) is acontraction. But then by part (ii) of the theorem, C is a bicontraction.

We note a norm estimate which will be useful in the commutant liftingtheorem to be proved later.

Corollary 2.2.3. Let H and K be Kreın spaces, and let T ∈ B(H,K)be a contraction. Let

C =

(

TE∗

)

∈ B(H,K ⊕ E)

be a contractive column extension of T with E a Hilbert space. If norms arecomputed relative to fixed fundamental decompositions ofH andK and the inducedfundamental decomposition of K ⊕ E , then

‖C‖2 ≤ 1 + 2‖T‖2.

40

Proof. Choose a defect operator for T and factor E = DG as in Theorem2.2.1 (i). Since T is a contraction, D is a Hilbert space by Corollary 1.4.2. For anyf ∈ H,

∥

∥

∥D∗f∥

∥

∥

2

=⟨

D∗f, D∗f⟩

D=

⟨

DD∗f, f⟩

H= 〈(1− T ∗T )f, f〉H

≤[

1 + ‖T‖2]

‖f‖2.

Since G is a Hilbert space contraction, for any f ∈ H,

‖Cf‖2 = ‖Tf‖2 + ‖E∗f‖2 = ‖Tf‖2 + ‖G∗D∗f‖2 ≤[

1 + 2‖T‖2]

‖f‖2,

as required.

2.3 Row Extensions

Let T ∈ B(H,K), H and K Kreın spaces. By a row extension of T wemean an operator of the form

R = (T E ) ∈ B(H⊕ E ,K),

where E is a Kreın space and E ∈ B(E ,K). If T is a contraction, then there existbicontractive row extensions of T . In fact, if D ∈ B(D,K) is a defect operator forT ∗, the operator

R = (T D) ∈ B(H⊕D,K)

is a bicontractive row extension of T by Corollary 1.4.2. We characterize all suchextensions. It is convenient to begin with the special case of an extension by aHilbert space.

Theorem 2.3.1. Let H and K be Kreın spaces, T ∈ B(H,K) a bicon-traction, and D ∈ B(D,K) a defect operator for T ∗. Let

R = ( T E ) ∈ B(H⊕ E ,K)

be a row extension of T with E a Hilbert space. Then R is a bicontraction if andonly if E = DG where G ∈ B(E ,D) is a contraction.

Note that D is a Hilbert space by the hypothesis that T is a bicontraction.

41

Proof. If R is a bicontraction, then the operators

R∗ =

(

T ∗

E∗

)

∈ B(K,H⊕ E)

and T ∗ are contractions. By Theorem 2.2.1 (i) applied to T ∗, E = DG whereG ∈ B(E ,D) is a contraction.

Conversely, assume that E = DG where G ∈ B(E ,D) is a contraction.By Theorem 2.2.1 (ii) applied to T ∗, the operator R∗ is a bicontraction, hence Ris a bicontraction.

A technical result is needed. It describes how bicontractive row exten-sions act with respect to the decompositions obtained in Theorem 2.1.2. Thisinformation will be used to determine the general form of any bicontractive rowextension of a contraction.

Theorem 2.3.2. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction. Let K− be a maximal uniformly negative subspace of K, and chooseclosed subspaces L− and L+ of K− such that

K− = L−+L+,

T ∗ maps L− in a one-to-one way onto a maximal uniformly negative subspace ofH, and T ∗L+⊥T ∗L−. Let

R = ( T E ) ∈ B(H⊕ E ,K)

be any bicontractive row extension of T , where E a Kreın space and E ∈ B(E ,K).Then E∗ maps L+ in a one-to-one way onto a maximal uniformly negative subspaceof E .

The existence of subspacesL− and L+ with the required properties followsfrom Theorem 2.1.2.

Proof. Choose fundamental decompositions E = E+ ⊕E− and H = H+ ⊕H− with H− = T ∗L− and H+ = (T ∗L−)⊥. Let K+ = K⊥

−.Since R∗ is a bicontraction, R∗K− is a maximal uniformly negative sub-

space of H ⊕ E . If b ∈ L+, then T ∗b ∈ H+ and for some positive number ǫ whichdoes not depend on b,

〈E∗b,E∗b〉E ≤ 〈T ∗b, T ∗b〉H + 〈E∗b,E∗b〉E = 〈R∗b,R∗b〉H⊕E

≤ −ǫ‖R∗b‖2 = −ǫ‖T ∗b‖2 − ǫ‖E∗b‖2

≤ −ǫ‖E∗b‖2,

42

where norms are computed in a suitable way. It follows that E∗ maps L+ in aone-to-one way onto a uniformly negative subspace of E .

We show that the projection of E∗L+ into E− is all of E−. Since R∗K−

is maximal uniformly negative in H⊕ E ,

PrH−⊕E−R∗K− = H− ⊕ E−.

Given e ∈ E−, we may therefore choose b ∈ K− such that

PrH−⊕E−R∗b = e.

Applying the projection onto E−, we obtain

PrE−E∗b = PrE−

R∗b = e.

Applying instead the projection onto H−, we see that

PrH−T ∗b = PrH−

R∗b = 0,

that is, T ∗b ∈ H+. Therefore if b = u + v, u ∈ L−, v ∈ L+, we have

T ∗u = T ∗b− T ∗v ∈ H− ∩H+ = {0}.

Since T ∗|L− is one-to-one, u = 0 and b = v ∈ L+. We have shown that theprojection of E∗L+ into E− is all of E−, and so E∗L+ is maximal uniformly negativeby the graph representation of negative subspaces.

The main result of this section follows.

Theorem 2.3.3. Let H and K be Kreın spaces, T ∈ B(H,K) a contrac-tion, and D ∈ B(D,K) a defect operator for T ∗. Let E ∈ B(E ,K), where E is aKreın space. Then the operator

R = ( T E ) ∈ B(H⊕ E ,K)

is a bicontraction if and only if E = DG, where G ∈ B(E ,D) is a bicontraction.

Proof of Sufficiency. Assume that E = DG where G ∈ B(E ,D) is abicontraction. Since 1−GG∗ ≥ 0,

1−RR∗ = 1− TT ∗ −DGG∗D∗ = D(1−GG∗)D∗ ≥ 0,

43

and R∗ is a contraction. By Theorem 1.2.4, T has a Julia operator of the form

U =

(

T DD∗ L

)

∈ B(H⊕D,K ⊕ D).

Thus

1−R∗R = 1−(

T ∗

G∗D∗

)

( T DG )

=

(

1− T ∗T −T ∗DG

−G∗D∗T 1−G∗D∗DG

)

=

(

DD∗ DLG

G∗L∗D∗ G∗L∗LG

)

+

(

0 0

0 1−G∗D∗DG−G∗L∗LG

)

=

(

DG∗L∗

)

( D∗ LG ) +

(

0 00 1−G∗G

)

≥ 0,

because D is a Hilbert space by Theorem 1.4.1 and 1 − G∗G ≥ 0. Thus R is abicontraction.

Proof of Necessity. Assume now that R is a bicontraction. Choose fun-damental decompositions

E = E+ ⊕ E−, D = D+ ⊕D−, K = K+ ⊕K−.

Define E− = E|E− ∈ B(E−,K). The operator

R− = (T E− ) ∈ B(H⊕ E−,K)

is a bicontraction by Lemma 2.1.3. Choose closed subspaces L− and L+ of K− asin Theorem 2.1.2. In particular, K− = L−+L+.

By Theorem 2.3.2, E∗ maps L+ in a one-to-one way onto a maximaluniformly negative subspace of E . The projection of E onto E− maps any maximalnegative subspace of E onto E− by the graph representation. Therefore E∗

− mapsL+ in a one-to-one way onto E−.

We show thatK = L++ kerE∗

−.

If k ∈ K, then E∗−k ∈ E− and there exists u ∈ L+ such that E∗

−k = E∗−u. Then

k = u + v where u ∈ L+ and v = k − u ∈ kerE∗−. If u ∈ L+ ∩ kerE∗

−, then u = 0because E∗

−|L+ is one-to-one and u is an element of L+, which is annihilated byE∗

−. Thus K is the direct sum of L+ and kerE∗−.

44

We next define a bicontractive operator G− ∈ B(E−,D) such that E− =DG−. To this end first define a mapping X on the dense set ran D∗ in D to E− by

XD∗f = E∗−f, f ∈ K.

The mapping is well defined. For suppose f ∈ K and D∗f = 0. Write f = u + v,u ∈ L+, v ∈ kerE∗

−. Then D∗u = −D∗v. By Corollary 1.4.2, (T D) is abicontractive row extension of T . Applying Theorem 2.3.2 to (T D), we obtain

〈D∗u,D∗u〉D ≤ 0,

with equality only for u = 0. Since R− is a bicontraction and E∗−v = 0,

0 ≤⟨

(1−R−R∗−)v, v

⟩

K=

⟨

(1− TT ∗ −E−E∗−)v, v

⟩

K

= 〈D∗v,D∗v〉D = 〈D∗u,D∗u〉D ≤ 0.

Therefore equality holds throughout and u = 0. Thus E∗−f = E∗

−v = 0, and X iswell defined.

As far as it is defined, X satisfies

〈Xg,Xg〉E−≤ 〈g, g〉D, g ∈ domX.

For if g = D∗f , f ∈ K,

0 ≤⟨

(1− R−R∗−)f, f

⟩

K=

⟨

(1− TT ∗ −E−E∗−)f, f

⟩

K

= 〈D∗f,D∗f〉D −⟨

E∗−f,E∗

−f⟩

E−

= 〈g, g〉D − 〈Xg,Xg〉E−.

Theorem 2.3.2 applied to (T D) implies that D∗L+ is a maximal uni-formly negative subspace of D. By the definition of X, D∗L+ ⊂ domX and Xmaps D∗L+ onto the maximal uniformly negative subspace E∗L+ of E . Thereforeby Theorem 1.4.4, X has an extension by continuity to a bicontraction which wedenote G∗

−. This completes the construction of a bicontractive operator G− ∈B(E−,D) such that E− = DG−.

Let DR−∈ B(DR−

,K) and DG−∈ B(DG−

,D) be defect operators forR∗

− and G∗−. Then

DR−D∗

R−= 1−R−R∗

− = 1− TT ∗ −E−E∗−

= DD∗ −DG−G∗−D∗ = D(1−G−G∗

−)D∗

= DDG−D∗

G−D∗.

Therefore there is a unitary operator V ∈ B(DR−,DG−

) such that

DR−= DDG−

V.

45

Write R = (T E− E+) where E+ = E|E+ ∈ B(E+,K). The 1×3 rowmatrix acts on H⊕E−⊕E+ to K. Its restriction to H⊕E− is the bicontraction R−,and so by Theorem 2.3.1, E+ = DR−

G+, where G+ ∈ B(E+,DR−) is a contraction.

Finally,

R = (T DG− DR−G+ ) = ( T DG− DDG−

V G+ ) = ( T DG ) ,

where G =(

G− DG−V G+

)

∈ B(E− ⊕ E+,D) = B(E ,D) is a bicontraction byTheorem 2.3.1.

A similar theorem on contractive row extensions follows as a consequence.

Theorem 2.3.4. Let H and K be Kreın spaces, T ∈ B(H,K) a contrac-tion, and D ∈ B(D,K) a defect operator for T ∗. Let E be a Kreın space, and letE ∈ B(E ,K). Then the operator

R = ( T E ) ∈ B(H⊕ E ,K)

is a contraction if and only if E = DG where G ∈ B(E ,D) is a contraction.

Proof of Sufficiency. If E = DG where G ∈ B(E ,D) is a contraction, weshow that 1− R∗R ≥ 0 exactly as in the sufficiency part of the proof of Theorem2.3.3.

Proof of Necessity. Assume that R is a contraction, and choose a de-fect operator DR ∈ B(DR,K) for R∗. By Corollary 1.4.2, R = (R DR) is abicontraction in B((H ⊕ E)⊕DR,K). In another view,

R = ( T ( E DR ) ) ∈ B(H⊕ (E ⊕ DR),K),

is a bicontractive row extension of T , and we may apply the necessity part ofTheorem 2.3.3. It follows that

(E DR ) = DΓ,

where Γ ∈ B(E ⊕ DR,D) is a bicontraction. Thus

R = R|(H⊕ E) = (T DG )

where G = Γ|E is a contraction.

46

2.4 Two-by-Two Matrix Completions

Let H and K be Kreın spaces, and let T ∈ B(H,K) be a contraction.We label all contractive two-by-two matrices with T in the upper left entry. Theextension space for the range is required to be a Hilbert space.

Theorem 2.4.1. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction with Julia operator

(

T DD∗ L

)

∈ B(H⊕D,K ⊕ D).

Let E be a Kreın space, and let E be a Hilbert space. If

Q =

(

T Q1

Q2 Q3

)

∈ B(H⊕ E ,K ⊕ E)

is a contraction, then

Q =

(

T DG1

G∗2D

∗ G∗2LG1 + DG2

G3D∗G1

)

,

where

(i) G1 ∈ B(E ,D) is a Kreın space contraction with defect operator DG1∈

B(DG, E),

(ii) G2 ∈ B(E , D) is a Hilbert space contraction with defect operator DG2∈

B(DG2, E), and

(iii) G3 ∈ B(DG1, DG2

) is a Hilbert space contraction.

Conversely, every operator Q ∈ B(H⊕ E ,K ⊕ E) of this form is a contraction.

Proof. Assume that Q is a contraction. Then so is its restriction

(

T

Q2

)

∈ B(H,K ⊕ E).

Since T is a contraction and E is a Hilbert space, we can apply Theorem 2.2.1(i) to write Q2 = G∗

2D∗ where G2 ∈ B(E , D) is a Hilbert space contraction. Let

DG2∈ B(DG2

, E) be any defect operator for G2.Since E is a Hilbert space, (T Q1) = PrK Q ∈ B(H ⊕ E ,K) is a con-

traction. By Theorem 2.3.4, Q1 = DG1 where G1 ∈ B(E ,D) is a Kreın spacecontraction. Let DG1

∈ B(DG1, E) be any defect operator for G1. Thus far we

have

Q =

(

T DG1

G∗2D

∗ Q3

)

.

47

After a short calculation, the inequality 1−Q∗Q ≥ 0 yields

0 ≤(

1− T ∗T − DG2G∗2D

∗ −T ∗DG1 − DG2Q3

−G∗1D

∗T −Q∗3G

∗2D

∗ 1−G∗1D

∗DG1 −Q∗3Q3

)

=

(

DD∗ − DG2G∗2D

∗ DLG1 − DG2Q3

G∗1L

∗D∗ −Q∗3G

∗2D

∗ 1−G∗1G1 + G∗

1(1−D∗D)G1 −Q∗3Q3

)

=

(

D 00 1

) (

1−G2G∗2 LG1 −G2Q3

G∗1L

∗ −Q∗3G

∗2 DG1

D∗G1

+ G∗1L

∗LG1 −Q∗3Q3

) (

D∗ 00 1

)

.

Since D is one-to-one, the range of D∗ is dense in D. Therefore(

1−G2G∗2 LG1 −G2Q3

G∗1L

∗ −Q∗3G

∗2 DG1

D∗G1

+ G∗1L

∗LG1 −Q∗3Q3

)

≥ 0

as operators in B(D ⊕ E). A straightforward calculation brings this to the formA∗A ≥ B∗B, where

A =

(

1 LG1

0 D∗G1

)

∈ B(D ⊕ E , D ⊕ DG1)

andB = ( G∗

2 Q3 ) ∈ B(D ⊕ E , E).Since D ⊕ DG1

and E are Hilbert spaces, there is a Hilbert space contraction

K = ( K0 K1 ) ∈ B(D ⊕ DG1, E)

such that B = KA, that is,

(G∗2 Q3 ) = ( K0 K1 )

(

1 LG1

0 D∗G1

)

= ( K0 K0LG1 + K1D∗G1

) .

Thus K0 = G∗2 and Q3 = K0LG1+K1D

∗G1

. Applying Theorem 2.3.1 in the Hilbertspace case, we obtain

( K0 K1 ) = (G∗2 DG2

G3 )

where G3 ∈ B(DG1, DG2

) is a Hilbert space contraction. Therefore

Q3 = K0LG1 + K1D∗G1

= G∗2LG1 + DG2

G3D∗G1

.

We have shown that Q has the required form if Q is a contraction. In the otherdirection, if Q has the stated form, we simply reverse the steps to show that1−Q∗Q ≥ 0.

A similar result characterizes bicontractive two-by-two matrices with Tin the upper left entry when T is a contraction and the extension space for therange is a Hilbert space.

48

Theorem 2.4.2. Let H and K be Kreın spaces, and let T ∈ B(H,K) bea contraction with Julia operator

(

T DD∗ L

)

∈ B(H⊕D,K ⊕ D).

Let E be a Kreın space, and let E be a Hilbert space. If

Q =

(

T Q1

Q2 Q3

)

∈ B(H⊕ E ,K ⊕ E)

is a bicontraction, then Q can be represented as in Theorem 2.4.1 with G1 abicontraction. Conversely, every Q ∈ B(H ⊕ E ,K ⊕ E) of the form described inTheorem 2.4.1 with G1 a bicontraction is a bicontraction.

Proof. Assume first that Q has the form described in Theorem 2.4.1 withG1 a bicontraction. By Theorem 2.3.3,

(T Q1) = (T DG1) ∈ B(H⊕ E ,K)

is a bicontraction. Now view

Q =

(

(T Q1 )( Q2 Q3 )

)

∈ B(H⊕ E ,K ⊕ E)

as a contractive column extension of the bicontraction (T Q1) with extensionspace E a Hilbert space. By Corollary 2.2.2, Q is a bicontraction.

Conversely, let Q be a bicontraction. If N is a maximal negative subspaceof H⊕ E , then QN is a maximal negative subspace of K⊕ E . Since E is a Hilbertspace, PrK QN = (T Q1)N is a maximal negative subspace of K. As in theproof of Theorem 2.4.1, (T Q1) ∈ B(H ⊕ E ,K) is a contraction, and hence itis a bicontraction by Theorem 1.3.6. By Theorem 2.3.3, Q1 = DG′

1 where G′1 ∈

B(E , D) is a bicontraction. Since D has zero kernel, G1 = G′1 is a bicontraction.

Corollary 2.4.3. Let H and K be Kreın spaces, and let T ∈ B(H,K)be a bicontraction. Let E be a Kreın space, and let E be a Hilbert space. If

Q =

(

T Q1

Q2 Q3

)

∈ B(H⊕ E ,K ⊕ E)

is a contraction, then E is also a Hilbert space and Q is a bicontraction.

Proof. Choose a Julia operator(

T DD∗ L

)

∈ B(H⊕D,K ⊕ D).

49

for T , and represent Q as in Theorem 2.4.1. Since T is a bicontraction,D is a Hilbertspace by Corollary 1.4.3. Since G1 ∈ B(E ,D) is a contraction, E is a Hilbert spaceand G1 is a bicontraction. So Q is a bicontraction by Theorem 2.4.2.

We recast the results of this section in a form that is convenient forapplication to the commutant lifting problem.

Theorem 2.4.4. LetH, K, E be Kreın spaces, E a Hilbert space. Assumethat

( C11 C12 ) ∈ B(H⊕ E ,K) and

(

C11

C21

)

∈ B(H,K ⊕ E)

are contraction operators. There exists an operator C22 ∈ B(E , E) such that

C =

(

C11 C12

C21 C22

)

∈ B(H⊕ E ,K ⊕ E)

is a contraction. If (C11 C12) is a bicontraction, C22 may be chosen so that C isa bicontraction.

Proof. Since (C11 C12) is a contraction, so is its restriction C11. WriteT = C11, and let

U =

(

T DD∗ L

)

∈ B(H ⊕D,K ⊕ D)

be a Julia operator for T . By Theorem 2.3.4, C12 = DG1, where G1 ∈ B(E ,D) is aKreın space contraction. By Theorem 2.2.1 (i), C21 = G∗

2D∗, where G2 ∈ B(E , D)

is a Hilbert space contraction. We may now invoke Theorem 2.4.1 with any choiceof G3, such as, G3 = 0 (which is a contraction since the underlying spaces areHilbert spaces), to produce an operator C22 such that C is a contraction.

If (C11 C12) is a bicontraction, then in the preceding argument we canuse Theorem 2.3.3 in place of Theorem 2.3.4 to choose G1 to be a bicontraction.In place of Theorem 2.4.1, we may use Theorem 2.4.2 to produce C22 such that Cis a bicontraction.

To conclude this section, we use Theorem 2.4.1 to show that Julia opera-tors for contractions, or operators whose adjoints are contractions, are essentiallyunique. See Appendix B for a more general result.

50

Theorem 2.4.5. Let H and K be Kreın spaces, and let T ∈ B(H,K)have two Julia operators

(

T Dj

D∗j Lj

)

∈ B(H⊕Dj ,K ⊕ Dj), j = 1, 2.

If either T or T ∗ is a contraction, then there exist unitary operators V ∈ B(D2,D1)and V ∈ B(D2, D1) such that

(

T D2

D∗2 L2

)

=

(

1 00 V ∗

) (

T D1

D∗1 L1

)(

1 00 V

)

.

Proof. Since the adjoint of a Julia operator for T is a Julia operator forT ∗, it is sufficient to give the proof in the case in which T is a contraction. Inthis case, Theorem 2.4.1 is applicable and yields contractions V ∈ B(D2,D1) andV ∈ B(D2, D1) such that

(

T D2

D∗2 L2

)

=

(

1 00 V ∗

)(

T D1

D∗1 L1

) (

1 00 V

)

+

(

0 00 DV GD∗

V

)

,

where DV ∈ B(DV ,D2) and DV ∈ B(DV , D2) are defect operators for V and V

and G ∈ B(DV , DV ) is a contraction. In particular, D2 = D1V and

D1D∗1 = 1− TT ∗ = D2D

∗2 = D1V V ∗D∗

1 .

Since D1 has zero kernel, V V ∗ is the identity on D1. Since D2 has zero kernel, Vis a unitary operator. In a similar way, D2 = D1V and

D1D∗1 = 1− T ∗T = D2D

∗2 = D1V V ∗D∗

1 .

Since D1 and D2 have zero kernels, V is a unitary operator. In particular, DV andDV each contain only the zero vector, and DV GD∗

V = 0.

51

Notes on Chapter 2

The methods of Chapter 2 follow Dritschel [30]. Theorems 2.1.1 and 2.1.4generalize a result of Kuzhel’ [48], but the treatment here is different.

Theorems 2.3.3 and 2.3.4 first appear in Dritschel [30] in the present generalitybut were derived in special cases in earlier works. In fact, for most of the extensiontheorems in §2.2–2.4, there have been versions worked out by Arsene, Constantinescu,and Gheondea [8]. The present work was done independently of the recent papers byConstantinescu and Gheondea [22]. These papers contain a great amount of detailedinformation on related extension problems which we do not treat. The problems aregenerally of a more precise nature and require stronger hypotheses. While there is overlapwith the present work in some cases, there is no inclusion of one theory in the other. Inthe bicontractive case, forerunners of the results in §2.2–2.4 also appear in Dritschel [29].

The method of proof of Theorem 2.4.1 is adapted from the account of theHilbert space result in Ptak and Vrbova [56]. In [17], de Branges uses complementationtheory to prove a version of Theorem 2.4.1 when both extension spaces are Hilbert spaces.Two-by-two extension theorems in the Hilbert space case are due to Arsene and Gheon-dea [7], Davis [24], Davis, Kahan, and Weinberger [26], and Shmul’yan and Yanovskaya[65]. Related papers are Davis [25] and Parrott [54].

The question of uniqueness of Julia operators is partially settled in Theorem2.4.5 and Appendix B. We feel that uniqueness will fail in general, but we do not knowan example.

52

Chapter 3: Commutant Liftingof Contraction Operators

3.1 Dilation Theory

The dilation properties of Kreın space operators are similar to the Hilbertspace case. The main problem is to finesse the fact that a densely defined Kreınspace isometry may fail to have a continuous extension: in some cases, this causesdifficulty with uniqueness. It turns out that minimal isometric dilations of contrac-tions are unique and have the familiar properties of the Hilbert space case. Thesame is true of minimal unitary dilations of bicontractions. For general operators,minimal isometric and minimal unitary dilations always exist with the propertiesof the Hilbert space case, but we cannot assert uniqueness nor that all minimalisometric and unitary dilations have desirable additional properties.

Definition 3.1.1. Let T ∈ B(H), whereH is a Kreın space. An isomet-

ric dilation of T is an isometry U ∈ B(H), where H is a Kreın space containingH isometrically as a regular subspace, such that

Tn = PrH Un|H, n = 1, 2, 3, . . ..

An isometric dilation U ∈ B(H) of T is minimal if

∞∨

n=0

UnH = H.

Minimal isometric dilations always exist.

Theorem 3.1.2. Let T ∈ B(H), H a Kreın space, and let D ∈ B(D,H)be a defect operator for T . Define H = H ⊕ D ⊕ D ⊕ · · · using fixed choices offundamental decompositions for H and D. Then the matrix

U =

T 0 0 0 · · ·D∗ 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·

· · ·

acts as an everywhere defined and continuous operator on H, and U is a minimalisometric dilation of T .

53

Proof. The infinite matrix determines an element U of B(H) becausethe shift operator is everywhere defined and continuous on D ⊕ D ⊕ · · · . If f =(f0, f1, . . .)

t is in H, then

〈Uf,Uf〉H =⟨

(T ∗T + DD∗)f0, f0

⟩

H+ 〈f1, f1〉D + 〈f2, f2〉D + · · ·

= 〈f, f〉Hbecause T ∗T + DD∗ = 1. Thus U is isometric. An inductive argument shows thatfor any element of H of the form f = (f0, 0, 0, . . .)t and any n = 1, 2, 3, . . .,

Unf =

Tnf0

D∗Tn−1f0

D∗Tn−2f0...

D∗f0

00...

.

Therefore Tnf0 = PrH Unf0 for any f0 ∈ H. The same identity implies that forevery positive integer n,

n∨

j=0

U jH = H⊕ D ⊕ · · · ⊕ D

with n copies of D. Hence U is a minimal isometric dilation of T .

The minimal isometric dilation of a Kreın space operator T ∈ B(H)constructed in Theorem 3.1.2 has special properties:

(a) H is invariant under U∗ and U∗|H = T ∗;(b) the closure L of {Uh− Th : h ∈ H} is a regular subspace of H, and

H = H⊕ L ⊕ U L ⊕ U2L ⊕ · · ·;(c) the positive and negative indices of L coincide with h±(1− T ∗T ).

Proofs are immediate from the construction in Theorem 3.1.2.Let T ∈ B(H), H a Kreın space, and let U ∈ B(H) be a minimal isomet-

ric dilation of T . If H′ is another Kreın space containing H as a regular subspace,and if U ′ = WUW−1 where W ∈ B(H, H′) is a unitary operator which coincideswith the identity on H, then the operator U ′ ∈ B(H′) is also a minimal isometricdilation of T . Two minimal isometric dilations of T related in this way are saidto be isomorphic. Isomorphic minimal isometric dilations are abstractly indistin-guishable. In particular, properties (a) - (c) are invariant under isomorphism.

The next result shows that minimal isometric dilations of contractionsare unique up to isomorphism, and therefore all have the properties (a)–(c).

54

Theorem 3.1.3. Let H be a Kreın space, and assume that T ∈ B(H) isa contraction.

(i) Any two minimal isometric dilations of T are isomorphic.(ii) If U ∈ B(H) is a minimal isometric dilation of T , then H⊖H is uniformly

positive, and hence any maximal negative subspace of H is maximal neg-ative in H.

(iii) The operator T is a bicontraction if and only if any minimal isometricdilation of T is a bicontraction.

Proof. To prove (i), it is sufficient to show that if U ∈ B(H) is a minimalisometric dilation of T of the form constructed in Theorem 3.1.2, and if U ′ ∈ B(H′)is any other minimal isometric dilation of T , then U and U ′ are isomorphic.

For any elements f0, . . ., fn and g0, . . ., gm of H,

⟨

n∑

j=0

U jfj ,m

∑

k=0

Ukgk

⟩

H

=

⟨

n∑

j=0

U ′jfj ,m

∑

k=0

U ′kgk

⟩

H′

by the definition of an isometric dilation. Since both dilations are minimal, ina routine way we obtain a well defined and densely defined isometry W0 on H,having dense range in H′, such that

W0

n∑

j=0

U jfj

=n

∑

j=0

U ′jfj

for any nonnegative integer n and vectors f0, . . ., fn in H. Both the domain andrange of W0 include H, and W0 coincides with the identity on H. The domain ofW0 is invariant under U , and W0Ug = U ′W0g for all g in the domain of W0.

Since T is a contraction, D is a Hilbert space by Corollary 1.4.2. ThereforeH⊖H is uniformly positive. We show that H′⊖H is also uniformly positive. SinceranW0 is dense in H′ and includes H, the intersection of ranW0 and H′ ⊖ H isdense in H′ ⊖H. In a similar way, the intersection of domW0 and H ⊖H is densein H ⊖ H. Since H ⊖ H is positive and W0 is isometric, H′ ⊖ H is positive andtherefore uniformly positive.

If M is a maximal negative subspace of H, by what we just proved Mis maximal negative in both H and H′. Now M is in the domain of W0 andW0M = M. Hence by Theorem 1.4.4, W0 has an extension to an operator W ∈B(H, H′). It is clear that W is unitary, W coincides with the identity on H, andU ′ = WUW−1. Therefore U and U ′ are isomorphic, and (i) is proved. A proof of(ii) is obtained incidentally from the argument.

For (iii), consider any minimal isometric dilation U ∈ B(H) of T . By part(ii) of the theorem, H ⊖ H is uniformly positive. Hence any maximal uniformly

55

negative subspace H− of H is maximal negative in H as well. Both T and Uare contractions, and UH− and TH− have the same projection onto H− by thedefinition of an isometric dilation. By Theorem 1.3.6, U and T are simultaneouslybicontractions or not.

Parallel results hold for unitary dilations.

Definition 3.1.4. Let T ∈ B(H), where H is a Kreın space. A unitary

dilation of T is a unitary operator U ∈ B(H), where H is a Kreın space containingH isometrically as a regular subspace, such that

Tn = PrH Un|H and T ∗n = PrH U−n|H

for all n = 1, 2, 3, . . .. A unitary dilation U ∈ B(H) of T is minimal if

∞∨

n=−∞

UnH = H.

We give two methods of constructing minimal unitary dilations. The firstiterates the construction of minimal isometric dilations.

Theorem 3.1.5. Let T ∈ B(H), H a Kreın space. Let V ∈ B(H′) beany minimal isometric dilation of T , and let U∗ ∈ B(H) be a minimal isometricdilation of V ∗ of the form constructed in Theorem 3.1.2. Then U is a minimalunitary dilation of T .

Proof. By assumption, U∗ has the form

U∗ =

V ∗ 0 0 0 · · ·E∗ 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·

· · ·

∈ B(H′ ⊕ E ⊕ E ⊕ · · ·),

where E ∈ B(E ,H′) is a defect operator for V ∗ ∈ B(H′). Since V is an isometry, Eis an isometry with range equal to kerV ∗. To see this, note that 1− V V ∗ = EE∗

is a projection operator, and ker E∗ = ker EE∗ = ker (1 − V V ∗) = VH′. Thenapply Theorem 1.1.6 to obtain the assertion. It follows that

V ∗E = 0 and E∗E = 1.

A matrix multiplication shows that U∗U = 1. Since U∗ is an isometry by con-struction, U is unitary. For any n = 1, 2, 3, . . .,

PrH V n|H = Tn and PrH V ∗n|H = T ∗n

56

andPrH′ U∗n|H′ = V ∗n and PrH′ Un|H′ = V n

because V and U∗ are isometric dilations of T and V ∗. The second pair of relationscan be further restricted to H. Applying the projection onto H and using the firstpair of relations, we then see that U is a unitary dilation of T .

It remains to show that U is minimal. From the matrix form of U , wehave U |H = V |H, and so ∨n

0 U jH = ∨n0 V jH. Since V is a minimal isometric

dilation of T by assumption, ∨∞0 U jH = H′. As in the proof of Theorem 3.1.2, aninductive argument shows that for any element of H of the form f = (f0, 0, 0, . . .)t,f0 ∈ H′, and any n = 1, 2, 3, . . .,

U∗nf =

V ∗nf0

E∗V ∗n−1f0

E∗V ∗n−2f0...

E∗f0

00...

.

The identity implies that ∨∞0 U∗jH′ = H′ ⊕ E ⊕ · · · ⊕ E with n copies of E . Since∨∞0 U jH = H′, this shows that U is a minimal unitary dilation of T .

The preceding method of constructing unitary dilations is useful for ap-plications, but it does not yield other desirable properties similar to the Hilbertspace case. These properties are obtained from a different construction.

Theorem 3.1.6. Let H be a Kreın space, and let T ∈ B(H) have Juliaoperator

(

T D

D∗ L

)

∈ B(H⊕D,H ⊕ D),

Define H = · · · ⊕ D ⊕ D ⊕ H ⊕ D ⊕ D ⊕ · · · using fixed choices of fundamentaldecompositions in the summands. Then the matrix

U =

· · ·· · · 1 0 0 0 0 0 0 · · ·· · · 0 1 0 0 0 0 0 · · ·· · · 0 0 D T 0 0 0 · · ·· · · 0 0 L D∗ 0 0 0 · · ·· · · 0 0 0 0 1 0 0 · · ·· · · 0 0 0 0 0 1 0 · · ·

· · ·

57

acts as an everywhere defined and continuous operator on H, and U is a minimalunitary dilation of T .

Proof. The continuity of shift operators implies that U exists as an ele-ment of B(H). For any element

h = (. . ., d2, d1, f, d1, d2, . . .)t

of H, with f as the element of H, we have

Uh =

...d3

d2

Dd1 + TfLd1 + D∗f

d1

d2...

and U∗h =

...d2

d1

D∗f + L∗d1

T ∗f + Dd1

d2

d3...

.

A straightforward calculation yields

〈Uh,Uh〉H = 〈U∗h,U∗h〉H = 〈h, h〉H,

and so U is unitary. If h = (. . ., 0, 0, f, 0, 0, . . .)t, where f ∈ H, then

Unh =

...00

TnfD∗Tn−1fD∗Tn−2f

...D∗f

00...

and U∗nh =

...00

D∗f...

D∗T ∗n−2fD∗T ∗n−1f

T ∗nf00...

.

The identities yieldn∨

j=0

U jH = H⊕ D ⊕ · · · ⊕ D,

n∨

j=0

U∗ jH = D ⊕ · · · ⊕ D ⊕H,

58

with n+1 summands on the right side in each case. It follows that U is a minimalunitary dilation of T .

As in the case of isometric dilations, the construction of a minimal uni-tary dilation in Theorem 3.1.6 for a given operator T ∈ B(H) immediately yieldsadditional properties:

(a′) the closures L and L of {Uh−Th : h ∈ H} and {U∗h−T ∗h : h ∈ H} areregular subspaces of H, and

H = · · · ⊕ U∗ 2L ⊕ U∗L ⊕ L ⊕H⊕ L ⊕ U L ⊕ U2L ⊕ · · ·;

(b′) the positive and negative indices of L are equal to h±(1−T ∗T ), and thoseof L are equal to h±(1− TT ∗).

Let T ∈ B(H), H a Kreın space, and let U ∈ B(H) be a minimal unitarydilation of T . If H′ is another Kreın space containing H as a regular subspace, andif U ′ = WUW−1 where W ∈ B(H, H′) is a unitary operator which coincides withthe identity on H, then the operator U ′ ∈ B(H′) is a minimal unitary dilation ofT . In this situation, we say that U and U ′ are isomorphic. The properties (a′)and (b′) are invariant under isomorphism.

Minimal unitary dilations of bicontractions are unique up to isomorphism.

Theorem 3.1.7. Let H be a Kreın space, and let T ∈ B(H) be a bicon-traction.

(i) Any two minimal unitary dilations of T are isomorphic.(ii) If U ∈ B(H) is a minimal unitary dilation of T , then H ⊖H is uniformly

positive, and hence any maximal negative subspace of H is maximal neg-ative in H.

Proof. The argument is similar to the proof of Theorem 3.1.3, parts (i)and (ii). In place of Theorem 3.1.2 we use Theorem 3.1.6. Since T is a bicon-traction, the spaces D and D in Theorem 3.1.6 are Hilbert spaces by Corollary1.4.3. Otherwise we proceed as before, except that in place of sums of the form∑n

j=0 U jfj with f0, . . ., fn ∈ H we now use sums of the form∑n

j=−n U jfj withf−n, . . ., fn ∈ H.

Theorem 3.1.3 (iii) has a variant form which does not require that T bea contraction. If T ∈ B(H), where H is a Kreın space, and if T ∗ is a contraction,then any minimal isometric dilation of T of the form constructed in Theorem 3.1.2is a bicontraction. Conversely, if such a dilation is a bicontraction, then T ∗ is acontraction. For U is a bicontraction if and only if U∗ is a contraction, and thecondition for this is that [T ∗ D] be a contraction. By Theorem 1.4.2, this holdsif and only if T ∗ is a contraction.

59

We give an example. Let H be the anti-space of a Hilbert space, and letT = −2V ∗, where V ∈ B(H) is an isometry whose range is not all of H. ThenT ∗ is a contraction, but T is not a contraction because its kernel is not uniformlypositive. Let D and D be H as a vector space, with

〈f, g〉D = −〈f, g〉H,

〈f, g〉D = −〈Pf, Pg〉H + 〈(1− P )f, (1 − P )g〉H,

for all f, g ∈ H, where P is the projection on VH. A Julia operator for T is givenby

(

T√

31− P −

√3P 2V

)

∈ B(H⊕D,H ⊕ D).

Its adjoint(

T ∗ 1− P +√

3P−√

3 2V ∗

)

∈ B(H⊕ D,H⊕D)

is a Julia operator for T ∗. Since T ∗ is a contraction, the minimal isometric dilationof T constructed by Theorem 3.1.2 is a bicontraction by the previous remarks.

3.2 Commutant Lifting

The background is now in place for a commutant lifting theorem forcontraction operators on Kreın spaces.

Theorem 3.2.1. Let H and K be Kreın spaces, and let T1 ∈ B(H)and T2 ∈ B(K) be contractions with minimal isometric dilations U1 ∈ B(H) andU2 ∈ B(K), respectively. Let A ∈ B(H,K) be a contraction satisfying

AT1 = T2A.

Then there exists a contraction A ∈ B(H, K) such that

AU1 = U2A

and A = PrK A|H.

Proof. Since any two minimal isometric dilations of a contraction areisomorphic by Theorem 3.1.3, we may assume that the minimal isometric dilationsof T1 and T2 have the form constructed in Theorem 3.1.2. In other words, we maytake

H = H⊕ D1 ⊕ D1 ⊕ · · · and K = K ⊕ D2 ⊕ D2 ⊕ · · ·

60

where D1, D2 are Hilbert spaces,

Uj =

Tj 0 0 0 · · ·D∗

j 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·

· · ·

, j = 1, 2,

and D1 ∈ B(D1,H), D2 ∈ B(D2,K) are defect operators for T1, T2. Let H =H+ ⊕H− and K = K+ ⊕ K− be fundamental decompositions, used, for example,to induce Hilbert space norms on H, H, K, K. Set K0 = K and

Kn = K ⊕ D2 ⊕ · · · ⊕ D2, n = 1, 2, . . .,

with n copies of D2 in the direct sum on the right. We construct contractionoperators

An : H → Kn, n = 0, 1, 2, . . .,

such that A = PrK An|H for all nonnegative integers n and

(α) AnU1 = U2An−1,(β) An−1 = PrKn−1

An

if n ≥ 1.Let A0 map any element (f, g1, g2, . . .)

t of H to the element Af of K0.Since A is a contraction and D1 is a Hilbert space, A0 is a contraction. By con-struction, A = PrK A0|H.

Assume that A0, . . ., Ar have the required properties. We determine Ar+1

as a matrix

Ar+1 =

(

C11 C12

C21 C22

)

,

relative to the decompositions

H = UH ⊕ (H ⊖ UH) and Kr+1 = Kr ⊕ D2.

The conditions (α) and (β) with n = r + 1 require that

(

C11

C21

)

U1 = U2Ar and ( C11 C12 ) = Ar

Define C11, C12, C21 by these equations. The two definitions of C11 are identicalin the case r = 0, because

A0U1 = PrK0U2A0.

61

In fact, this identity reduces to AT1 = T2A, which holds by hypothesis. In the caser ≥ 1, the two definitions of C11 are identical because

ArU1 = U2Ar−1 = U2PrKr−1Ar = PrKr

U2Ar.

Here the first two equalities are by (α) and (β) with n = r, and the third is anelementary property of the minimal isometric dilation.

Since Ar is a contraction and U1 and U2 are isometries, the operators

(

C11

C21

)

and (C11 C12 )

are contractions. By Theorem 2.4.4, there exists an operator C22 such that

Ar+1 =

(

C11 C12

C21 C22

)

is a contraction. By construction, (α) and (β) hold with n = r + 1. Finally,

PrK Ar+1|H = PrK PrKrAr+1|H = PrK Ar|H = A.

This completes the inductive construction of operators A0, A1, . . . with the statedproperties.

We show that ‖An‖2 ≤ 1+2‖A‖2 for all n. This is true for n = 0 because

‖A0‖2

= ‖A‖2 by the definition of A0. For n ≥ 1, we obtain A0 = PrK An byrepeated application of (β). Hence we may write

An =

(

A0

X

)

∈ B(H,K ⊕ (Kn ⊖K)),

where X ∈ B(H,Kn ⊖ K). Since A0 and An are contractions and Kn ⊖ K is aHilbert space, Corollary 2.2.3 yields

‖An‖2 ≤ 1 + 2‖A0‖

2= 1 + 2‖A‖2,

which proves the assertion.By (β), A∗

n−1 = A∗n|Kn−1 for all n ≥ 1. Hence there is an operator A ∈

B(H, K) such that ‖A‖2 ≤ 1 +2‖A‖2 and A∗n = A∗|Kn, equivalently, An = PrKn

Afor every nonnegative integer n. It follows that limn→∞ Anf = Af, f ∈ H,and from this we verify without difficulty that A has the required properties.

62

The norm estimate obtained in the proof of Theorem 3.2.1, namely,

‖A‖2 ≤ 1 + 2‖A‖2,

holds for any operator A having the properties stated in the theorem (it is notspecial to the construction in the proof). Moreover, arbitrary fundamental decom-positions of H and K may be used in the computation of norms. To see this, recallthat by Theorem 3.1.3 (ii),

H = H⊕ (H ⊖ H) and K = K ⊕ (K ⊖ K),

where H ⊖H and K ⊖ K are Hilbert spaces. Thus fundamental decompositions ofH and K determine fundamental decompositions of H and K. The norm estimatefollows from Corollary 2.2.3 by an argument used in the proof of Theorem 3.2.1.

Theorem 3.2.1 has a companion for bicontractions, which is deduced asan immediate consequence.

Theorem 3.2.2. Let H and K be Kreın spaces, and let T1 ∈ B(H)and T2 ∈ B(K) be contractions with minimal isometric dilations U1 ∈ B(H) andU2 ∈ B(K), respectively. Let A ∈ B(H,K) be a bicontraction satisfying

AT1 = T2A.

Then there exists a bicontraction A ∈ B(H, K) such that

AU1 = U2A

and A = PrK A|H.

Proof. Let N1 be a maximal negative subspace of H, and let N2 be amaximal uniformly negative subspace of K. Since A is a bicontraction, AN1 is amaximal negative subspace of K by Theorem 1.3.6, and so

PrN2AN1 = N2.

By Theorem 3.1.3 (ii), N1 is a maximal negative subspace of H andN2 is a maximaluniformly negative subspace of K. Since A is a contraction, AN1 is a negativesubspace of K. It is maximal negative in K because

PrN2AN1 = PrN2

PrK AN1 = PrN2AN1 = N2.

Therefore A is a bicontraction by Theorem 1.3.6.

With stronger hypotheses, it is also possible to lift an intertwining relationto minimal unitary dilations.

63

Theorem 3.2.3. Let H and K be Kreın spaces, and let T1 ∈ B(H)and T2 ∈ B(K) be bicontractions with minimal unitary dilations U1 ∈ B(H) andU2 ∈ B(K), respectively. Let A ∈ B(H,K) be a bicontraction satisfying

AT1 = T2A.

Then there exists a bicontraction A ∈ B(H, K) such that

AU1 = U2A

and A = PrK A|H.

Proof. By Theorem 3.1.7, minimal unitary dilations of bicontractions areunique up to isomorphism. Therefore we may assume that U1, U2 are constructedas in Theorem 3.1.5. Let V1 ∈ B(H′), V2 ∈ B(K′) be minimal isometric dilationsof T1, T2. We take U∗

1 , U∗2 to be minimal isometric dilations of V ∗

1 , V ∗2 .

By Theorem 3.2.2, there exists a bicontraction A′ ∈ B(H′,K′) such that

A′V1 = V2A′

and A = PrK A′H. The operators

T ′1 = V ∗

2 ∈ B(K′) and T ′2 = V ∗

1 ∈ B(H′)

are contractions by Theorem 3.1.3 (iii), and A′∗ ∈ B(K′,H′) is a bicontractionsatisfying

A′∗T ′1 = T ′

2A′∗.

Since U∗2 and U∗

1 are minimal isometric dilations of T ′1 and T ′

2, respectively, asecond application of Theorem 3.2.2 produces a bicontraction A∗ ∈ B(K, H) suchthat

A∗U∗2 = U∗

1 A∗

and A′∗ = PrH′ A∗|K′. Thus A ∈ B(H, K) is a bicontraction satisfying AU1 =U2A. Finally, we have A′ = PrK′ A|H′, and restricting this relation further to H,we obtain

A = PrK A′|H = PrK A|H.

Thus A has the required properties.

64

3.3 Characterization of Extensions

We do not solve the problem of labeling all operators produced by thecommutant lifting theorem, but we give some information in this direction.

Let H and K be Kreın spaces, and let T1 ∈ B(H) and T2 ∈ B(K) becontractions with minimal isometric dilations U1 ∈ B(H) and U2 ∈ B(K). Assumethat A ∈ B(H,K) is a contraction such that

AT1 = T2A.

By Theorem 3.1.3, H⊥ = H ⊖ H and K⊥ = K ⊖ K are uniformly positive in theKreın spaces H and K.

Lemma 3.3.1. If A ∈ B(H, K), AU1 = U2A, and PrK A|H = A, thenAH⊥ ⊂ K⊥ and therefore A∗|K = A∗.

Proof. By Theorem 3.1.3, we may choose U1 and U2 as in Theorem 3.1.2.Thus

H = H⊕ D1 ⊕ D1 ⊕ · · · and K = K ⊕ D2 ⊕ D2 ⊕ · · ·where D1, D2 are Hilbert spaces,

Uj =

Tj 0 0 0 · · ·D∗

j 0 0 0 · · ·0 1 0 0 · · ·0 0 1 0 · · ·

· · ·

, j = 1, 2,

and D1 ∈ B(D1,H), D2 ∈ B(D2,K) are defect operators for T1, T2. WritingA = (Ajk)∞j,k=1 relative to the same decompositions and comparing entries in the

relation AU1 = U2A, we find that A11 = A and A1n = 0 for n ≥ 1. In particular,AH⊥ ⊂ K⊥.

Let L be the direct sum of the anti-space of H together with K, so L isthe space of pairs

(

fg

)

with f in H and g in K, and

⟨(

fg

)

,

(

fg

)⟩

L

= −〈f, f〉H + 〈g, g〉K, f ∈ H, g ∈ K.

Since A ∈ B(H,K) is a contraction, its graph

G(A) =

{(

fAf

)

: f ∈ H}

65

is a closed negative subspace of L.Given a contraction A ∈ B(H, K), we define L and G(A) in a parallel

way, so L is the space of pairs(

fg

)

with f in H and g in K, and

⟨(

fg

)

,

(

fg

)⟩

L

= −〈f, f〉H + 〈g, g〉K, f ∈ H, g ∈ K.

The graph

G(A) =

{(

fAf

)

: f ∈ H}

of A is a closed negative subspace of L. If

U =

(

U1 00 U2

)

∈ B(L),

the relation AU1 = U2A holds if and only if UG(A) ⊂ G(A).Choose fundamental decompositions H = H+ ⊕ H−, K = K+ ⊕ K−,

L = L+ ⊕ L−, L = L+ ⊕ L− in a consistent way such that

L+ = K+ ⊕H−, L− = H+ ⊕K−,

andL+ = L+ ⊕K⊥ = K+ ⊕H− ⊕K⊥,

L− = L− ⊕H⊥ = H+ ⊕K− ⊕H⊥.

Notice that L⊥ = L ⊖ L is given by L⊥ = H⊥ ⊕K⊥.

Theorem 3.3.2. In the preceding situation, the relation

M = G(A)

establishes a one-to-one correspondence between all contractions A ∈ B(H, K) suchthat AU1 = U2A and PrK A|H = A and all closed negative U -invariant subspacesM of L satisfying

(i) M⊂ G(A) + L⊥, and(ii) PrL−

M = PrL−

[

G(A) + L⊥]

.

Proof. Assume that A ∈ B(H, K) is a contraction, AU1 = U2A, andPrK A|H = A, and defineM = G(A). ClearlyM is a closed negative U -invariantsubspace of L. For any f ∈ H,

(

f

Af

)

=

(

(PrH + PrH⊥)f

(PrK + PrK⊥)A(PrH + PrH⊥)f

)

=

(

PrHf

APrHf

)

+

(

PrH⊥f

PrK⊥Af

)

66

by Lemma 3.3.1, and (i) holds.Let P± be the projections of H onto H±, Q± the projections of K onto

K±. For any f ∈ H,

PrL−

(

f

Af

)

=

(

P+PrHf + PrH⊥f

Q−APrHf

)

= (P+ + Q−A)PrHf + PrH⊥f

by Lemma 3.3.1. Therefore

PrL−M = ran (P+ + Q−A) +H⊥.

In a similar way,

PrL−

[

G(A) + L⊥]

= PrH+⊕K−

{(

fAf

)

: f ∈ H}

+H⊥

= ran (P+ + Q−A) +H⊥,

and (ii) follows.Conversely, let M be a closed negative U -invariant subspace of L satis-

fying (i) and (ii). If g ∈ K and(

0g

)

∈M, then by (i) we can write

(

0g

)

=

(

uAu

)

+ h + k

with u ∈ H, h ∈ H⊥, k ∈ K⊥. Projecting onto H and H⊥, we find that u = 0and h = 0. Therefore g = k ∈ K⊥. But M is a negative subspace of L, and K⊥

is uniformly positive in L, so g = 0. It follows that M = G(A) is the graph of alinear transformation A with domain in H and range in K.

We use conditions (i) and (ii) to show that dom A = H. For any f ∈dom A, by (i) there exist u ∈ H, h ∈ H⊥, k ∈ K⊥ such that

(

f

Af

)

=

(

uAu

)

+ h + k.

Projecting onto H, H⊥, K⊥, we find that u = PrHf , h = PrH⊥f , k = PrK⊥ Af ,and so

(

f

Af

)

=

(

PrHfAPrHf

)

+ PrH⊥f + PrK⊥ Af (3.3.1)

and

PrL−

(

f

Af

)

= PrH+⊕K−⊕H⊥

{(

PrHfAPrHf

)

+ PrH⊥f + PrK⊥ Af

}

= (P+ + Q−A)PrHf + PrH⊥f.

67

But since

PrL−M = PrL−

[

G(A) + L⊥]

= ran (P+ + Q−A) +H⊥

by (ii), it follows that dom A = H.Since M is closed, A is continuous by the closed graph theorem. That

is, A ∈ B(H, K). Since M is negative, A is a contraction. The invariance of Munder A implies that AU1 = U2A. The identity (3.3.1) implies that PrK A|H = A.

The correspondence between operators and their graphs is obviously one-to-one, and so the result is proved.

The result takes a simpler form when A is a bicontraction. As above,T1 ∈ B(H) and T2 ∈ B(K) are contractions with minimal isometric dilationsU1 ∈ B(H) and U2 ∈ B(K).

Theorem 3.3.3. Assume that A is a bicontraction. Then the relation

M = G(A)

establishes a one-to-one correspondence between all bicontractions A ∈ B(H, K)such that AU1 = U2A and PrK A|H = A and all maximal negative U -invariantsubspacesM of L satisfyingM⊂ G(A) + L⊥.

Proof. Let A ∈ B(H, K) be a bicontraction such that AU1 = U2A andPrK A|H = A. By Theorem 3.3.2, G(A) is a negative U -invariant subspace of LsatisfyingM⊂ G(A) + L⊥. In addition,

PrL−M = PrL−

[

G(A) + L⊥]

= ran (P+ + Q−A) +H⊥,

where the last equality follows as in the proof of Theorem 3.3.2. Since A is abicontraction, ran (P+ + Q−A) = L− by Theorems 1.3.3 and 1.3.4. ThereforePrL−

M = L− +H⊥ = L−, andM is maximal negative.

Conversely, letM be a maximal negative U -invariant subspace of L sat-isfyingM⊂ G(A) + L⊥. ThenM is closed and

PrL−M = L− = ran (P+ + Q−A) +H⊥ = PrL−

[

G(A) + L⊥]

.

By Theorem 3.3.2,M = G(A) is the graph of a contraction A ∈ B(H, K) such thatAU1 = U2A and PrK A|H = A. But A is a bicontraction, and so

PrK−AH− = PrK−

AH− = K−.

Since K− is maximal uniformly negative in K, the operator A is a bicontractionby Theorem 1.3.6.

68

3.4 Abstract Leech Theorem

We use a theorem of Shmul’yan on the factorization of bicontractions.

Theorem 3.4.1. If A, B, C are Kreın spaces and A ∈ B(A, C) andB ∈ B(B, C) are bicontractions, then A = BC for some bicontraction C ∈ B(A,B)if and only if AA∗ ≤ BB∗.

Proof. The condition is necessary: if A = BC with C a bicontraction,then AA∗ = BCC∗B∗ ≤ BB∗.

Conversely, assume AA∗ ≤ BB∗. We show that kerB∗ ⊂ kerA∗. Supposef ∈ kerB∗ and f 6= 0. Let C− be a maximal uniformly negative subspace of C.Then f /∈ C− because kerB∗ is a uniformly positive subspace of C by Theorem1.3.1. Let C− be the span of f and C−, so C− is properly contained in C−. Notethat A∗C− is a negative subspace of A. For if g = αf + h where α is a complexnumber and h is in C−, then since AA∗ ≤ BB∗ by assumption,

〈A∗g,A∗g〉A ≤ 〈B∗g,B∗g〉B = 〈B∗h,B∗h〉B ≤ 0. (3.4.1)

Since A∗C− ⊃ A∗C− and A∗C− is maximal negative by Theorem 1.3.6, A∗C− =A∗C−. Therefore A∗ annihilates some nonzero element g = αf + h of C−, whereα is a complex number and h ∈ C−. For such g, equality holds in (3.4.1), and soh = 0 and α 6= 0. Hence f ∈ kerA∗. It follows that kerB∗ ⊂ ker A∗.

The inclusion on kernels allows us to construct a linear transformationX0 on ranB∗⊕kerB into A such that A∗ = X0B

∗ and X0 annihilates kerB. SinceAA∗ ≤ BB∗ and kerB is uniformly positive by Theorem 1.3.1, we obtain

〈X0u,X0u〉A ≤ 〈u, u〉B, u ∈ domX0.

The domain of X0 is dense in B and contains B∗C−, which is maximal uniformlynegative in B by Theorem 1.3.6. Moreover, X0 maps B∗C− onto A∗C−, which ismaximal uniformly negative inA. By Theorem 1.4.4, X0 extends to a bicontractionC∗ ∈ B(B,A). Then C ∈ B(A,B) is a bicontraction and A = BC.

Leech’s theorem is a structured form of Shmul’yan’s theorem.

Theorem 3.4.2. Let A, B, C be Kreın spaces, SA ∈ B(A), SB ∈ B(B),SC ∈ B(C) isometries, and A ∈ B(A, C), B ∈ B(B, C) bicontractions. Assume that

ASA = SCA and BSB = SCB.

Then A = BC for some bicontraction C ∈ B(A,B) satisfying

CSA = SBC

if and only if AA∗ ≤ BB∗.

69

The case in which SA, SB, SC coincide with the identity operators isShmul’yan’s theorem.

Proof. Necessity follows from Theorem 3.4.1.Conversely, assume AA∗ ≤ BB∗. Set H = A∗C ⊂ A and K = B∗C ⊂ B.

Since A and B are contractions,H⊥ = kerA andK⊥ = kerB are uniformly positiveby Theorem 1.3.1. In particular, H and K are regular subspaces of A and B andmay be viewed as Kreın spaces in the scalar products of A and B. By Shmul’yan’stheorem there is a bicontraction X ∈ B(A,B) such that A = BX. Define

T1 = PrH SA|H ∈ B(H), T2 = PrK SB|K ∈ B(K),

andY = PrK X|H ∈ B(H,K).

Then T1 and T2 are contractions, and Y is a bicontraction. Since S∗AH ⊂ H and

S∗BK ⊂ K, we have T ∗

1 = S∗A|H and T ∗

2 = S∗B|K. We also have X∗|K = Y ∗ because

X∗B∗ = A∗, and so

Y ∗T ∗2 B∗ = X∗S∗

BB∗ = X∗B∗S∗C

= A∗S∗C = S∗

AA∗ = S∗AX∗B∗ = T ∗

1 Y ∗B∗.

Therefore Y T1 = T2Y .Next note that

H =∞∨

0

SnAH and K =

∞∨

0

SnBK

are regular subspaces of A and B and hence Kreın spaces in the scalar products ofA and B. For example, we have

H ⊂ H ∨ SAH ⊂ H ∨ SAH ∨ S2AH ⊂ · · ·,

where at each stage the extension is obtained by forming a direct sum with aHilbert space. Therefore each subspace in the chain is regular. By Lemma 1.1.9,H is a regular subspace of A. In a similar way, K is a regular subspace of B.Note that H is invariant under SA and S∗

A, and K is invariant under SB and S∗B.

Therefore U1 = SA|H and U2 = SB|K are minimal isometric dilations of T1 andT2.

By Theorem 3.2.2, there is a bicontraction Y ∈ B(H, K) such that Y U1 =U2Y and PrK Y |H = Y . By Lemma 3.3.1, Y ∗|K = Y ∗. Set

C = Y PrH ∈ B(A,B).

70

Then C = PrKY PrH, C∗ = PrHY ∗PrK, and

X∗|K = Y ∗ = Y ∗|K = C∗|K.

So from A∗ = X∗B∗ we obtain A∗ = C∗B∗, hence A = BC. Since Y is abicontraction and the orthogonal complements of H and K inA and B are uniformlypositive, C is a bicontraction. Finally,

CSA|H = Y U1 = U2Y = SBC|H,

whereasCSA|H⊥ = 0 = SBC|H⊥.

Thus CSA = SBC.

Notes on Chapter 3

The dilation theory in §3.1 follows the methods of the Hilbert space case as givenin Sz.-Nagy and Foias [66], with some modifications. An extension of dilation theory toarbitrary operators was first made by Davis [23]; see Bognar [12]. In the indefinitesetting, dilation theory is used by Bruinsma, Dijksma, and de Snoo [20], Constantinescuand Gheondea [21,22], and Dritschel [29].

The commutant lifting theorem in the Hilbert space case is given in Sz.-Nagyand Foias [66]. It was inspired by applications in interpolation theory due to Sarason [60].There have been numerous accounts and applications; references are given in Rosenblumand Rovnyak [59]. In the indefinite case, extensions of the commutant lifting theoremare proved in Alpay [1], de Branges [14,16,17], Constantinescu and Gheondea [21,22],and Dritschel [29,30] in various settings. The general case of Theorem 3.2.1 first appearsin Dritschel [30]. The corresponding result on bicontractions, Theorem 3.2.2, is due toDritschel [29]; different proofs and extensions are given in Constantinescu and Gheondea[22] and de Branges [17]. Theorem 3.2.3 is new.

The use of matrix completions in commutant lifting problems is a known wayto organize calculations in the Hilbert space case. The idea seems to be due to Parrott[54]. It has been adopted by other authors, including Frazho [35] and Ptak and Vrbova[56]. The method is also used in Dritschel [29,30] and Constantinescu and Gheondea[21,22].

Theorems 3.3.2 and 3.3.3 appear in Dritschel [30] and [29], respectively. Thegraph approach is due to Ball and Helton [11]. Much has been written on the labelingproblem in the Hilbert space case. See Arsene, Ceausescu, and Foias [6], Foias and Frazho[34], and Helton et al. [39]. A comparison of methods is given in Frazho [35].

Leech’s original theorem [51] concerns the factorization of power series withoperator coefficients in the form A(z) = B(z)C(z). Helton [39, p. 52] gives a version forbounded measurable matrix valued functions. Rosenblum [58] proves an abstract formof Leech’s theorem using the commutant lifting theorem (see also [59]). Leech’s theoremwas generalized to Kreın spaces by de Branges [14,16]. The version in Theorem 3.4.2appears in Dritschel [30] and extends the abstract result in Rosenblum [58].

71

Appendix A: Complementation Theory

The operator methods in this paper are related to a theory of comple-mentation in Kreın spaces due to de Branges [13,15]. The Kreın space version ofcomplementation extends Hilbert space notions which appear in de Branges andRovnyak [18,19]. We state without proof some results on an operator approachto complementation, with a full account to appear in a later paper.

A Kreın space P is contained continuously , contractively , or isometrically

in a Kreın space H if P is a linear subspace of H and the inclusion mapping iscontinuous, contractive, or isometric, respectively. Let P be a Kreın space whichis contained continuously in a Kreın space H, and let A be the inclusion mapping.We associate with P the selfadjoint operator P ∈ B(H) given by

P = AA∗.

The operator P plays the role of a generalized projection for P. Viewed as map-pings on H, the operators P and A∗ have the same action. In the terminology ofde Branges, P is the selfadjoint operator on H which coincides with the adjointof the inclusion of P in H. The positive and negative indices of the Kreın spaceP coincide with the hermitian indices h±(P ) of the operator P by Theorem 1.2.1.The range P0 of P is a scalar product space with scalar product defined by

〈Pf, Pg〉P0= 〈Pf, g〉H, f, g ∈ H.

The space P0 is contained in P as a dense subspace, and the P0 and P scalarproducts coincide on P0. Every selfadjoint operator P arises in this way.

Theorem A1. LetH be a Kreın space, and let P ∈ B(H) be a selfadjointoperator. Write P in any way in the form

P = EE∗,

where E ∈ B(E ,H) for some Kreın space E and E has zero kernel. Let PE be therange of E viewed as a Kreın space in the scalar product which makes E a Kreınspace isomorphism of E onto PE . Then PE is contained continuously in H, andthe adjoint of the inclusion of PE in H coincides with P .

Special properties hold whenever the operators P and Q for two spaces Pand Q satisfy P +Q = 1. In the special case of isometric inclusion, these propertiesreflect the fact that P and Q are regular subspaces of the larger Kreın space H,and H decomposes into the orthogonal direct sum of P and Q.

72

Theorem A2. Let P and Q be Kreın spaces which are contained con-tinuously in a Kreın space H, and let P and Q be the selfadjoint operators on Hwhich coincide with the adjoints of the inclusions of P and Q in H. Assume thatP + Q = 1.

(i) The mapping U : (f, g)→ f + g is a partial isometry from P ×Q onto Hwith adjoint U∗ : h→ (Ph,Qh).

(ii) The intersection L of P and Q is a Kreın space in the scalar productdefined by

〈f, g〉L = 〈f, g〉P + 〈f, g〉Q, f, g ∈ L.

The Kreın space L is called the overlapping space for P and Q. It iscontained continuously in H, and the adjoint of the inclusion coincideswith PQ.

(iii) The following conditions are equivalent: (a) P is contained contractivelyin H, (b) Q is contained contractively in H, (c) P 2 ≤ P , (d) Q2 ≤ Q,(e) U is a contraction, and (f) the overlapping space L is a Hilbert space.

(iv) The following conditions are equivalent: (a) P and Q are containedisometrically in H as regular subspaces with Q = P⊥, (b) P 2 = P ,(c) Q2 = Q, (d) U is an isometry, and (e) the overlapping space L con-tains no nonzero element.

The ideas in Theorem A2 go back to Schwartz [63], who created a theoryof operator ranges both in the Hilbert space and Kreın space settings. The authorsthank Daniel Alpay for calling their attention to Schwartz’s paper; a related workis Alpay [2]. It is of interest to know when a unique Kreın space P is associatedwith a given selfadjoint operator P . The following uniqueness condition, using anextra hypothesis, is similar to one given by Schwartz [63].

Theorem A3. LetH be a Kreın space, and let P ∈ B(H) be a selfadjointoperator. Let P1 and P2 be Kreın spaces which are contained continuously in Hsuch that the adjoints of the inclusions each coincide with P . If P1 is containedcontinuously in P2, then P1 and P2 are equal isometrically.

Uniqueness holds under a condition of a different nature.

Theorem A4. LetH be a Kreın space, and let P ∈ B(H) be a selfadjointoperator. Let P1 be a Kreın space which is contained continuously in H such thatthe adjoint of the inclusion of P1 in H coincides with P . Assume that the rangeof P contains a subspace M which is maximal uniformly definite in P1. Then ifP2 is any Kreın space which is contained continuously in H such that the adjointof the inclusion of P2 in H coincides with P , P1 and P2 are equal isometrically.

73

Corollary A5. Let P be a Kreın space which is contained continuouslyin a Kreın space H, and let P ∈ B(H) be the selfadjoint operator which coincideswith the adjoint of the inclusion of P in H. Assume that the range of P containsa subspaceM which is maximal uniformly definite in P. If

P = EE∗,

where E ∈ B(E ,H) for some Kreın space E and E has zero kernel, then E is aKreın space isomorphism of E onto P.

The condition for uniqueness in Theorem A4 is always satisfied in theimportant special case of contractive inclusion. This yields a new derivation of aresult of de Branges [13].

Theorem A6. Let P1 and P2 be Kreın spaces which are contained con-tinuously and contractively in a Kreın space H such that the adjoints of the in-clusions each coincide with the selfadjoint operator P on H. Then P1 and P2 areequal isometrically.

Combining Theorem A6 and Corollary 5, we see that the Kreın spaces inde Branges’ theory of complementation [13] are operator ranges. In the indefinitecase, this was first shown by Heinz Langer (private communication, 1988) using histheory of definitizable operators and spectral functions [49]. In Hilbert spaces, thishas been known for a long time. It was shown, for example, in seminar lectures byMarvin Rosenblum at the University of Virginia in the 1960’s. In Hilbert spaces,Sarason has used the operator view in applications to function theory in a seriesof papers including [61,62].

The choice of a contraction operator leads to an example of complemen-tation Kreın spaces.

Theorem A7. Let H and K be Kreın spaces, and let T ∈ B(H,K) be acontraction. DefineM(T ) to be the range of T in the scalar product which makesT a partial isometry of H onto M(T ). Let D ∈ B(D,K) be any defect operatorfor T ∗. Define H(T ) to be the range of D in the scalar product which makes D aKreın space isomorphism of D onto H(T ). ThenM(T ) and H(T ) are Kreın spaceswhich are contained continuously and contractively in K, and the adjoints of theinclusions coincide with TT ∗ and 1− TT ∗, respectively. The definition of H(T ) isindependent of the choice of defect operator D for T ∗. An element g of K belongsto H(T ) if and only if

supu∈H

[〈g + Tu, g + Tu〉K − 〈u, u〉H] <∞,

in which case the value of the supremum is 〈g, g〉H(T ).

74

Appendix B: More on Julia Operators

The existence of a Julia operator was proved in §1.2 by an argumentbased on the factorization of any selfadjoint operator in the form AA∗, where Ais an operator having zero kernel. We present an alternative proof following theoriginal method due to Arsene, Constantinescu, and Gheondea [8]. The methodgives the additional information that a particular choice of Julia operator satisfiesnorm estimates (Theorem B3). We also show that uniqueness holds under weakerassumptions than that of Theorem 2.4.5 (Theorem B4).

Two preliminary Hilbert space results are needed. As usual, we write ∗for Kreın space adjoint and × for Hilbert space adjoint. If H and K are Hilbertspaces and T is a continuous everywhere defined operator on H to K, then by thepolar factorization of T we mean the representation T = RU , where U is a partialisometry on H to K with kernel equal to the kernel of T and R is a nonnegativeoperator on K which is zero on the orthogonal complement of the range of U .

Lemma B1. Let H be a Kreın space, and let H ∈ B(H) be a selfadjointoperator. Assume that H = H+ ⊕ H− is a fundamental decomposition. Thenthere is a Kreın space A with fundamental decomposition A = A+ ⊕ A− and anoperator A ∈ B(A,H) with these properties:

(i) A has zero kernel and H = AA∗;(ii) if A = RW is the polar factorization of A as an operator on A+ ⊕ |A−|

to H+ ⊕ |H−|, then AA∗ = R2WW ∗ and R2 = AA∗(WW ∗)×.

Proof. The operator A constructed in the proof of Theorem 1.2.2 satisfies(i). We show that A also satisfies (ii).

The representation A = RW in the proof of Theorem 1.2.2 is the polarfactorization of A as an operator on A+ ⊕ |A−| to H+ ⊕ |H−|. The operatorWJAW× is 1 onM+, −1 onM−, and 0 on the orthogonal complement ofM+ +M− in H+ ⊕ |H−|. Hence R(WJAW×)R = R2(WJAW×), and

AA∗ = RWW ∗R∗ = R(WJAW×)RJH = R2(WJAW×)JH = R2WW ∗.

Since WW ∗(WW ∗)× = WW× is the projection of H+ ⊕ |H−| onto ranR,

AA∗(WW ∗)× = R2WW ∗(WW ∗)× = R2,

and so (ii) holds.

We use a result due to Kreın [41] and Reid [57] and rediscovered by Lax[50] and Dieudonne [27]. We include Reid’s proof for completeness.

75

Lemma B2. Let H be a Hilbert space and A,X ∈ B(H). If A is nonneg-ative and AX is selfadjoint, then for all f ∈ H,

|〈AXf, f〉| ≤ ‖X‖〈Af, f〉.

Proof. It is enough to give the proof when ‖X‖ = 1. Since A is nonneg-ative, for any f and g in H,

|〈Af, g〉| ≤ 〈Af, f〉1/2〈Ag, g〉1/2 ≤ 1

2[〈Af, f〉 + 〈Ag, g〉] .

Since AX = X∗A, for any positive integer n,

|〈AXnf, f〉| ≤ 1

2[〈AXnf,Xnf〉+ 〈Af, f〉] =

1

2

[⟨

AX2nf, f⟩

+ 〈Af, f〉]

.

Iteration of this inequality yields

|〈AXf, f〉| ≤ 2−n⟨

AX2n

f, f⟩

+

(

1

2+

1

4+ · · ·+ 1

2n

)

〈Af, f〉,

and we obtain the result on letting n tend to ∞.

Theorem B3. LetH and K be Kreın spaces with fundamental decompo-sitions H = H+⊕H− and K = K+⊕K−. For any T ∈ B(H,K), there exist Kreınspaces D and D with fundamental decompositions D = D+⊕D− and D = D+⊕D−

and a Julia operator

U =

(

T DD∗ L

)

∈ B(H⊕D,K ⊕ D).

such that

‖L‖ ≤ ‖T‖

and

max (‖D‖, ‖D‖) ≤[

1 + ‖T‖2]1/2

.

Proof. By Lemma B1, there exist Kreın spacesD and D with fundamentaldecompositions D = D+⊕D− and D = D+⊕D−, and operators D ∈ B(D,K) andD ∈ B(D,H) with zero kernels, such that

1− TT ∗ = DD∗, 1− T ∗T = DD∗, (B− 1)

76

and the polar representations D = RW and D = RW as Hilbert space operatorssatisfy

1− TT ∗ = R2WW ∗, R2 = (1− T ∗T )(W W ∗)×.

In particular, ‖D‖ and ‖D‖ are bounded by[

1 + ‖T‖2]1/2

.

We construct L ∈ B(D, D) with the aid of Lemma B2, applied to A = R2

and X = (WW ∗)T (W W ∗)×T× viewed as operators on K+ ⊕ |K−|. Clearly A isnonnegative, and

AX = R2(WW ∗)T (W W ∗)×T× = (1− TT ∗)T (W W ∗)×T×

= T (1− T ∗T )(W W ∗)×T× = TR2T×

is selfadjoint. Since ‖X‖ ≤ ‖T‖2, Lemma B2 yields

∣

∣

∣

⟨

R2Xf, f⟩

K+⊕|K−|

∣

∣

∣≤ ‖T‖2

⟨

R2f, f⟩

K+⊕|K−|, f ∈ K.

In other words, TR2T× ≤ ‖T‖2R2 as operators on K+⊕|K−|. Therefore TR = RCwhere C ∈ B(H,K) and ‖C‖ ≤ ‖T‖. We can choose C so that ranC ⊂ ranR, andthen TRW = RWW×CW . We obtain

TD = −DL∗, (B− 2)

where L = −(W×CW )∗ ∈ B(D, D) and ‖L‖ ≤ ‖T‖.It remains to prove the six identities in (1.2.1) and (1.2.2). The relations

(1.2.1a) and (1.2.2a,b) hold by (B-1) and (B-2). By (1.2.2b),

DLD∗ = −DD∗T ∗ = −(1− T ∗T )T ∗ = −T ∗(1− TT ∗) = −T ∗DD∗.

Since kerD = {0}, (1.2.1b) follows. Similarly,

DL∗L = −TDL = TT ∗D = (1−DD∗)D = D(1−D∗D),

DLL∗ = −T ∗DL∗ = T ∗TD = (1− DD∗)D = D(1− D∗D),

and (1.2.1c) and (1.2.2c) hold.

We proved uniqueness of Julia operators in Theorem 2.4.5 under the as-sumption that either T or T ∗ is a contraction. The conclusion can be obtainedwith a weaker hypothesis, namely, that one of the operators 1− T ∗T or 1− TT ∗

has at most a finite number of negative squares.

77

Theorem B4. Let H and K be Kreın spaces, and let T ∈ B(H,K) havetwo Julia operators

(

T Dj

D∗j Lj

)

∈ B(H⊕Dj ,K ⊕ Dj), j = 1, 2.

If one of the indices h−(1− T ∗T ), h−(1− TT ∗), h+(1− T ∗T ), or h+(1− TT ∗) isfinite, then there exist unitary operators V ∈ B(D2,D1) and V ∈ B(D2, D1) suchthat

(

T D2

D∗2 L2

)

=

(

1 00 V ∗

) (

T D1

D∗1 L1

)(

1 00 V

)

.

Lemma B5. Any dense subspaceM of a Pontryagin space H contains amaximal uniformly negative subspace.

Proof of Lemma B5. We prove the assertion by showing that if a uniformlynegative subspace N ofM is not maximal, then there exists a uniformly negativesubspace N ′ of M which properly contains N . Granting this, then starting withN = {0} one can construct in a finite number of steps a maximal uniformly negativesubspace ofM.

Assume that N is a uniformly negative subspace which is contained inMand that N is not maximal. Choose a fundamental decomposition H = H+ ⊕H−

such that N ⊂ H−. Norms are to be computed in the associated Hilbert spaceH+ ⊕ |H−|. Note that dimH− <∞ because H is a Pontryagin space.

Since N is not maximal, there is a unit vector e ∈ H− which is orthogonalto N . Since M is dense in H, we can choose φ ∈ M such that ‖e − φ‖ ≤ 1/3.Then ‖φ‖ ≥ 2/3. We show that the span N ′ of N and φ is uniformly negative.For any g ∈ N ′ we can write g = f + αφ where f ∈ N and α is a scalar. Then

〈g, g〉H = 〈f, f〉H + |α|2〈φ, φ〉H + 2 Re [α〈f, φ〉H]

= −‖f‖2 − |α|2‖φ‖2 + 2 Re [α〈f, φ− e〉H]

because e⊥f . Thus

〈g, g〉H ≤ −‖f‖2 − |α|2‖φ‖2 + 2|α|‖f‖‖φ− e‖

≤ −‖f‖2 − 4

9|α|2 + 2·1

3|α|‖f‖

= −[

‖f‖ − 1

3|α|

]2

− 1

3|α|2

≤ 0

with equality only if g = 0. Thus the form 〈g, g〉H is strictly negative on the unit

sphere of N ′, which is a compact set. Hence 〈g, g〉H ≤ −δ‖g‖2 for all g ∈ N ′ and

78

some δ > 0. Therefore N ′ is a subspace of M which is uniformly negative andproperly contains N .

Proof of Theorem B4. We assume that h−(1 − T ∗T ) < ∞. For the caseh−(1−TT ∗) <∞, we obtain the conclusion by replacing T by T ∗. The other casesare handled by easy modifications as noted below.

The hypothesis h−(1−T ∗T ) <∞ implies that D1 and D2 are Pontryaginspaces each with finite negative index equal to h−(1 − T ∗T ) (see the remarksfollowing Definition 1.2.3).

Since D1D∗1 = 1− T ∗T = D2D

∗2 , for any f, g ∈ H,

⟨

D∗1f, D∗

1g⟩

D1

=⟨

D∗2f, D∗

2g⟩

D2

.

It follows that there is a well defined and densely defined isometry X from D1 toD2 with domX = ran D∗

1 and ranX = ran D∗2 such that

XD∗1f = D∗

2f, f ∈ H.

Since D1 is a Pontryagin space, Lemma B5 implies that domX contains a maximaluniformly negative subspace D−

1 . But X is an isometry and D2 is a Pontryaginspace with the same negative index as that of D1. So XD−

1 is a maximal uniformlynegative subspace D−

2 of D2. Therefore by Theorem 1.4.4, X has an extension bycontinuity to a unitary operator V ∗ ∈ B(D1, D2). By construction, D∗

2 = V ∗D∗1 .

If we had assumed instead that one of the positive hermitian indices isfinite, then the same conclusion could be drawn. We simply replace the defectspaces by their anti-spaces and note that the densely defined isometry X is still anisometry on the new spaces. Then we apply the argument given above to concludethat X may be extended to a unitary operator.

We show next that(

D1

V ∗L1

)

∈ B(D1,K ⊕ D2)

is a defect operator for the adjoint of(

TD∗

2

)

∈ B(H,K⊕D2). The operator has zero

kernel because D1 has zero kernel. We have

1−(

TD∗

2

)

(T ∗ D2 ) = 1−(

TV ∗D∗

1

)

( T ∗ D1V )

=

(

1− TT ∗ −TD1V−V ∗D∗

1T ∗ 1− V ∗D∗1D1V

)

=

(

D1D∗1 D1L

∗1V

V ∗L1D∗1 V ∗L1L

∗1V

)

=

(

D1

V ∗L1

)

( D∗1 L∗

1V ) ,

79

which proves the assertion.Now to complete the proof, think of

(

TD∗

2

D2

L2

)

as (A B), where A =(

TD∗

2

)

∈ B(H,K ⊕ D2) and B =(

D2

L2

)

∈ B(D2,K ⊕ D2). The operator A is an

isometry and hence a contraction. The operator A∗ has defect operator

(

D1

V ∗L1

)

∈ B(D1,K ⊕ D2)

by what was shown above. Since (A B) is unitary and hence a bicontraction,Theorem 2.3.3 yields a bicontraction V ∈ B(D2,D1) such that

(

D2

L2

)

=

(

D1

V ∗L1

)

V.

We see easily that V is unitary. In fact,

D1V V ∗D∗1 = D2D

∗2 = 1− TT ∗ = D1D

∗1 ,

and since D1 has zero kernel, this implies that V V ∗ = 1, that is, V ∗ is an isome-try. But D2 = D1V and therefore V has zero kernel. By elementary properties ofisometries and partial isometries described in §1.1(C), V is unitary. By construc-tion,

(

T D2

D∗2 L2

)

=

(

T D1VV ∗D∗

1 V ∗L1V

)

,

as required.

After this paper was completed, the authors obtained a copy of AgnesYang’s thesis [68]. This includes a stronger form of Kreın ’s theorem, Lemma B2,which is due to Dijksma, Langer, and de Snoo (“Unitary colligations in Krein spacesand their role in the extension theory of isometries and symmetric linear relationsin Hilbert spaces,” Functional Analysis, II (Dubrovnic 1985), pp. 1–42, LectureNotes in Math., Vol. 1242, Springer, Berlin-New York, 1987; MR 89a:47055). Thestronger form of Kreın ’s theorem is essentially no more difficult to prove, and byappropriate choices of operators one obtains a somewhat more direct proof of theconstruction of Julia operators in Theorem B3.

80

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Department of Mathematics Department of MathematicsMathematical Sciences Building Mathematics-Astronomy BuildingPurdue University University of VirginiaWest Lafayette, Indiana 47907 Charlottesville, Virginia 22903

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