Structure of Hilbert Space Operators

s t r u c t u r e o f

Operators

Chunlan Jiang Zongyao Wang

s t r u c t u r e o f Hilbert Space Operators

Chunlan Jiang Hebei Normal University, China

Zongyao Wang East China University of Science and Technology, China

s t r u c t u r e o f

Hilbert Space Operators

Y ^ World Scientific NEWJERSEY • LONDON • SINGAPORE • BEIJING • SHANGHAI • HONGKONG • TAIPEI • CHENNAI

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STRUCTURE OF HILBERT SPACE OPERATORS

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Preface

In the matrix theory of finite dimensional space, the famous Jordan Standard Theorem sufficiently reveals the internal structure of matrices. Jordan Theorem indicates that the eigenvalues and the generalized eigenspace of matrix determine the complete similarity invariants of a matrix. It is obvious that the Jordan block in matrix theory plays a fundamental and important role. When we consider a complex, separable, infinite dimensional Hilbert space H and use C{7i) to denote the class of linear bounded operators on H, we face one of the most fundamental problems in operator theory, that is how to build up a theorem in /3(H) which is similar to the Jordan Standard Theorem in matrix theory, or how to determine the complete similarity invariants of the operators. Two operators A and B in C(H) are said to be similar if there is an invertible operator X, XA is equal to BX. The complexity of infinite dimensional space makes it impossible to find generally similarity invariants. The main difficulty behind this is that it is impossible for people to find a fundamental element in C(H), similar to Jordan's block, so as to construct a perfect representation theorem. We appreciate such a mathematical point of view as people's being not powerful enough to deal with a complicated mathematical problem then that reflects their lack of sufficient knowledge and understanding of some fundamental mathematical problems. It is because of the sufficient study of the *-cyclic self-adjoint (normal) operators that people have set up the perfect spectral representation theorem for self-adjoint (normal) operators and commutative C*-algebra. It is also because of the introduction of the concept of irreducible operators by Halmos, P.R. in 1968 that Voiculescu, D. obtained the well-known Non-commutative-Weyl-von Neumann Theorem for general C*-algebra. But irreducibility is only a unitary invariant and can not reveal the general internal structure of operator algebra and non-self-adjoint op-

V

VI Structure of Hilbert Space Operators

erators. Since the 1970s, some mathematicians have showed their concern for the problem on Hilbert space operator structure in two aspects. In one aspect, the mathematicians, such as Foias, C , Ringrose, J.R., Arveson, W.B., Davidson, K.R. etc. have made great efforts to study the structures of different classes of operators or operator algebras, such as Toeplitz operator, weighted shift operator, quasinilpotent operator, triangular and quasitriangular operators, triangular and quasitriangular algebras etc. In the other aspect, they have set up the approximate similarity invariants for general operators by introducing the index theory and fine spectral picture as tools. One of the most typical achievements, made by Apostol, C , Filkow, L.A., Herrero, D.A. and Voiculescu, D. is the theorem of similarity orbit of operators. This theorem suggests that the fine spectral picture is the complete similarity invariant as far as the closure of similarity orbit of operators are concerned. Besides, in the 1970s, Gilfeather, F. and Jiang, Z.J. proposed the notion of strongly irreducible operator ((SI) operator) respectively. And Jiang, Z.J. first thought that the (SI) operators could be viewed as the suitable replacement of Jordan block in £("H). An operator will be considered strongly irreducible if its commutant contains no non-trivial idempotent. In the theory of matrix, strongly irreducible operator is Jordan block up to similarity. Through more than 20 years' research, the authors and their cooperators have founded the theorems concerning the unique strongly irreducible decomposition of operators in the sense of similarity, the spectral picture and compact perturbation of strongly irreducible operators, and have formed a theoretical system of (SI) operators preliminarily. But, with the research going deeper and deeper, the authors and their cooperators are badly in need of new ideas and new tools to be introduced so as to further their research. In the 1980s, Elloitt, G. classified AF-algebra successfully by using of K-theory language, which stimulated us to apply the K-theory to the exploration of the internal structure of operators, which features this book. In 1978, Cowen, M.J. and Douglas, R.G. denned a class of geometrical operators, Cowen-Douglas operator, in terms of the notion of holomorphic vector bundle. They, for the first time, applied the complex geometry into the research of operator theory. Cowen, M.J. and Douglas, R.G. have proved the Clabi Rigidity Theorem on the Grass-man manifold, defined a new curvature function and indicated that this curvature is a complete unitary invariant of Cowen-Douglas operators. It is these perfect results that have inspired us since 1997, to combine K-theory with complex geometry in order to seek the complete similarity invariants of Cowen-Douglas operators and their internal structures. Cowen-Douglas

Preface vn

operator is a class of operators with richer contents and contains plenty of triangular operators, weighted shift operators, the duals of subnormal operators and hypernormal operators. Its natural geometrical properties support a very exquisite mathematical structure. Based on some of our successful research on Cowen-Douglas operators, we have made headway in the study of other operator classes.

This monograph covers almost all of our own and our cooperators' research findings accumulated since 1998. The book consists of six chapters. Chapter 1 provides the prerequisites for this book. Chapter 2 explains the Jordan Standard Theorem again in i^o-group language, and gives readers a new point of view to understand the complete similarity invariants in the theory of matrix. And this chapter also helps readers get well-prepared for the study of operator structure in terms of K-theory in later chapters. Chapter 3 mainly discusses how to set up the theorem on the approximate (SI) decomposition of operators by using the (SI) operators as the basic elements. Meanwhile, to meet the needs of the study of the structure, it also reports the relationship between (SI) operators and the compact perturbation of operators, and proves that each operator is a sum of two (SI) operators. Chapter 4 describes the unitary invariants and similarity invariants of operators in .Ko-group language by observing the commutants. This chapter contains the following four aspects: (1) Gives a complete description of the unitary invariants of operators using ifo-group and lists some properties of lattices of reducing subspaces of operators. (2) Illustrates the establishment of the relationship between the unique (SI) decomposition of operator up to similarity and the A"o-group of its commutant, and at the same time, carefully states the complete unitary invariants and complete similarity invariants, and the uniqueness of (SI) decomposition of the operator weighted shift and analytic Toeplitz operators using the results of (1) and (2). (3) Makes a concrete description of the commutant of (SI) Cowen-Douglas operators by using complex geometry. (4)Discusses Sobolev disk algebra, the internal structure of the multiplication operators on it and their commutants by using Sobolev space theory, complex analysis and the results in (3). Chapter 5 focuses the discussion mainly on the complete similarity invariants of Cowen-Douglas operator and proves that the A'o-group is the complete similarity invariant of it. In addition, our discussion is extended to the other classes of operators which are related to Cowen-Douglas operators. Chapter 6 concerns some applications of operator structure theorem, including the determination of i^o-group of some Banach algebras, the distribution of zeros of analytic functions in the

Vlll Structure of Hilbert Space Operators

unit disk and a sufficient condition for a nilpotent similar to an irreducible operator.

We would hereby like to give sincere thanks to all the following professors: Davidson, K.R., Douglas, R.G., Elloitt, G., Gong, G.H., Lin, H.X., Yu, G.L., Zheng, D.C., Ge, L.M. etc. For their many years' encouragement and support. We would like to give special thanks to Gong, G.H., Yu, G.L. and Ge, L.M., for, since 2000, both of them have enthusiastically lectured on K-theory and geometry in our seminar, which have enabled us to make greater progress with our research. We are also grateful to Academician Gongqing Zhang and professor Zhongqin Xu at Beijing university and professor Yifeng Sun at Jilin university. They have given us enormous concern and encouragement since the early days of our research. It is their encouragement and support that have encouraged us to unshakably finish the course of research. We also wish to thank Mr. Xianzhou Guo for the technical expertise with which he typed the manuscript of this monograph.

C.L. Jiang Z.Y. Wang

Contents

Preface v

1. Background 1

1.1 Banach Algebra 1 1.2 K-Theory of Banach Algebra 3 1.3 The Basic of Complex Geometry 4 1.4 Some Results on Cowen-Douglas Operators 5 1.5 Strongly Irreducible Operators 7 1.6 Compact Perturbation of Operators 9 1.7 Similarity Orbit Theorem 9 1.8 Toeplitz Operator and Sobolev Space 10

2. Jordan Standard Theorem and Ko-Group 13

2.1 Generalized Eigenspace and Minimal Idempotents 13 2.2 Similarity Invariant of Matrix 14 2.3 Remark ! 18

3. Approximate Jordan Theorem of Operators 19

3.1 Sum of Strongly Irreducible Operators 19 3.2 Approximate Jordan Decomposition Theorem 29 3.3 Open Problems 42 3.4 Remark 42

4. Unitary Invariant and Similarity Invariant of Operators 43

4.1 Unitary Invariants of Operators 44

ix

x Structure of Hilbert Space Operators

4.2 Strongly Irreducible Decomposition of Operators and Similarity Invariant: i^o-Group 57

4.3 (SI) Decompositions of Some Classes of Operators 69 4.4 The Commutant of Cowen-Douglas Operators 80 4.5 The Sobolev Disk Algebra 94 4.6 The Operator Weighted Shift 126 4.7 Open Problem 147 4.8 Remark 147

5. The Similarity Invariant of Cowen-Douglas Operators 149

5.1 The Cowen-Douglas Operators with Index 1 149 5.2 Cowen-Douglas Operators with Index n 154 5.3 The Commutant of Cowen-Douglas Operators 157 5.4 The Commutant of a Classes of Operators 169 5.5 The (57) Representation Theorem of Cowen-Douglas

Operators 176 5.6 Maximal Ideals of The Commutant of Cowen-Douglas

Operators 189 5.7 Some Approximation Theorem 192 5.8 Remark 201 5.9 Open Problem 201

6. Some Other Results About Operator Structure 203

6.1 Ko-Group of Some Banach Algebra 203 6.2 Similarity and Quasisimilarity 206 6.3 Application of Operator Structure Theorem 237 6.4 Remark 239 6.5 Open Problems 239

Bibliography 241

Index 247

Chapter 1

Background

In this chapter, we review briefly some of the facts about operator algebra and operator theory which will be needed to read this book. Most of the material can be found in books or papers such as [Admas (1975)], [Apostal, C , Bercobici, H., Foias, C. and Pearcy, C. (1985)], [Blanckdar, B. (1986)], [Conway, J.B. (1978)], [Cowen, M.J. and Douglas, R. (1977)], [Douglas, R.G. (1972)], [Herrero, D.A. (1990)], [Herrero, D.A. (1987)], [Jiang, C.L. and Wang, Z.Y. (1998)] and [Rudin, W. (1974)].

1.1 Banach Algebra

A Banach algebra is a Banach space A over C which is also an (associative) algebra over C such that ||ob]|<||a||||6|| for all a, b in A. When A has a unit e, the spectrum of a a(a) (or (7.4(a) if A needs to be clarified) is the set {AGC : Ae — a is not invertible in A }. The left spectrum of a ai(a) is the set {AeC : Ae —a is not left invertible in A }; the right spectrum of a ar(a) is the set {AGC : Ae — a is not right invertible in A }. The resolvent set of a p(a) := C\<r(a). The left and right resolvent set of a are pi{a) := C\<7;(a) and pr{a) := C\<rr(a) respectively. a(a) is a non-empty compact subset of C and a(a) = cr;(a)Uoy(a). Let / be holomorphic in a neighborhood fi of a (a) and let c be a finite union of Jordan curves such that indc(X) = 1 for every A in a(a). Define

f{a)=^-ff{z){ze-a)-ldz.

Let Hoi (a (a)) denote the set of all functions which are holomorphic in a neighborhood of 17(a). We have the following theorems.

1

2 Structure of Hilbert Space Operators

Riesz Functional Calculus Let a be an element of a Banach algebra A with identity, then for every f£Hol(a(a)), f{a) is well defined independent of the curve c. The mapping f*—>f(a) is an algebra homomorphism and

n n maps each polynomial p(z) = ^2 CkZk to c^e + J^ c^ak.

k=o k=\

Spectral Mapping Theorem For feHol(cr{a)), cr(f(a)) = f(cr(a)).

U p p e r Semi-continuity of t he Spec t rum Let a be an element of a Banach algebra A with identity. Given a bounded open set Q, flDo-(a), there exists S > 0 such that a(b)cQ, provided \\a — b\\ < 5 and b&A-

Let H be a complex, separable, infinite dimensional Hilbert space and let £(H) denote the algebra of linear bounded operators on H. For each Te£(W), a{T),(ri(T), ar(T),p(T),pi(T),pr(T) and f(T) are defined as above, where f£Hol(a{T)).

Riesz Decomposition Theorem Assume that cr(T) = aiL)o~2, o~\C\a2 = 0, where ai,a2 are non-empty compact sets, thenH is the direct sum of two invariant subspaces Hi and Ti.2 ofT, such that o~(T\-ni) = o~i and "Hi is the range of Riesz idempotent corresponding to o~i(i = 1,2), where T\-ni is the restriction of T onTii-

Let A be an abelian Banach algebra with identity. A multiplicative linear functional <j> is an algebra homomorphism <f> : A—>C with \\(f>\\ = 1. The collection J^ of all maximal ideals of A is a compact Hausdorff space in the sense of weak-* topology. The Gelfand transform a of a is the function a : ^2—>C defined by a{tb) = 4>{a).

Gelfand's Theorem If A is an abelian Banach algebra with identity, a&A and C(^2) is the space of continuous functions on Y^,, then the Gelfand transform a of a belongs to C(J^) and a (a) = {&(<p) : 4>&Y2}- The mapping a—>d is a continuous homomorphism of A into C(^2).

Let A be a Banach algebra with identity. A two-sided ideal radA of A is the Jacobson Radical if it is the intersection of all maximal left (right) ideals of A- Equivalently, radA = {a : a(ab) = a(ba) = {0} for all b£A}.

A C*-algebra C is a Banach algebra with a conjugation operator * such that (a*)* = a,(ab)* = b* a*, (aa + fib)* = aa* +~J3b* and ||a*a|| = ||a||2

for all a,bEC and a, /3GC. A *-homomorphism p of a C*-algebra C is a •-homomorphism from C into C(HP), where Hp is a Hilbert space. If C has an identity e and p(e) = I, then p is unital; if kerp = {0}, p is faithful. It is obvious that p is faithful if and only if p is a *-isometric isomorphism

Background 3

from C onto p(C).

Gelfand-Naimark-Segal Theorem Every abstract C*-algebra C with identity admits a faithful unital * -representation p in C(Ti.p) for a suitable Hilbert space fip, i.e., C is isometrically ^-isomorphic to a C*-algebra of operators. Furthermore, ifC is separable, then Hp can be chosen separable.

von-Neumann Double Commutant Theorem Let AcC(H) be a unital C*-algebra. Then the closure of A in any of weak operator, strong operator and weak-* topologies is the double commutant A", where

A" = {A'}'

and

A'{T) = {TeC(H) :AT = TA for all A<=A}.

We call A'(T) commutant ofT.

1.2 K-Theory of Banach Algebra

.Ko-group. Let A be a Banach algebra with identity, and let e and / be idempotents in A. e and / are said to be algebraic equivalent, denoted by e ~ a / , if there are x, y£A such that xy = e and yx = / ; e and / are said to be similar,denoted by e ~ / , if there is an invertible z&A such that zez"1 = / . It is obvious that e ~ a / and e ~ / are equivalent relations. Let M^A) be the set of all finite matrices over A, Proj(A) be the set of algebraic equivalent classes of idempotents in A. Set \/(A) = Proj(M'00(A)), then \J(Mn(A)) is isomorphic to \J(A). If Pi Q a r e idempotents in Proj(A), p~ sq if and only if p®r~aq®r for some rGProj(A), then "~3" is called stable equivalence. KQ{A) is the Grothendieck group generated by \J{A) [B. Blackadar [1]]. The pair (G,G+) is said to be an ordered group if G is an abelian group and G+ is a subset of G satisfying

i. G+ + G+CG+; ii. G+n(~G+) = {0}; iii. G+ - G+ = G. An ordered relation "<" can be defined in G by x<y if y — x€G+ for

x,y€G.

Isomorphism Theorem Let A, Ai and Ai be Banach algebras and

A = Ai®A2,

then

\J(A)* \J(At)® \f(A2), K0(A)cK0(A1)®K0(A2),

\J(Mn(A))~ \J(A) and K0(Mn(A))~K0(A),

where "~" means isomorphism.

Let A and B be two Banach algebras and let a be a homomorphism from A into B. Then there is a homomorphism a* induced by a from K0(A) into K0(B).

Six-Term Exact Sequence Let A be a unital Banach algebra and let J be its ideal, then we have the following standard exact sequence

and the following exact cyclic sequence

Ko(J) ^K0(A)^K0(A/J) d] d[

ffiCA/J) «—ff i (4) «— Kx{J),

where K\(B) is the Ki-group of Banach algebra B.

1.3 The Basic of Complex Geometry

Let A be a manifold with a complex structure and let n be a positive integer. (E, 7r) is called a holomorphic vector bundle of rank n over A if n is a holomorphic map from E onto A such that each fibre Ex = 7r-1(x) is isomorphic to Cn(a;eA) and such that for each zo£A, there exist a neighborhood A of zQ and holomorphic functions e\{z), e2{z)^ • • • , en(z) from A to E such that e\{z), e2(z), • • • , en(z) form a basis of Ez = ir~1(z) for each zeA. The functions e\, e2, • • • ,en are said to be a holomorphic frame for E on A. The bundle is said to be trivial if A can be assigned to A.

Let E and F be two holomorphic bundles over a complex manifold A. A map (p from E to F is a bundle map if F\ is a linear transformation for every AeA.

A Hermitian holomorphic vector bundle E over A is a holomorphic bundle such that each fibre E\ is an inner product space. Two Hermitian

Background 5

holomorphic vector bundles E and F over A are said to be equivalent if there exists an isometric holomorphic bundle map from E onto F.

Let H be a separable complex Hilbert space and let n be a positive integer. Denote Gr(n,H), the Grassmann manifold, the set of all n- dimensional subspaces of H. For an open connected subset A of Cfc, a map / : A—>Gr(n,H) is said to be holomorphic if at each AoGA there is a neighborhood A of Ao and n holomorphic W-valued functions

n ^i(z),e2(z),-" >e„(z) such that f(z) = V ( e j (2)}- I f / : ^—>Gr(n,K) is

3=1 a holomorphic map, then an n-dimensional Hermitian holomorphic vector

bundle Ef over A and a map <f> can be induced by / , i.e.,

Ef.= {(x,z)eHxA:xef(z)}

and

(j): Ef—>A,<j){x,z) = zeA.

Given two holomorphic maps / and g : A—>Gr(n,'H), we have two vector bundles Ef and Eg over A. If there exists a unitary operator U on H such that g = Uf, then / and g are said to be unitarily equivalent. If there is an open subset A of A such that Ef\& is unitarily equivalent to •Eg I A! then Ef and Eg are said to be locally unitarily equivalent.

Rigidity Theorem Let A be an open connected subset of Cfc and let f and g be holomorphic maps from A to Gr(n, TCj such that

V /(*) = V aw = H-zeA zeA

Then f and g are unitarily equivalent if and only if Ef and Eg are locally unitarily equivalent.

1.4 Some Results on Cowen-Douglas Operators

Let Q be a connected open subset of C, n is a positive integer, the set Bn(Q) of Cowen-Douglas Operators of index n is the set of operators T&JC(H)

satisfying (i) Oca(r); (ii) r a n ( z - T ) := {(z - T)x : x£H} = H for each z€fi; (iii) V/ fcer(* -T) = H\

(iv) dimker(z — T) = n for each zefi.

It can be proved that if Qo is a nonempty open subset of fi, then Bn(£i)cBn{yio). For an operator T£Bn(fl), the mapping z\—>ker(z — T) defines a Hermitian holomorphic vector bundle of rank n. Let (ET,TT) denote the subbundle of trivial bundle QxH given by

ET := {(z,x)eflxH : xeker(z-T) and n(z,x) = z}.

Let A'(T) be the commutant of T, i.e., A'{T) := {A&C{H) : TA = AT}, then for TeS„(fi), there is a monomorphism TT from ^4'(T) into H™,E JQ) satisfying TTX = X\ker{z_T) for XeA'(T) and zeft, or r T X ( z ) =

ker(z-T) '•— X(z), where -ff^(j5;T)(^) is the set of all bounded bundle endomorphisms from E? to ET-

To summarize the above and Section 1.3, we can find a holomorphic frame (ei(z), • • • ,en(z)) such that

n

ker(z-T) = \/ ek(z), ZGQ for TeB„(fi).

Fix a zoSfi, denote

Hi = ker(z0 - T ) , H2 = ker(z0 - T)2Qker(z0 - T),

Hm = ker(zQ - T)mQker(zQ - T)"1"1 .

We have:

Theorem CD1 [Cowen, M.J. and Douglas, R. (1977)] m . . .

(i) E ©Wfc = V ( e (*o) : l < i < « , 0<fc<m - 1}; fc=i

00

r«; E®-Hk = n; k=l

(Hi) {el- (zo) : l<j<n,0<k<m — 1} is a basis of ker(zo — T)m,m — 1,2, • • •, where e^ (zo) denote the k-th derivative ofej(z) at z = ZQ.

The following theorems will often be used in the chapters hereafter.

Theorem H [Herrero, D.A. (1990)] IfTeBn(Q), then ap{T*) = 0, where T* is the adjoint ofT and o~p(T*) is the point spectrum ofT*.

Theorem JW1 [Jiang, C.L. and Wang, Z.Y. (1998)] Let Te6„(fi) and let Pz be the orthogonal projection from H onto ker(z — T) for z£fl, then (I — Pz^heriz-T)1- *s similar to T.

Background 7

Theorem JW2 [Jiang, C.L. and Wang, Z.Y. (1998)] Let TeC(H). For given numbers > 0, there exist a positive integer n and Cowen-Douglas operators {Ai}\=1, {bj}?=i+1 such that

F-(®^)©(® s*)||<£. j=i i=i+i

Theorem JW3 [Jiang, C.L. and Wang, Z.Y. (1998)] Given TeBi(fl), there exist compact operators K\,K2,--- ,Kn,--- with \\Ki\\ < 7 - such that T + Ki&Bi(Q) and kerTT+K,T+K, = {0},i ^ j , where TA,B is the Rosenblum operator from C(H) to C(H) given by TA,B(X) = AX — XB forXeC{H).

1.5 Strongly Irreducible Operators

Operator T is strongly irreducible if there is no nontrivial idempotent in A'(T) ([Gilfeather, F. (1972)], [Jiang, Z.J. (1979)], [Jiang, Z.J. (1981)]). Operator T is irreducible if there is no nontrivial orthogonal projection in A'(T) ([Halmos, P.R. (1968)]). It is obvious that strongly irreducibility is invariant under similarity while irreducibility is just unitarily invariant. Denote (57) and (IR) the set of all strongly irreducible operators and irreducible operators, respectively, on H.

Let K{l-L) be the ideal of compact operators on H and let

•K : £(H)^A(H) := C{H)/1C(H)

be the canonical quotient mapping, A(H) is called the Calkin algebra. The essential spectrum of operator T is ae (T) = {AsC : A—ir(T) is not invertible in A{H)} and the Predholm domain of T is pF(T) = C\ae(T). It is well known that

o-e{T) = ale{T)U<rre(T),

where

CTU{T) := aifr(T))

and

are(T) := O-T(TT{T)).

Operator T is a Fredholm operator if 0€PF(T). T is a semi-Fredholm operator if the range of T, ranT, is closed and either

nulT := dimkerT

or

nulT* := dimkerT*

is finite. In this case the index indT of T is defined by

indT := nulT - nulT*.

The Wolf spectrum aire (T) of T is given by

alre(T) := are(T)nale(T)

and PS-F{T) := C\oire(T) is the semi-Fredholm domain of T. The spectral picture A(T) of T consists of the compact set <7;re(T) and the index ind{T— A) on the bounded connected components of ps-F(T).

Spectral picture theorem of strongly irreducible operators [Jiang, C.L. and Wang, Z.Y. (1996b)] Let T be in C(H) with connected spectrum cr(T). Then there exists a strongly irreducible operator L satisfying

(i) A (L)=A(T) ; (ii) TeS(L)-; (Hi) If there is another strongly irreducible operator L\ with A(Li) =

A(T), then L\ES(L)~, where S(L) is the similarity orbit of L, i.e.,

S(L) := {XTX-1 : Xe£(H) is invertible}

and S{L)~ is the norm closure ofS(L).

Spectral picture theorem of Cowen-Douglas operators [Jiang, C.L. and Wang, Z.Y. (1998)] Let T be in C(H) with connected a(T) and a(T)\p®_F(T). If pp{T) ^ 0, then there exists a Cowen-Douglas operator A&(S I) such that

A(T) = A(A)

and if there is another operator B&S(T)~, then BGS(A)~, where

P°S-F(T) := {\£ps-F(T)na(T) : ind(X - T) = 0}.

Background 9

Commutant theorem of strongly irreducible operators [Fang, J.S. and Jiang, C.L. (1999)] Operator T is strongly irreducible if and only if a (A) is connected for each A in A'(T).

The following theorem will be used frequently in this book.

Theorem CD2 [Cowen, M.J. and Douglas, R. (1977)] Each operator in B$fl) is strongly irreducible.

1.6 Compact Perturbation of Operators

We introduce only two famous theorems on compact perturbation of operator here.

Brown-Douglas-Fillmore theorem / / T\ and T2 are essentially normal operators on Ti, then a necessary and sufficient condition that T\ be unitarily equivalent to some compact perturbation of T2 is that

o-e(Ti) = o-e(T2)

and

ind{\ — T$ ~ ind{\ — T2)

for ««A^cre(Ti).

An operator T is essentially normal if T*T — TT* is compact.

Voiculescu's theorem Let T£C(H) and p be a unital faithful *-representation of a separable C* -subalgebra of the Calkin algebra A(H) containing the canonical images n(T) and n(I) on a separable space Ti.p. Let A •= p(w(T)) and k be a positive integer. Given e > 0, there exists KeK,(H), with \\K\\ < e, such that

T - K^T®A^^T®A^k\

where "=" means unitarily equivalent.

1.7 Similarity Orbit Theorem

Complex number A is a normal eigenvalue of T if A is an isolated point of cr(T) and the dimension of H(X,T), the range of the Riesz idempotent

corresponding to A, is finite. Denote the set of all normal eigenvalues of T by ao(T). The minimal index of A — T, AS/0s_ir(T), is defined by

min-ind{\ — T) := min{nul(X — T), nul(X — T)*}.

Similarity orbit theorem Given T, Ae.C(H) satisfying (i) <T0(A)Ca0(T),dimH(\,A) = dimH{\,T) for all \£a0(A); (ii) Each component of aire(A) meets ae(T); (Hi) ps-F(A)cps-F(T),

ind{\ - A) = ind{\ - T), min-ind(X — A)k>min-ind(X — T)k for all \£ps-F(A) and k>l;

(iv) There is no isolated point in ae(T); then AeS(T)~.

Note that this is only a sufficient condition of the similarity orbit theorem. The reader is referred to [Apostal, C , Fialkow, L.A. Herrero, D.A. and Voiculescu, D. (1984)] for the general form of the theorem.

1.8 Toeplitz Operator and Sobolev Space

Let D and C be the open unit disk and unit circle in the complex plane respectively, and let p, be the Lebesgue measure on C, normalized so that fi(C) = 1. If e„(z) = zn for ZGC and n = 0, ±1 , ±2, • • •, then {en, —oo < n < +00} is an orthonormal basis (ONB) for L2(C, p). Let H2 := span{en : n>0} and H°° = L°°(C,fi)nH2. If P is the orthogonal projection from L2(C,p) onto H2 and if ^eL 0 0(C,p), then the Toeplitz operator T$ with symbol <j> is defined by T+f = P{cpf) for all feH2. If </.eff°°, T+ is called an analytic Toeplitz operator.

A function m£H2 is inner if |m(z)| = 1 a.e. on C. If <j> is a positive measurable function on C such that logcpGL1 (C', p) and if Q(z) = c-exp{fc ^^log<p(iij)dfj,(u>)} for z€C, then Q is called an outer function, where c is a constant and \c\ — 1.

The Blaschke product is a class of inner functions with the form

where k>0, ceC, with \c\ = 1 and {A.,} is a sequence of nonzero numbers 00

in D satisfying J2 (1 — l^jl) < +00.

Background 11

Factorization theorem If f&H2 and / ^ 0, then f is the product of an inner function m and an outer function Q, i.e., f = mQ.

Beurling's theorem A subspace H\ of H2 is invariant under the operator Tz if and only if Hi = 4>H2 := {<j)f : f&H2} for some inner function 4>.

Let Q be an analytic Cauchy domain in the complex plane and let W22(Q.) be the Sobolev space

„ , „ , „ , f „ Tn,^ , x the distributional derivative of first and ] W22M:=[feL2(n,dm): s e c o n d o r d e r o f / b e l o n g t o j L 2 ( A d m ) )

where dm denotes the planar Lebesque measure. For f,geWn(tt), we define (/,<?) = £ jDafD^g~dm, then W22(fl)

\a\<2

is a Hilbert space and a Banach algebra with identity under an equivalent norm. By Sobolev embedding theorem, f&W22(fl) implies that f£C(Q) and

ll/llc(n)<M|l/llw»(n)

for some M > 0. Thus a sequence of functions {fn}^Li converges to / in W22(Q) implies that /„ converges to / uniformly on Q. For f£W22(Q,), the multiplication operator Mf on W22(Q) is defined as follows

Mfg = fg, geW22(Q).

Let W(ti) := {Mf : feW22(Q)}, then W(fl) is a strictly cyclic operator algebra with strictly cyclic vector e(s,t) = 1. An operator algebra A on a Hilbert space 7i is said to be strictly cyclic if there exists a separating vector e such that

Ae := {Ae : AGA} = H.

Theorem JW4 [Jiang, C.L. and Wang, Z.Y. (1996a)] (i) a(Mz) = aire(Mz) = H ; (it) A'(Mz) = W(fl); (Hi) Aa(Mz) = R(Q), where Aa(Mz) is the algebra generated by the

rational function of Mz with poles outside fi and R(i}) is the closure in W22(Q) of all rational functions with poles outside fl.

Chapter 2

Jordan Standard Theorem and K0- Group

2.1 Generalized Eigenspace and Minimal Idempotents

Recall that a kxk Jordan block

Jk(X) =

has the following properties (i) Jfc(A) is strongly irreducible on Ck, i.e., there is no nontrivial idem-

potent in A'(Jk(X))\ (ii) nul(X- Jfc(A)) = l,ker(X-Jk(X)y g ker(X- Jk{X))i+l,l<j<k-l

and ker(X - Jk(X))j = ker(X - Jk{X))k = Ck for all j>k; (iii) A'(Jk(X)) consists of all kxk lower triangular matrices B

X 1

0

A

1

0"

A.

B =

0

o-k • • • a-2 a\_

where di£C,i = 1,2,- , fc\

It follows from (iii) directly that (iv) A'(Jk(X))/radA'(Jk(X))~C. For j4eMn(C) and XGCT(A), ker(X — A) is called the eigenspace of A

related to A. If there is a positive integer m,m < n, such that

ker(X - A)m = ker(X - A)m+1,

then ker(X — A)m is called the generalized eigenspace of A related to A. By

13

elementary matrix theory, we have the following proposition.

Proposition 2.1.1 Let Ae.Mn(C) and \£a(A) with nul(X - A) = 1 and let M be the generalized eigenspace of A related to X, then A\tf~Jk{X) for some k,k < n.

A nonzero idempotent PGA'(T) is minimal, if for each idempotent Q&A'(T), ranQcranP implies Q = 0.

Proposition 2.1.2 Let A&Mn(C) and let P be a minimal idempotent in A'(A), then there exists a number AsC, such that ^4|ranP~«^fc(A), where k = dimranP. Proof Since PEA'(T), ranP and ran(I — P) are invariant subspaces of A. {Claim 1} B :— A\ranp has a unique eigenvalue XB- Otherwise, it follows from the theory of linear algebra that P is not minimal. {Claim 2} B is strongly irreducible and UUI{XB — B) = 1. Since P is minimal, BG(SI). Using the basic theory of linear algebra and the fact that P is minimal again, we have HUI{XB — B) = 1. Thus the proposition is a conclusion of Proposition 2.1.1.

Let {Afc}£=1 be all of the eigenvalues of AeM n (C) , the multiplicity is included, by Proposition 2.1.2 and the theory of matrix, we have the next proposition.

Proposition 2.1.3 Let A€Mn{C), then there exist minimal idempotents

{PxiliLi such that

(i) E Pxk = / c - and PXk Px3 = 0 for k ± j ; fc=i

(ii) A\ ranP\.

~Jmfc(Afc), where rrik = dimranP\k; n

(Hi) A = Y, +A\ranP^ •

2.2 Similarity Invariant of Matrix

Jordan standard theorem Let A£Mn(C), then A is similar to a direct sum of finitely many Jordan blocks, i.e.,

i

A~Q)Jmk(Xk). k=i

Jordan Standard Theorem and Ko-Group 15

By Propositions 2.1.2, 2.1.3 and Jordan standard theorem, the similarity of two nxn matrices A and B depends completely on their eigenvalues and generalized eigenspaces. A quantity (or quantities) or property (or properties) V is similarity invariant if operator A has V and A~B imply B has V. For a subset 11 of C(H), similarity invariant (or invariants) V is completely similarity invariant if AGTZ, then A~B if and only if BE.%

and A and B have same V. One of our aims is to find or determine complete similarity invariants of operators. But if H is an infinite dimensional separable Hilbert space, it is very difficult to obtain complete similarity invariant of operators in £(7i). As a matter of fact, some operators in £(Ji) even have no eigenvalues. To provide a new idea for finding of the similarity invariants of Hilbert space operators, we interpret the Jordan standard theorem in the view point of i^o-theory.

The propositions below are from [Blanckdar, B. (1986)] and [Aupetit, B. (1991)].

Proposition 2.2.1 K0(Mn(C))^Z and V(Mn(C))^N, where N = {0,1, 2, • • • } andZ = {0, ±1 , ±2, • • • }.

Proposition 2.2.2 Let A be a unital Banach algebra and let P be an idempotent in A and R&adA. If P + R is still an idempotent in A, then there exists an invertible element XGA such that X{P + R)X_1 = P.

We know that A'{Jk(X))/radA'{Jk{X))=C for each Jordan block Jfc(A). If P is an idempotent in A'(Jk(ty), it follows from the structure of A'(Jk(X)) that P = ICk + R, where R is a lower triangular idempotent. Thus P~A'{Jk(\))Ic>< by Proposition 2.2.2.

From Proposition 2.2.1 and the definition of .RVgroup, we have the following proposition.

Proposition 2.2.3 \J{A'{Jk{X)))^N and K0(A'{Jk(X)))=Z.

Lemma 2.2.4 Let A\, A2€(SI)<l£(H) satisfying

A'(Ai)/radA'(Ai)*iC, i = 1,2,

then at least one of the following is true (i) AX~A2; (ii) If X,Ye£{H) with AXX = XA2 and YAx =• A2Y, then

XYeradA'(Ai) and YXGradA'(A2). Proof If A\ is not similar to A2 and AXX = XA2, YA\ = A2Y. Thus

AXXY = XA2Y = XYAX

and

XYtA'iAi).

If XY&radA'iAx), then XY = A + R for some AeC,A ^ 0 and R&radA'{Ai), since ^'(A^/racM'(;4i)5=!C. Therefore XY is invertible. Similarly, YX is invertible. This implies that X and Y" are both invertible, i.e., Ai~A2. The contradiction indicates that XYGradA'{A\). Similarly, YX&adA'{A2).

i Lemma 2.2.5 Let A£(Mn(C)), then A'{A)/radA'(A)^ *£ Mki(C),

i=l where

k\ + k2 H \-ki—n.

Proof By Proposition 2.1.3 and Proposition 2.1.2, we have

m

A ~ ^ ® J f c i ( A i ) . i = l

For simplicity we prove the lemma for m = 2 and consider the following two cases. {Case 1} k\ = k2 and Ai = A2, therefore Jfc^Ai) = Jk2{^2)- A simple computation indicates that

A'(A) = <

from which we have

An A12

A21 A22

Oil

a2

Oik

Oil

012

0 "

Oil.

radA'(A) = < Ru -R12

R21 R22 Ri

0 0 a 0

/?••• a 0

> .

Thus A'(A)/radA'(A)^M2{C). {Case 2} fci ^ k2 or Ai ^ A2. If k\ ^ k2, without loss of generality

we assume that k\ > k2. By Lemma 2.2.4, if XGkerTj ( hJ (A } and

YekerTj .. w .. ., then XY&adA'(JkM),YX&adA'{Jk2(\2)).

Jordan Standard Theorem and KQ-Group 17

Set

J = Xn X±2

X21 X22

XneradA'(JkM), X12X21&adA'(Jkl^i)) X22eradA'(Jk2(X2)), X21X12£radA'(Jk2(\2)) }•

It is easily seen that J is a two sided ideal of A'(A). We claim that

Xu - A X\2

X2\ X22 — A

is invertible for each A 7 0 and

Xu Xi2

X2i X22 &J.

In fact, observing that

1 0 - ( X n - A ) - 1 !

Xu — A XX2

X2\ X22 — A

X\\ — A X\2

0 ( X 2 2 - A ) - X 1 2 ( X 1 1 - A ) - 1 X 2 i .

where Xl2(Xlx - \yxX2X&adA\J^{\2)). Thus

(X22 — A) - X\2{Xii — X)~ X2\

is invertible. Therefore is invertible. X\\ — A X12

X21 X22 — A

The claim above indicates that CF{X) — {0} for each X&J. Thus

J = radA'(A)

A'(A)/radA'(A)^C®C.

and

This complete the proof. 1

Theorem 2.2.6 Let AeMn(C) and i 4 ~ E ® 4 W ( n i ) .

i/ien V(-^'(^))=N (Z) and K0(A'(A))^Z«\ where Jki{Xi)(ni) denotes the

orthogonal direct sum of rii copies of Jki{Xi).

Proof By Lemma 2.2.5, A'(A)/radA'(A)S* £ ©M„,(C). By Proposition i = l

2.2.2 and Proposition 2.2.3 we have

i

\/(A'(A))= \j{A'{A)/radA'(A)}^ \ / { E ® ^ ( C ) } S N ( " 2 = 1

and

K0(A'(A))=iZ^.

Theorem 2.2.6 is another form of Jordan standard theorem. The following result gives the complete similarity invariant for matrices in terms of ifo-grc-up.

m Theorem 2.2.7 Let A,BeMn{C) and A = £ © 4 ? with Akie{SI)

i= i and Aki is not similar to Akj for i ^ j . Then A~B if and only if there exists an isomorphism h such that

h(K0(A'(A®B)))^ZW

and h[IA,(A@B)} = 2niei + 2n2e2 + ••• + 2nkek, where 0 ^ n i £ N , i = 1, 2, • • • , k, {e,}^=1 are the generators of the semigroup N ' m ' of Z^"1' and IA'(A®B) is the identity of A'(A®B).

m ,„ . Proof If A~B, then A®B~ £ ®A£ni). The "necessary" part follows

i= l from Theorem 2.2.6.

Proof of "Sufficient" part. Since B G M „ ( C ) , B ~ £ ®Bk > w h e r e

j=l

Bkj£(SI), Bkj + Bkj, if j rf, f. It follows from h(K0(A'(A®B)))^Z^ and

h[lA'(A<BB)] — 2niei + 2n2e2 H + 2nkek

that I = m. By Lemma 2.2.4, for each i?fc. there exists an Ak. such that Bkj~Aki and m^ = ns. This implies that A~B.

2.3 Remark

Theorem 2.2.6 and Theorem 2.2.7 are different forms of Jordan standard theorem and can be obtained in different ways. Lemma 2.2.4 is due to [Cao, Y., Fang, J.S. and Jiang, C.L.(2002)].

Chapter 3

Approximate Jordan Theorem of Operators

In the operator theory of finite dimensional space, or in matrix theory, Jordan canonical theorem is one of the core contents. The Jordan theorem gives the complete similarity invariant of matrices. But in the infinite dimensional Hilbert space case, it is very difficult to find complete similarity invariant for operators. We can only give an approximate Jordan theorem, or obtain complete similarity invariant for some special class of operators. In this chapter, we give some different kinds of approximate Jordan theorems considering strongly irreducible operators as the replacement of Jordan blocks in matrix theory.

3.1 Sum of Strongly Irreducible Operators

[Radjaval, H. and Rosenthal, P. (1973)] proved that every operator in C(Ti.) is a sum of two irreducible operators. In this section, we will prove the following result.

Theorem 3.1.1 Every bounded linear operator on H. is a sum of two strongly irreducible operators.

In order to prove the theorem, we need the following lemmas.

Lemma 3.1.2 Assume that Te£(H) with indT = —1 and min-indT = 0. If B\ and B2 are two left inverses ofT and eo is a unit vector in (ranT)-1, then there exists an f£H such that B\ — B2 + /<8>eo. Proof Set f =(B1- B2)e0 and A = B1-B2- /<8>e0. Since indT = - 1 and since min-indT = 0, ranT is closed and dim^anT)1- = 1. Thus for each XGH, there is a number a€C and x\€H such that x = ae^ + Tx\.

19

Since BiT = B2T = I,

eo - (f®e0)Tx! Ax = a{Bi - B2)e0 + B{Txi - B2Txx - a = af + X1-X1- af- < Txi,e0 > f = 0.

Thus, A = 0 and B\ = B2 + f®e0.

Lemma 3.1.3 Let TGC(H) with indT = — 1 and min-indT = 0. Let eo be a unit vector in (ranT)1- and B be a left inverse of T with Beg = 0. ThenTB = 1 -e0®e0. Proof Set A = TB + e0(g>eo — / . Since for each XGH there exist a e C and y€7i such that x = aeo + Ty,

Ax = aTBeo + TBTy + a(e0®eo)e0 4- (e0®e0)Tj/ — ae0 — Ty = 0 + Ty + ae0 + 0-ae0-Ty = 0.

Thus A = 0 and TB = I - e0<g>e0.

Proposition 3.1.4 Given T, eo and B as that in Lemma 3.1.3. If BG(SI), thenTe(SI).

Proof If there is a nontrivial idempotent PGA'(T), kerT* is an invariant subspace of P*. Since indT = —1 and min-indT = 0, dimkerT* = 1. Therefore, P*eo = Aeo and A = 0 or A = 1. Assume that A = 0 (otherwise consider I — P), then

P = 0 Q 0 IranP'

Set

x = IkerP Q

0 IranP'

kerP ranP*

kerP ranP* '

and T = Ti T12

0 T2

kerP ranP*.

p = X-'-PX 0 0 0 IranP*

kerP ranP*

and

f = X - 1 T X = T : 0 0 T2

kerP ranP*.

Note that P* = P , PeA'(f),indf = - 1 and min-indf = 0. Thus Ti and T2 are injective with closed ranges and indT\ + indT2 = indT = — 1. Assume that e0 = ei + e2, where eiGkerP and e2€ranP*. Since P*eo = 0 and e0 ^ 0,ei ^ 0 and Q*ei = - e 2 . Since Tj*ei + T^ex + T2*e2 = 0, Ti*ei = 0. Thus indTi = - 1 and indT2 = 0. This implies that T2 is

Approximate Jordan Theorem of Operators 21

invertible and T\ is left invertible. Suppose that B2 is the inverse of T2 and B\ is a left inverse of Ti satisfying kerB\ = (ranTi)1-. Thus Biei = 0. Set

B = Bt 0 0 B2

kerP ranP* '

then B is a left inverse of f and BP = PB. Set B = X ^ R X " , then B

is also a left inverse of T. It follows from T*ei = T,*ei = 0 that jrhi is a unit vector in (ranT)1-. By Lemma 3.1.2, there exists a vector / such that J5 = B + /®ijU|. From-BX^eo = X ^ B e o = 0,Xeo = e 0 + Q e 2 + e 2 and BXeo = 0, we have:

0 = 6(ei + Qe2 + c2)

= 5 i e i + 5 iQe 2 + 5 2 e 2 + < e2, ^ > / + < Qe2, p ^ > /

= 5! Qe2 + £2e2 + | | e 1 | | / -g | r / .

If e2 ^ 0, it follows from BiQe2±B2e2 and B2e2 7 0 that BiQe2+B2e2 ^ 0. Thus

/ : e i

| e 2 | | 2 - | | e 1 | | 2 BxQe2 + e i

Ml2-INI2 B2e2

and

B

D 1 BiQe20ei n " 1 " r l l « . „ l | 2 _ n „ , 112 <J

I|e2|r-||ei||

j?2£2®ei

IMP-IK 5 2

kerP

ranP*

Moreover,

TB = herP - $$• + Qe2®el - < 2 | a ^ e i ® e i 0

e?<»ei e2<»

By Lemma 3.1.3,

TB = I — eo®eo = I — ei®ei — e\

Therefore,

(3.1.1)

i-ranP*

e\ - e2<g)e2.

TB -e2®ei *

kerP ranP*

(3.1.2)

Compare (3.1.1) with (3.1.2) we have

INI2 - INI2 = 1.

Since | |e i | |2 + ||e2 | |2 = ||e0 | |2 = 1, ||e2|| = 0 and e2 = 0. Thus, Pe0 = Pex =

0. Therefore,

T{PB - BP) = P(TB) - (TB)P

= P(I - e0®e0) - (/ - e0<8e0)P

= - (Pe 0 )®e 0 + e0®P*e0

= 0.

But T is injective, thus PB - BP = 0, i.e., PB = BP. This contradicts B&(SI). Therefore TG(SI) and the proof of the proposition is now complete.

Lemma 3.1.5 Let A, S&C(H), \\A\\ < \ and let S be the froward unilateral shift. Then for each AGC with |A| < | , S + A — A is a Fredholm operator and

ind(S + A - A) = - 1 , min-ind(S + A - X) = 0.

Proof Note that for each x£H,

\\(S + A- X)x\\>\\Sx\\ - \\Ax\\ - | A | H > ( 1 - piDllzll.

Thus S + A — A is bounded below, ran(S + A — A) is closed and

dimker(S + A - A) = 0,min-ind(S + A- X) = 0.

On the other hand, it follows from | | (^ - A)5*|| < 1 that I + (A - X)S* is invertible. Therefore,

ind{S + A - A) = ind[S + {A- X)S*S]

= ind[(I + (A - X)S*)S]

= ind[I +{A- X)S*} + indS

= 0 + ( - l ) = - l .

Thus S + A — A is a Fredholm operator.

Lemma 3.1.6 Assume that Te£(H), X0eC,ind(T - A0) = - 1 and

min-ind(T — Ao) = 0.

Then

T~T\ran(T_\0).

Proof Define X : H —> ran(T - A0) by Xx = (T - A0)x for x€H. Since ran(T — Ao) is closed and T — Ao is injective, X is invertible. Note that

T\ran{T-x0)Xx = T(T - X0)x = (T- \0)Tx = XTx,

thus

T\ran(T-\o)X = XT,

i.e.,

T\ran(T-\0)~T.

Lemma 3.1.7 Given AGC(H) that admits the following lower triangular matrix representation

"Ao 0 a2\ 0

A = a32 0

with respect to the ONB {efc}£L0. 7/||A|| < | , then (S + A)*e#i(fi), where S is the unilateral shift. Proof Since ||A|| < \ and |A0| < \, ind(S + A - A) = - 1 and

min-ind(S + A - A) = 0 for |A| < - .

Thus (S + A - A)* is surjective and nul(S + A - A)* = 1. In particular,

ker(S + A- A0)* = {ae0 : aGC}.

Therefore ran(S + A - A0) = \J{ek : fc>l} := Hx. Set T = (S + A)\Hl. By Lemma 3.1.6, S + A~T. Thus

r a n ( T - A ) * = Hi

eo e i

e2 (3.1.3)

and

nul{T - A)* = 1

for all AGfi := {A : |A| l} = Hi.

This implies that T*eBi(ft). Thus (S + 4)*eSi(fi) .

Lemma 3.1.8 Given BGC(H) that admits the following upper triangular matrix representation

eo ei

e2

with respect the ONB {ek}f=0. If \\B\\ < 1, then S + BG(SI), where S is the unilateral shift. Proof Since \\BS*\\ < 1, (I + BS*) is invertible. Set

A = S*(l + BS*)-\

Thus A - A = S*(I + BS*)-1^ - \{I + BS*)S) for |A| < \\I + BS*\\-\ Note that (/ — X(I + BS*)S) is invertible. Therefore, A — A is surjective and nul(A — A) = 1. Since A admits a strict upper triangular matrix representation with respect to the ONB {efc}£L0, \J{kerAk,k>l} = H. Denote D, = {AGC : |A| < (1 + HSU)-1}, then AeB^Q.) and AG(SI).

Because of A{S + B) = A(I + BS*)S = I, A is a left inverse of S + B and

Ae0 = A(e0 + BS*e0)

= A(I + BS*)e0

= S*(I + BS*)-1(I + BS*)e0

= S*e0 = 0.

It is easy to see that eo£[ran(S + B)}1-. By Proposition 3.1.4, S + Be(SI).

B =

0 0 0

&22

0

Lemma 3.1.9 Let BEC(?{) admit the following upper triangular matrix representation

B

0 bu b13 •••

&22 ^23 *

0

eo ei

with respect the ONB {ek}kL0- If\\B\\ < JQ, then there exist BGC(H) and a S C satisfying

(ii) \\B\\ < I and B admits the following matrix representation

B =

0 0 0 •••

^22 &23 ' • '

&33

0

eo ei

e2 ;

(Hi) S + B + ae0®e0~S + B and S + B + ae0®eo£(SI), where S is the unilateral shift. Proof Denote H 0 = {Ae0 : AeC}, Hi = V{e™ : n > l } . Then

S = 0 0

ei<g>e0 Si Ho Hi

and B 0 e 0 ® / 0 B±

Ho Hi

under the decomposition of the space H = Ho ©Hi. Note that Si is uni-tarily equivalent to S, i.e., Si=S. Set

g = Si(IHl + SiB*i)~lf = J2(-±)nSi(SiB*i)nf. 71=0

Then 5-Lei and

n*n< f > r 11/11 = !!% i = 0

10

Since / = B*e0, | |/ | |<||fl*|| < ^ . Therefore ||5|| < §. Let

a = - < ei,g >,

n | a |< | | / | | < £,i.e. Set

a satisfies (i).

B = "0 0 0 Bi +e iOg

Ho Hi

26 Structure of Hilberi Space Operators

i.e., B = B - e 0 ® / + eiC Then

\B\\ = \\Bx+e&g\\<\\B\\ + \\g\\<l,

and

i.e., B satisfies (ii). Set

B =

"0 0 0 •••"

&22 ^23 • ' •

.0

eo e i

X = 0 JWl

i.e., X = I + eo®g. It is obvious that X is invertible and X x = I — eoi Note that

X-1(S + B + ae0®e0)X

ae0®e0 e 0 ®/ eoQs>e0 — e0<x> 0 I W l

eoQjeo e 0 ®g 0 I W l

a e 0 ® e 0 - (e0®9)(e1<8>e0) e 0 ® / - (eo®g)(Si + Bx) ei<8»eo 5 i + B i

a e 0 ® e 0 - < ei,g > (e0®e0) e 0 ®/ - e0®(5i + B\)*g ei®e0 Si + B\

0 /Wl J

eo®e 0 e 0 ®g

0 / W l .

0 0 ei<8>e0 S i + £?i

eo®eo e 0 ®g L 0 /Hl J

0 0 ei®e0 ei®g + Si + Bx

= S + B,

i.e., S + B + ae 0 ®e 0 ~5 + B. By Lemma 3.1.8, S + Be(SI). Therefore,

S + B + ae0®e0£(SI).

The following lemma is due to [Jiang, C.L. and Wu, P.Y. (1998)].

Lemma 3.1.10 For given TeC(H), there exist ONB { e f c } ^ 0 and A,B£C(H) satisfying

(i) T = A + B; (ii) A admits a strict lower triangular matrix representation with re

spect to {efc}£L0;

(Hi) B admits an upper triangular matrix representation with respect

to {ek}%0.

We are now in a position to prove Theorem 3.1.1.

Proof of Theorem 3.1.1 By Lemma 3.1.9, we can find ONB {efc}£L0

and A\, B\ &£(H) such that T = A\ -f B\, A\ is strict lower triangular and B\ is upper triangular with respect to {efc}^:1. Denote b =< Bieo,eo >, then b is the upper left entry of the upper triangular matrix of Bi. Set Ai = A\ + 6eo®e0, Bi = B\ - be0®e0 and r =

10(||A2|| + | |B2 | | + 1) B3 rB2.

Then || — -S3 j| < 375 a n d —B3 admits the following matrix representation

-B3

0 * *

0

eo ei

e2

with respect to {efc}£L0. By Lemma 3.1.9, there is an aGC with |a| < § such that S — B3 + aeo®eo£(SI), where S is the unilateral shift. Denote

Ti 1

(S - Bi + ae0

then

TiG(57).

Let A3 = rAi + aeo®eo and AQ = rb + a, then

||^4.i||<T-||>l2|| + ]«| < -

and A3 admits the matrix representation (3.1.3) with respect to ONB

{e/c}£°=o-Since

\b\<\\As

|Ao|<|r-6| + H e0 + rA2 + ae0®e0)

= B2 + A2

= B\— 6e0®e0 + A\ + 6eo®e0

= A1 + B1

= T.

Theorem 3.1.1 indicates that every bounded linear operator acting on an infinite dimensional, separable Hilbert space can be expressed as a sum of two strongly irreducible operators. In the proof, we used the strongly irreducibility of Cowen-Douglas operators of index 1 frequently. As a matter of fact, in finite dimensional Hilbert space we have the following theorem.

Theo rem 3.1.11 Every operator on a finite dimensional Hilbert space is a sum of two strongly irreducible operators. Proof Let T e £ ( C n ) and A = T - (£)trT, where t rT is the trace of T, i.e., the sum of diagonal elements in the matrix representation with respect to some ONB. If rankA<l,

'Oa 0 ••• 0" 0 aO • • • 0

A^

0 0 0 0

Let b = ( ^ ) t r T , c ^ 0 and c ^ a, then

T =

6c 0 ••• 0 6 c ••• 0

0 b c

b

+

b a — c 0 b a — c

0

... o

... o

b a — c b

Since Ti~T2~Jn(6)€(SI) , T is the sum of two (57) operators. If rank A > 1, by the matrix theory, we have

0 a12 • • • a l n _ i ai„ a-21 0 a2n

fln-n 0 a„_i„ &nl a n 2 • • • O-nn-l 0

0 a i 2 a J 3 • • • a l n

0 a23 • • • a2„

+ 0 a„_i„

0

where a^ ^ 0(i ^ j ) [Choi, M.D., Lausee, C. and Radjavi, H. (1981)]. It is easily seen that T is the sum of two (SI) operators.

0 021

a-n-n O-nl

0

O-n-12 •

O n 2 •

• 0

^ n n -

0

- i 0

3.2 Approximate Jordan Decomposition Theorem

If cr(T) is disconnected for T£L(7i), by Riesz decomposition theorem T£(SI), and by upper semi-continuity of spectrum, there exists an e > 0 such that T + Ag(SI) for all AeC(H) with \\A\\ < e. If a(T) is connected, we have the following theorem stated in section 1.5.

Spectral picture theorem of (SI) operators Given T€JC(H) with connected spectrum <J(T), there exists a (SI) operator L satisfying

(i) A(L) = A(T); (ii) TeS(L)-; (Hi) If there is another Lj6(57) such that A(Li) = A(T), then

L I G 5 ( L ) " .

By the above spectral picture theorem, we get the following theorem.

First approximate Jordan decomposition theorem [Jiang, C.L. and Wang, Z. Y. (1996b)] Let T&C(H), then there exists A&C(H) such that A is the direct sum of at most infinitely many (SI) operators and satisfying

(i) TGS(A)-;

(ii) IfB€£(H) andBeS(T)-, then BGS(A)~.

The interested reader is referred to [Jiang, C.L. and Wang, Z.Y. (1998)] for the proof of the theorem. In general, we call this theorem the unique decomposition theorem with respect to similarity orbit.

In 1988, D.A. Herrero raised the following conjecture in personal communication: Every operator with connected spectrum is a small compact perturbation of (SI) operator. Or precisely, given T€C(H) with connected a(T) and given an e > 0, there exists a compact operator K with \^K\\ < e such that T + Ke(SI). Following theorem confirms Herrero's conjecture.

Theorem 3.2.1 Let T&C(H) (or T&Bn(9)) with connected a(T). Then given e > 0, there is a compact operator K, \\K\\ < e, such that

T + Ke(SI) (orT + KeBn(n)n(SI)).

We will prove the theorem only in the Bn(Cl) operator case. Reader is referred to [Ji, Y.Q. and Jiang, C.L. (2002)], for the general case, which is a little more complex. Before we begin our proof, we need the following lemmas.

Lemma 3.2.2[Fialkow, L.A. (1981)] Let A, B^C(H), then the following are equivalent:

(i) ar(A)nat(B) = 0; (ii) TA,B is surjective; (Hi) ranTA,B contains IC(H).

Lemma 3.2.3 [Herrero, D.A. and Jiang, C.L. (1990)] Let

A,BeC(H). Assume that H = \J{ker(X - B)k : Aer,fc>l} for a certain subset T of the point spectrum o~p(B) of B, and ap(A)r\T — 0, then TA,B is infective.

Let 0 be a non-empty open connected subset of C such that (Q)° = Q, where (Q)° denotes the interior of the closure Q of Q. Given zo&Q, there exists a probability measure /z supported by T := d£l, the boundary of Q, satisfying f(zo) = fr fd^i for every function / analytic on fi [Herrero,

D.A. (1990)]. Let M(T) = "multiplication by A" on L2(T,/i). The subspace H2(T) of L2(T,fj,), spanned by the functions analytic on Q, is obviously invariant under M(T), i.e.,

M(r) = M+(T) Z o M_(r)

#2(r) L2(i»e#2(ry

Lemma 3.2.4[Herrero, D.A. (1990)] Let M{T),M+{T) and M_(r ) 6e as above, then

(i) M(T) is normal and both M+(F) and M-(T) are essentially normal; (ii) a{M{T)) = o-e{M{T))= <re(M+(T)) = <re(M_(r)) = T; a(M + ( r ) )=a(M_(r)) = fi; *nd(A - M+(T)) = -nul(X - M+(T)) = -nul{\ - M+(T))* = - 1 for

all \€il; (Hi) If Q is simply connected, then \\Z\\<m^ ', where m denotes the

planar Lebesgue measure.

It is obvious that M + ( r )*€# i (£ r ) and M-(T)eBi(Q), where Q* := {A : Aeft}.

An operator is quasitriangular if ind(\ — T)>0 for all A€/9s_ir(T). The Weyle spectrum aw{T) of TeC{H) is defined by aw{T) := D M ? 1 + # ) •' K&K(H)}. It is well known that aw(T) := {AGC : A - T is not a Fredholm operator of index 0}.

L e m m a 3.2.5[Herrero, D.A. (1990)] Suppose that T£C(H) is quasitriangular and a(T) = aw(T). Let T = {A„}^L1Ccr(T) satisfy that each clopen subset of o~(T) intersects the closure ofT. Given e > 0, there exist a compact operator K, \\K\\ < e, and an ONB {e^})^ such that

T+K

~ai

. 0

02

* e i

e2

where the set {a-i}^ll = T and card{j : aj = ai} = oo for each i.

Lemma 3.2.6[Herrero, D.A. (1990)] Given T£C{H), nonempty subset rco"; re(T) and e > 0, there exists a compact operator K, \\K\\ < e, such that

T + K = 0 AJ H2'


where N is a diagonal operator of uniformly infinite multiplicity with

o-(N)=T,a(A) = a(T),alre(A) = alre(T)

and

ind{A - A) = ind(T - A)

for each A£ps_ir(T).

Lemma 3.2.7 Suppose that

j , _ T i T12

~ [O T2

satisfies the following conditions: (i) T2eBn(n)n(SI); (ii)

nul{T\ — Afc) =.n, k = 1, 2, • • •

and

\J{ker(Xk - Ti) : k = 1,2, •••} = Hi;

(in) Bk := Pfcer(Ti-Afc)'?i2|fce7-(T-Afc) is injective, where Pfcer(Ti-Afc)* *«

t/ie orthogonal projection from H onto ker(Ti — Afc)*. Then Te£ n(f t )n(SJ) .

Proof {Claim} fcer(T - Afc) = ker{Tx - Xk). Assume that

Ti - Afc T12 0 T2 - Afc

for xGHi and y£H2, then (Ti - Afc)x + T i2y = 0 and (T2 - Afc)y = 0. Since

Pfcer(T1-A t)*(Tl - Afc) = 0,

Pfcer(Ti-Afc)*Ti22/ = 0.

Since Pfcer(Ti-Afc)*Ti2 is injective, y = 0. This implies that

ker(T - Xk) = ker(T\ - Xk)

n2

and

\ / { & e r ( r - A f e ) : * > l } = W i .

Since cr0(Ti)nfi = { fc}fcLi> there exists connected open subset fiiCnnp(Ti). Since T2eS„(fi) , \ f{ker{T2- \) : Aefti} = W2. Furthermore,

\/{ker(T - A) : Xe{Xk}^=1UQi} = Hi®H2.

Thus Te£ n ( f t ) . Now we will prove that TG(SI). Assume that PGA'(T) is an idempo-

tent. Since ker(T — Xk) = ker(T\ — Xk), k = 1,2, • • • , Hi is an invariant subspace of P, i.e.,

P = PiPn 0 P2

Hi H2

Since T2&{SI), P2 = SI-^2,S = 0 or 1. Without loss of generality, we can assume that 5 = 0, i.e., P2 = 0. Thus ranPcHi. Note that PS-A'(T) and T€Bn(fl). If P ^ 0, then T'|ranp6^ rn(0) for some m, l < m < n . Since

T| ranP = Ti\ • a n P i

Tii = Ti\ranp£Bm(Q),

It is easy to see that ranP is an invariant subspace of T\. Therefore,

Ti = 0 T^

ranP HiQranP'

By the condition (ii) of the lemma, Ti — A is invertible for AGfi\{Afc}^=1. Thus we can find an invertible XGC(H),

such that

Simple computations indicate that Xu(Tu — A) = I Thus Tn — A is invertible. This contradicts Tn&Bm(n). Therefore Px = P2 = 0 and TG(SI).

x =

Xn X\2 X2i X22_

^ 1 1 -X"l2

_X 2 i X 2 2 .

ranP HiQranP

'Tn - A £ ' 0 2 "22 ~ ^.

7o" 0 /

Lemma 3.2.8[Ji, Y.Q., Jiang, C.L. and Wang, Z.Y. (1996)] Suppose that B is essentially normal and B(=Bn(Cl). Then for given e > 0, there exists a compact operator K with \\K\\ < e, such that T + K£Bn(Cl)C\(SI).

Lemma 3.2.9[Herrero, D.A. (1987)] Suppose that R&C(7i) and satisfies the following conditions:

(i) o~(R) and o~w(R) are connected and contain a connected open set Cl;

(ii) ind{\ - R)>0 for all \€ps-F(R); (Hi) PS-F(R)DQ and ind(X — R) = n for all AGH.

Then given e > 0, there exists a compact operator K, \\K\\ < e, such that R-KeBn(fl).

Lemma 3.2.10[Herrero, D.A. (1984)] LetT€C(H) be quasitriangular operator and \£ae(T). If o~w{T) is connected, then given e > 0 and natural number m, there is a compact operator K, \\K\\ < e, such that nulT^ — km, k>\, and \J{kerT^ : k>l} = H, where Te = T + K - A.

Lemma 3.2.11 Let T&C{H) be quasitriangular operator and let Cl be a connected component of ps_F(T) such that crw{T)L)Cl is connected. Given e > 0 and natural number n, there exist a sequence {^k}k^=i of complex numbers in Cl and a compact operator K with ||K|| < e, such that {Afc}2L1C<r0(T + K) and

T + K^

Ai *

A2

0

cn

C"

where pflF{T) := {\£ps-F{T) : ind{\ - T) = 0}. Proof Note that dClCo-ire(T). We choose a dense subset {/ifc}^ of dCl. Then each clopen subset of o-w{T) intersects {pk}%K'=i- By Lemma 3.2.5, there exists a compact C\, ||Ci|| < §, such that

and

\J{ker{T + d - pky : j>l, k>l} = H

dim \J{ker(T + Cx - pk)j : j > l } = oo

for all k>l. Without loss of generality, we can assume that ap{T* + Cl) = 0. Since

ClCP^F(T) = p%(T + C1),

nncr(T + C 1 ) = 0 .

Denote

M1=\J{ker(T + C1-fi1)j:j>l},

then M. is an invariant subspace of T + Cx and dimM.\ = oo. Let

iwr + dju,,, then /UiGcre(Ti) and cr(Ti) = aw{Ti) is connected. If

dim\f {ker(T + Ci - fnY : i = 1,2,-•• , j > l } e A 1 i < oo,

then

fcer(T+ d - H i ) ' C M

for all j>l. Denote

Mk = yikeriT+d-fii)' : l<i<k}e\J {ker{T + Cx-fnY : l<i<k-l},

where k>l,j>l. Without loss of generality, we can assume that dimMk = oo. Thus

T + d

0

Since \J{ker(Tk — Hk)j '• j>l} = Mk, crw(Tk) = cr(Tfc) is connected and Cp(Tk*)c{Jik}- Thus Tk is a triangular, /ifeG<re(Tfc) and £lC\a(Tk) = 0 for each fc>l. By Lemma 3.2.10, there exists a compact operator Kk, \\Kk\\ < fz, such that

Tk + Kk

Mfc * " (J-k

0

c n

C" cn> fc = l , 2 , - .

(fc) . j

fc>l, j > l } in ft such that \fXk — A^KJ| < -^ for all j and fc. Thus there

Since {/Ujtj^iCcftl, we can choose pairwise distinct numbers {A

is a compact operator Kk with \\Kk\\ < pr such that

Ak = Tk + Kk + Kk*

iW

\ ( 2 )

i<3>

C"

c C"i ^ = 1>2, •••

Furthermore, there is a compact operator C2,11C21 i < § satisfying

T + Cx + Co =

'J4I •

0

M i M 2

M3'

where C2 = ^ ©(-Kk + ^fc)- Set if = C\ + C2, then K is compact and fc=i

i ( *> H^ll < e. Rearrange {A - J : J>1, /s>l} as {^k}kLi- ^ is n o t ; difficult to see that K and {Afcj^j satisfy all the requirements of the lemma.

Now we are in a position to prove Theorem 3.2.1.

Proof of Theorem 3.2.1 For T&Bn(fi), assume that dQ.Caire(T) (otherwise, replace Q with the component of PS-F(Q) containing fi). Denote $ = (Q)° and T = d$. Then rccr, r e(T). By Lemma 3.2.6 and Lemma 3.2.9, we can find a compact operator K\, \\Ki\\ < | such that

T + K^ Tx * 0 iV W2

:

where Ti€#n(fi) , N is a diagonal operator of uniformly infinite multiplicity and CT(JV) = T. Let M(T) be given as in Lemma 3.2.4, then M(T)^ =

n 0 M(T) is normal and

M{T) (n) 0 M+(T) 0 Z fc=i fc=i

0 0 M_(r ) fc=i

It is clear that a(M(T)W) = T.

By Voiculescu theorem [Voiculescu, D. (1976)] there exists a compact operator Fx such that \\Fi\\ < § and N + F^MIT)^. Thus

T + Ki+ K2^

Ti * *

o 0M+(r) J z k=l k=l

0 0 0 M_(T) fc=i

where K2 is compact, K2 = 0©i<2 and H-ft H < f • Set

o ©M+(r) fc=l

It is obvious that a(Bi) = CT(T), Q C p ^ B i ) and ind(Bx - A) > 0 for all A in cr(Si)npir(-Bi)\f2. Therefore B\ is quasitriangular, fi is a connected component of pp{B\) and Q,r)aw(Bi) = cr(T) is connected. By Lemma 3.2.11 we can find a sequence {Afc}^ °f pairwise distinct complex numbers in Q and a compact operator E, \\E\\ < JQ such that {Afe}^=1Ccr0(yl1) and

Ai = B + E^

Ai C"

c

Summarizing the arguments above, we can find a compact operator if3, ||#31| < y | such that

T + Kt + K2 + K3* A\ *

n 0 © M _ ( T )

fc=i

n3

Denote B2 = 0 M_(r) . Note that B 2 is essentially normal and k=i

B2€Bn{$)cBn(£l) [Cowen, M.J. and Douglas, R. (1977)]. By Lemma 3.2.9,

we can find a compact operator F2, \\F21| < ^ such that

A2 = B2 + F2eBn(n)n(Sl).

Thus there is a compact operator K4, \\K^\\ < ^ such that

T + Kx + K2 + K3 + X 4 = [ ^ A.12

U A2

According to Lemma 3.2.7, it is sufficient for us to find a compact operator E12 such that | |£1 2 | | < fi and Pfcer(A!-Afc)*(^12 + •£;i2)Uer(A1-Afc) is injective. Since

{Xk}k^L1Cap(A1)nPF(A2)

and

nul(A\ — Xk) — n,

X\ G\2

A2

cn

C"

1 - 0

"Gi * " 0 G2 Woo

J n o

where

Gi

Xi G\2 G13 • •

X2 G23 • •

A3 '•

C" C"

and G2 = A»- Since A2eBn(Q),ker(A2 - Xk) = ker(Gx - Xk) and

G = Ax Ex E2

0 Gi * L 0 0 G2

©C" fc=l

0 0

©C" fc=l

We need only to find a compact operator i*3, H-F3JI < < , such that

Pker(A-\k)*(El + i i3)Uer(G 1 -A J , )

is injective. Since \k£0o(Ai) for k>l,A\ has the following expression:

Ax = A3 C23 C13

A2 C12

Ai

"Ho

C" c cn

Thus Afc £ a(Aoo) for all &>1. Under the above decomposition,

£1 =

* * *

E31 E32 E33 • • •

E21 E22 E23 • • •

En E12 E13 • • •

: *

: *

: *

where E^ is an operator from one n-dimensional space to another for i, j>l. It is obvious that we can choose F n with \\Fn\\ < | such that En + Fu is invertible. Inductively, we can choose Fjj with \\Fjj\\ < ^ such that Ej+ij+i + WjX~1Vj + Fj+i:j+i is invertible, where

Wi = (Gj,j+i> Gj-i,j+i, • • • . Gitj+i, Ej+iti, • • • , Ej+ij),

Xj — Xj

-\? cj,j-i ' •• c j , i Ej,i

Xj-i " • Cj-1,1 Ej-iti

Ai En + Fn A2

Aw + Au

Xj

and

Vj =

Ej-i,j+i

Fi,j+i

Set

F2 = 0

Fn

F22

F33 '•

0 :

then F2 is compact and j|JF211 < -§i-From the construction we can see PkeriAt-x^'iEi + Fi)\ker(A2-\k)

ls

injective for all k>l. Therefore the proof of the theorem is now complete.

Definition 3.2.12 A sequence {Pj : l<j<l}(l<l<oo) of nonzero idem-potents in C{Ti) is called a spectral family if there exists an invertible operator X such that {XPjX~l : l<j<l} is a sequence of pairwise orthog

onal projections and ^ Pj = L n-i = i

Definition 3.2.13 Let TeC(H) and V = {Pj : l<j<l}(l<l<oo) be a spectral family. V is called a strongly irreducible decomposition of T if the following conditions are satisfied for all j :

(i) PJT = TPJ;

(ii) T\ranPje(SI).

Note, if there exists a strongly irreducible decomposition of T, then we say that T has a strongly irreducible decomposition. In other words, T has a strongly irreducible decomposition means that T is the topological direct sum of at most countably many (SI) operators, or, T is similar to

the orthogonal sum of at most countably many (57) operators.

Definition 3.2.14 Let Vx = {Pj • l<j<h} and V2 = {Qj • l<j<h} be two strongly irreducible decomposition of T. V\ and V2 are said to be similar about T if

(i) h=h = l;

(ii) there exist a permutation ir of {j : 1<7</} and invertible operator

Xi£C(ranPj, ranQv(j))

such that

5 U P { | | X j | | , | | X - 1 | | , l < i < 0 < + C O

and

XjT\ranpj = TranQ^u)Xj, l<j<l.

In Chapter 4, we will show that V\ and Vi are similar about T if and only if there exists an invertible operator X&A'(T) such that

{XPjX-1 : l<j<l] = {Qj : l<j<l}.

Definition 3.2.15 Suppose that T has strongly irreducible decomposition. T is said to have a unique strongly irreducible decomposition up to similarity if any two of the (SI) decomposition of T are similar.

The Jordan canonical theorem in finite dimensional space means essentially that each nxn matrix has a unique (SI) decomposition up to similarity. For operators in £(H), it is very difficult to obtain a "Jordan canonical theorem". In Chapter 5 of this monograph we will prove the following theorem.

Theorem FJ Every Cow en-Douglas operator has a unique (SI) decomposition up to similarity.

Using Theorem FJ, we obtain the following two approximate Jordan canonical theorems.

Theorem 3.2.16 Given T&C(H) and e > 0, there exists Ae£(H) such that A has a unique (SI) decomposition up to similarity and \\T — A\\ < e.

Theorem 3.2.17 Let TGJC-(H) be quasitriangular. Assume that cr(T) consists of finitely many components and each component intersects pp(T).

Given e > 0 there exists a compact operator K with \\K\\ < e such that T + K has a unique strongly irreducible decomposition up to similarity.

Theorem 3.2.16 can be proved by using of Theorem JW2 in Chapter 1 and Theorem FJ, and Theorem 3.2.17 can be proved by Theorem 3.2.1 and Lemma 3.2.4.

3.3 Open Problems

1. Given TGC(H) and e > 0, does there exist a compact operator K with \\K\\ < e such that T + K has unique (SI) decomposition up to similarity? 2. Is every operator in C(H) the sum of two Cowen-Douglas operators of index 1? 3. Given TGC(H) and e > 0, does there exist an integer p, l<p < oo and KeCP(H) with \\K\\P < e such that T + Ke(SI)? where

OO

CP(H) = {KeK{H) : ^ A £ < oo, \n£ap(K*K)±} n=l

and

3.4 Remark

The concept of unique strongly irreducible decomposition up to similarity appeared first in [Jiang, C.L. and Wang, Z.Y. (1998)]. Theorem 3.1.1 is given by [Yue, H. (2002)]. Before that, [Jiang, C.L. and Wu, P.Y. (1998)] proved that each operator in £(H) is the sum of three (SI) operators, and each triangular or compact operator is the sum of two (SI) operators. Theorem 3.1.11 is given by [Jiang, C.L. and Wu, P.Y. (1998)]. Theorem 3.2.1 is due to [Ji, Y.Q. and Jiang, C.L. (2002)]. Theorem 3.2.16 and Theorem 3.2.17 are both proved by [Jiang, C.L.(l) ].

Chapter 4

Unitary Invariant and Similarity Invariant of Operators

In this chapter % always denotes a complex, separable infinite dimensional Hilbert space. One of the basic problems in operator theory is to determine when two operators A and B in C(H) are unitarily equivalent or similar. In infinite dimensional Hilbert space, this problem has no general solution. What we can do is to find the answer for some special classes of operators. A quantity (quantities) or a property (properties) P is unitary (or similarity) invariant (invariants) if A has P and A=B(A~B) implies that B has P. For example, reducibility and strong reducibility are unitary invariants while strong reducibility is only the similarity invariant. For a subset R of C(H), unitary (similarity) invariant (invariants) P is completely unitary (or similarity) invariant (or invariants) if AGR, then A=B (or A~B) if and only if BGR and A and B have same P. From this point of view, one of the basic problem in operator theory mentioned above is to determine the completely unitary or similarity invariants. We have seen in Chapter 2 that eigenvalues and generalized eigenspaces are completely similarity invariants of nxn matrices. [Conway, J.B. (1990)] showed that two *-cyclic normal (or subnormal) operators A and B in £(H) are similar if and only if the scalar-valued spectral measures induced by them a equivalent, while they are unitary equivalent if and only if they are similar. Here, an operator A is normal if A* A — AA*, and A is subnormal if there exist a normal operator JV and an invariant subspace M. of N such that N\M = A. For two injective unilateral weighted shifts, the boundedness of the ratios of the products of their weights is the completely similarity invariant [Shields, A.L. (1974)]. There have already been a lot of results on the similarity invariants of operators, especially that of non-adjoint operators, which can be found in, for example, [Herrero, D.A. (1987)], [Herrero, D.A. (1990)], [Conway, J.B. (1990)].

43

In this chapter, we will discuss further the unitary invariants and similarity invariants of non-self-adjoint operators.

4.1 Unitary Invariants of Operators

We begin with a famous theorem.

Schur theorem Each nxn matrix X is unitary equivalent to the orthogonal direct sum of irreducible matrices.

For T££(H), let W*(T) denote the von-Neumann algebra generated by T. By von-Neumann double commutant theorem we can easily prove the equivalence of the following conditions.

(i) T is irreducible; (ii) A'(W*(T))=I;

(hi) W*(T) = C(H). The following proposition tells us that the Schur Theorem can not be

generalized to C{Ji).

Proposition 4.1.1 Let Ne£(H) be a self-adjoint operator with crp(N) = 0, then N is not orthogonal direct sum of irreducible operators. Furthermore, N is not the topological direct sum of (SI) operators.

i

Proof For the first part of the proposition, if N = Yl (BNi,l<l<oo, i = i

where NiG(RI), then Ni is self-adjoint and the the dimension of the space Hi, on which Ni acts, is just 1. Thus o~p(N) ^ 0. It is a contradiction.

For the second part of the proposition, if PGA'(N) is a nontrivial idem-potent, then by the spectral theorem of self-adjoint operators, there exists a orthogonal projection P' such that ranP = ranP' and P'GA'(N) [Putnam I.[l]]. If N is the topological direct sum of sum (SI) operators, then by the fact stated above, we can find an orthogonal projection P'GA'(N) and N\ranP' is irreducible. By the argument used in the first part we conclude that crp(N) j= 0. It is also a contradiction.

Although it is not every operator to be the direct sum of irreducible operators, we have the following proposition.

Proposition 4.1.2 Every operator T&C(7i) is the direct integral of irreducible operators. Proof In fact this is a corollary of Theorem 3.6 of [Azoff, E.A., Fong, C.K. and Gilfeather., F. (1976)]. The weakly closed algebra A(T), generated by

Unitary Invariant and Similarity Invariant of Operators 45

T and / , can be expressed as /A ®A\d[j,(\), where A is a separable metric space, fi is a regular a-finite Borel measure on A and A\ is a weakly closed irreducible operator algebra for almost all AsA. Recall that an operator algebra is irreducible if it has no nontrivial reducing subspace. Therefore, we have

T= / ® T A ^ ( A ) , T A e A , A e A ./A

a.e..

Thus A(T)cfA®A(Tx)diJ,(\)cfA®Axd[i(\) = A(T). This implies that A{T\) = A\ for almost all AeA. Thus the irreducibility of ,4A implies that T\£(RI). Therefore T = fA®T\dn(\) is the asserted irreducible integral decomposition of T.

For C*-algebra A and natural number n, let M„(A) denote the set of all nxn matrices with entries in A. The following theorem gives the number of reducing subspaces of an operator.

Theorem 4.1.3 The number of reducing subspaces of any operator Te£(W) is either finite or uncountably infinite. The former case occurs if and only ifT is the direct sum of finitely many irreducible operators, i.e., T = Ti©T2© • • • ®Tn, and Ti,Tj are not unitarily equivalent ifi^j. In this case, the number of reducing subspace is 2™.

Theorem 4.1.3 has a similar pattern but in a different context with the following result of [Ong, S.C. (1987)].

Theorem Ong Given T € £ ( C n ) , the number of invariant subspaces ofT is either finite or uncountably infinite. The former case occurs if and only if T has a cyclic vector.

To prove Theorem 4.1.3, we need four lemmas. The first is a structure theorem for two orthogonal projections, and has been quoted in many literatures before. The reader is referred to [Halmos, P.R. (1968)].

Lemma 4.1.4 Given two orthogonal projections P and Qg£(W), there is a unitary operator U such that

U*PU

and

U*QU =

/ i 0 0 0 ®h®h®0®0

A B B h-B

©/2©0©/4©0

under the decomposition of the space H = Hi®H2®H3®Hi®Hs, where J 4 G £ ( W I ) is a positive contraction and B = [A(Ii — A)] 2. We may assume that 0 < A<^Ii and A is unique up to unitary equivalence.

Using Lemma 4.1.4, we can prove the next lemma.

Lemma 4.1.5 IfTGC(H) has countably many reducing subspaces, then A'(W*(T)) is abelian. Proof Let P , Q&A'(W*(T)) be two orthogonal projections represented as in Lemma 4.1.4. Since PT — TP and QT = TQ, a simple computation

5 indicates that T = T\@T2® ^ ©T, under the decomposition of the space

i=3 5

and T\A = AT\. For each complex number A, denote

M\ = {\BX®X®Q®Q®Q : XeHi}.

It is easily seen that M\ is a reducing subspace of T, and if Hi ^ {0}, then

Mx ± My (A ^ A').

Since T has only countably many reducing subspace, Hi = {0}. Thus

P = /2©/3®o©0

and

Q = /2©0©/4©0.

This implies that PQ = QP. Since von-Neumann algebra A'(W*(T)) is generated by the projections in it, A'(W*(T)) is abelian.

Recall that a projection p in a C*-algebra is minimal if there is no projection q other than 0 and p such that pq = q.

L e m m a 4.1.6 Let PeA'(W*(T)) be a projection, then P is minimal if and only ifT\ranpG(RI).

The proof of this lemma is an easy consequence of the definition of minimal projection.

Lemma 4.1.7 Let A, BGC(H)C\(TZT), then A and B are unitarily equivalent if and only if there exists a nonzero operator X such that XA = BX and XA* = B*X.


Proof If XA = BX and XA* = B*X for some nonzero X, then kerX and ranX are reducing subspaces of A and B. If kerX ^ {0}, by the irreducibility of A kerX = Ti., i.e., X = 0, a contradiction. Thus kerX — {0}. Similarly, we can conclude that ranX = "H, i.e., X has a dense range. Let X = UP be the polar decomposition of X, where U is unitary and P = (X*X)2>0. Since

X*X,4 = X * £ X = AX*X,

PA = AP.

Therefore

UAP = UP A = XA = BX = BUP.

Since P is also range dense, UA = BU. Thus A=B.

We are now in a position to prove Theorem 4.1.3.

Proof of Theorem 4.1.3 Assume that T has countably infinite many reducing subspaces. By Lemma 4.1.5, A'(W*(T)) is abelian. Thus A'(W*(T)) is generated by some Hermitian operator A [Radjaval, H. and Rosenthal, P. (1973)].

Note that cr(A) can not be finite. Otherwise,

n

A = Y^ ®*Ji i=l

and n

W*(A) = {J2®aiIi:ai€C}. »=i

This implies that W*(A) = A'(W*(T)) contains only finitely many projections, and contradicts the assumption. Thus a{A) can be decomposed into countably infinitely many pairwise disjoint Borel subsetslcrj}?^, each of which has a strictly positive spectral measure. Since a (A) has un-countably many different decompositions, so is the spectral projections of A. Therefore, there are uncountably many orthogonal projections in W*(A) = A'(W*(T)). This is also a contradiction. Thus the number of reducing subspaces of T can not be countably infinite.

Now we assume that T has finitely many reducing subspaces. By Lemma 4.1.6 A'(W*{T)) is abelian. Let Pi , - - - , P n be minimal projections in

A'(W*(T)). Since PP,- = P , P , it is not difficult to prove that PiPj = 0 n n

for i ^ j , and ^ P* = / . Let T — S 0 ^ under the decomposition 1 = 1 2 = 1

n

H = Yl ®ranPi, where T{ = T\ranPi- By Lemma 4.1.7, Ti£(RI). We are i=l

now to prove that T is not unitarily equivalent to Tj for i ^ j . Otherwise, if Ti=Tj, there is a unitary operator £/ such that UTi = TjU, where l if A 7 A' and A4\ is a reducing subspaces of T. Note that there are infinitely many of .MA'S, this contradicts our assumption of T.

Conversely, if T = £ ®Tt on H = £ ©"Hi, T^RI) and Tj is not i = l i= l

unitarily equivalent to Tj for i ^ j . Let P = {Py}".,=i be orthogonal projections commuting with T. Then PyTj = TiPij for all l<z, j<n. Thus

pyT/ = P^T; = (TjPjy = ( P , ^ ) * = T ;P^ = i-p,-.

Since T, and Tj are irreducible and not unitarily equivalent for i 7 j , by Lemma 4.1.7, Py = 0 and so Pji = 0. Thus Pjj is an orthogonal projection commuting with T,, which implies that Pa = 0 or I-nt (i = 1,2, ••• , n). Therefore T has only 2™ reducing subspaces.

In the following, we will characterize the decomposibility of an operator into direct sum of irreducible operators in terms of C*-algebra language. For TG£(H) and an integer n, l<n<oo, let T^ denote the direct sum of n copies of T.

Theorem 4.1.8 An operator T£C(H) is the direct sum of irreducible " ( •) operators (i.e., T = ^ © T ; , l<Tii<oo, l<n<oo,TiG(PJ^ one! are pair-

i=l wise not unitary equivalent) if and only if A'(W*(T)) is ^-isomorphic to n

^2 ©M n i(C). Moreover, the (RI)-decomposition ofT is unique in the sense i= l

m . . of unitary equivalent. Precisely, if T = Yl © fc is another direct sum

fc=i of irreducible operators {Sk}™=1, which are pairwise not unitary equivalent, then n = m and there exist a permutation n of {1,2, • • • ,n) and a unitary operator U in A'(W*(T)) such that n, = rnw^ and UTi = Sn^U,i =

1,2,--- ,n. Note that [Takesaki, M. (1979)] showed that every finite dimensional

C*-algebra is *-isomorphic to the direct sum of finitely many full matrix algebras. Thus we have the following corollary.

Corollary 4.1.9 T is the direct sum of finitely irreducible operators if and only ifdimA'(W*(T)) < oo.

To prove Theorem 4.1.8, we need the following lemma.

Lemma 4.1.10 If T is irreducible on H and X££(H) satisfying XT = TX and XT* = T*X, then X = dl for some deC. Proof Since X*X commutes with T, T commutes with any spectral projection P of X*X. Since T&{RI),P = 0 or P = I. Thus a{X*X) is a singleton {a} and X*X = al. On the other hand, it follows from XT = TX and XT* = T*X that kerX is a reducing subspace of T. Since T€(RI),kerX = {0} or H. Similarly, ranX = H or ranX~ = {0}. This implies that X = 0 or X is injective with dense range. Thus

where U is unitary. If a ^ 0, then UT = TU and UT* = T*U. Repeating the arguments above, we have U = /3I or X = dl, where d = \/a/3. Thus, the proof is complete.

Proof of Theorem 4.1.8 Let T = £ ®T}ni) satisfy the conditions of i= l

the theorem. For any X€A'(W*(T)), by Lemma 4.1.10,

n

X = J^(BXi,Xi€A'(W*(T^)). i = l

Let

then

YjkzA'(W(Tt)).

By Lemma 4.1.10, Yjk = ^)kh, where 7j is the identity on Wj.Thus

n It is obvious that the mapping J n ^ ®[^}fc-^]?fc=i defines a *-

i=l isomorphism from A'(W*(T)) onto £) ©Mn i(C).

Conversely, if 4> is a ^-isomorphism from A'(W*(T)) onto A := n

X)©Mn j(C) and let Etj denote the element 0® • • • ®etj® • • • ffiO in A, i=l where e^ is an riiXTii matrix whose (i,j)-entry is 1 and the others are 0. Then 4>~l(Eij)£A'(W*(T)) are pairwise orthogonal minimal projections with £</>_1(.EV,) = ln- Obviously, by Lemma 4.1.7, ^{E^H is one of

•J the reducing subspaces of T with

Ta := TU-HEii)nZ{BI)

and

T = £©!;,.

Since Eij^Eik for all (j, k), Tij^Tik and T ^ £ ®TJni). »=i

m . .

In order to prove the uniqueness, let T = X) ©S^. be another ex-

pression of T, where {Sk}™^ are pairwise not unitary equivalent and m .

H — ^2 ®L(™k • Let Pki be the orthogonal projection from H onto the fc=i

Z-th subspace of £j, > then F ^ = <t>{Pki) are pairwise orthogonal in n

.4 := £ ® M n i ( C ) , and Y,Fki = ^ Since T | r a n p u e ( i ? / ) , P H is minimi fc,i

mal. Thus Fki is minimal in A and there exists an integer n, such that Fki€Mni(C) with rankFu = 1. Therefore X).FM = 7ni and so mk = rii.

I Prom .Fjfez = / , we can conclude that m — n. The remainder of the theorem can be proved directly from the *-isomorphism 4>.

Next we will consider when two operators have isomorphic reducing subspaces lattices. [Conway, J.B. and Gillespie, T.A. (1985)] solved this problem in the case of normal operators. Using their result, we can characterize isomorphism of the reducing subspace lattices of two operators if they can be expressed as direct sums of irreducible operators.

Proposition 4.1.11 Let

j = i

anc?

* = £«*£" fc=l

where Aj,Bk&(RI), {Aj} and {Bk} are pairwise not unitarily equivalent, l<n,m<oo, l<nj,mfc<oo. Then RedA is isomorphic to RedB if and only if n = m and there exists a permutation n of {1, 2, • • • , n} such that rij = mT(j) for j = 1,2, ••• ,n, where RedA and RedB denote the lattices of reducing subspaces of A and B respectively.

To prove the proposition, we need the following lemma.

Lemma 4.1.12 IfTe(RI), then RedT^ is isomorphic to Redln for all n, l<n<oo, where In is the identity on an n-dimensional space. Proof If V = {Pij}lj=1 is a projection in A'(T^), by Lemma 4.1.10, Pij = \jl f° r some AySC, l{A,J}"_7-=1

induces a lattice isomorphism from RedT^ onto Redln.

Proof of Proposition 4.1.11 By Lemma 4.1.7, RedA is isomorphic to ^ © i Z e d A ^ 0 , and by Lemma 4.1.12 £@#e<L4^ni) is isomorphic to J i

RedJ2®jInj- Thus ite<L4 is isomorphic to Red^2®\lnj. i i

Similarly, RedB is isomorphic to Red ^ ®\lnk • Therefore if RedA is iso-k

morphic to RedB, then n — m and there is a permutation 7r of {1, 2, • • • , n} such that rij = ^ ( j ) [Conway, J.B. and Gillespie, T.A. (1985)]. This proves the necessity part. The sufficiency is obvious.

Proposition 4.1.13 IfT^ is a direct sum of irreducible operators, then so is T. Proof Without loss of generality, we assume that

n

t= i

{Tj}™=1 is a sequence of irreducible operators, which are pairwise unitarily inequivalent. Then there are pairwise orthogonal projections Pj,j =

1,2, • • • , n, each of which commutes with T^ and J2Pj = !•> s u c n t n a t

r^fc^Irani3,,i = 1,2, • • • ,k are pairwise unitarily inequivalent. By Lemma 4.1.7,

Pj = 2_^ ®Qij,

where Qij£A'(T> ) , {Qij} are pairwise orthogonal, Y^®Qij = ^ anc^ j

Tt \ranQij, j = 1,2, • • • , A; are pairwise unitarily equivalent. Therefore we need only to prove that if A^^B^, l<n<oo, Be(RI), then A is a direct sum of irreducible operators. We may also assume that n = oo. Otherwise A'(W*(AW)) = Mk(A'(W*(A))) is finite dimensional by Corollary 4.1.9. This implies that A'(W*(A)) is also finite dimensional. Therefore A can be expressed as a direct sum of irreducible operators. Since A^=B^°°^ = C, A(k)^C(k\ It follows that A^C by [Kadison, R.V. and Singer, I.M. (1957)] and the proof is complete.

Now we will describe the necessary and sufficient condition for an operator to be sum of irreducible operators in terms of if-theory language. The main result is the following theorem.

Theorem 4.1.14 Given T£C(H). T is the direct sum of irreducible operators if and only if \J (A'(W* (T)))^^ ®{N ^{00})^ for some ki,k2,0<k1,k2<oo, where \J(A'(W*(T))) is the semigroup of A' (W* (T)).

The proof of Theorem 4.1.14 needs the following lemmas.

Lemma 4.1.15 Let P and Q be two projections in A'(W*(T)). If there exists a unitary operator UGA'(W*(T)) such that UP = QU, then

J- \ranP~-L \ranQ-

Proof Set W = U\ranp- Then W is a unitary operator from ranP onto ranQ and satisfies W(T\ranP) = (T\ranQ)W.

Lemma 4.1.16 Let TeC(H) and

\J(A'(W*{T)))^N{fl]®(N+u{<^}){k2\

0<ki,k2<oo and let I = k\ + &2,{e,}'=1 be I free generators in \/(A'{W*{T))) and P 7 0 be a projection in A'(W*(T)). Then T\ranP e (RI) if and only if [P] = ei for some i.

1 Proof Assume that T|ranf> is irreducible and [P] = J2 ©aje,, where a* is

an integer, CKOJJ < oo. If there are at least two nonzero a^s, say a\, a? ^ 0.

Then / = a.\e\ and g = J2 ®aiei are nonzero elements in \J(A'(W*(T))). i=1

Thus there exist a natural number m and mutual orthogonal projections Q and RGA'(W*(T^)), such that [Q] = f and [R] = g. Set S = Q + R, then

/ [5] = [Q] + [R} = f + g = J2 ^id = [P].

Therefore S is unitarily equivalent to P©0( m - 1 ) in A'(W*(T^m^)), where 0 denotes the zero operator on H.

By Lemma 4.1.15, rr(rn)] „ /^> /T>(m)j -t '\ranS = J- " r a n P © 0 ( " - ! )

^ T\ranP€(RI).

This contradicts the fact that T\ranp€(SI). Thus [P] = a»ei. Similarly, we can prove a* = 1.

Conversely, assume that [P] = ei and T\ranp is reducible. Then there are nonzero projections Q^GA'CW^T^)) such that QR = 0 and P = <2 + P. Let

[Q] = 53 ®aiei

i = l

and

[P] = 53©ftei, i = l

where 0<aj,/3i < oo for all i. Thus

ei = [P] = [Q] + [P] = ^ ©(a, + &>)<* i = i

and

for all i>2. Therefore ax = 0 and j3x = 0 and on = # = 0 for i>2. This implies that [Q] = 0 or [R] = 0. A contradiction. Therefore T\ranp£(RI)-

Lemma 4.1.17 Let AGC(H) be a direct sum of irreducible operators and B&C{K) have no reducing subspace on which B is irreducible. If there exists an operator X such that XA = BX, XA* = B*X, then X = 0.

oo

Proof Without loss of generality, we assume that A = £ ®An with n = l

oo respect to H = £ ©W„, where Ane(RI). Then X* = [X{, X | , • • • ]*. Now

n = l

we prove X\ = 0. In fact, from XA — BX and XA* = 5*X, we have XXB = AtXi and XXB* = A*X. Thus {X^^Ai = Ai(XiX{) and (XiXj*)^ = A\(XiX{). Since ^ is irreducible, by Lemma 4.1.10 XXX{ = XIHl for some AeC. If A ^ 0. Let U = \~lXi. Then UU* = IHl and Q := U*U is a projection in C{K.) with Q-B = BQ. Set

p = IHl®0, Q = 0©Q,

then p, ge£(Wi ©£) • Set

p' = pffiO, ?' = q®0,

where

Set C = J 4 I © B . We claim that p' and g' are unitarily equivalent in A'{W*(C^)). To prove this, we define

v=^QU^££(H1+JC).

It is easy to see that v is a partial isometry, vv* = p and v*v = q. Then the assertion follows from the Proposition 5.2.12 of [Wegge-Olsen, N.E. (1993)]. By Lemma 4.1.15, C^\ranp^C^\ranql. But C^\ranp, = A^RI). Thus

C \ranq'=B\ranQe(RI),

which contradicts our assumption on B. Hence X\ = 0. By the same arguments we can prove that Xn = 0 for n>2. Thus X = 0.

Proof of Theorem 4.1.14 The necessity follows from the analysis above. We only prove the sufficiency. Assume that

\/{A'(W*{T)))^l)@{N+U{oo}Yk2\ 0<fci, fc2<oo.

Let P be a projection in Mk(A'(W*(T))) = A'{W*{T^)) such that [P] is a free generator of \/(A'{W*(T))). By Lemma 4.1.16, T^\ranPe(RI) (here we embed A'(W*(T)) into Mk(A'(W*(T))) with the embedding

\A 0] map A>-» ). Using Zorn's lemma, we can find a maximal fam

ily in A'(W*(T^)) of pairwise orthogonal projections {Pj}n=l, l<n<oo

such that T^\ ranPj is irreducible, j = 1,2, ••• . Set Q, — ^2,Pj, we j

will prove that Q = 7^fc', the identity operator on H^. Otherwise, set 7\ := T^\ranQ,T2 := ^ ( fc) | r<m(/«-Q)- Since <? is a projection in .4'(W*(T(fe))), T\ is a direct sum of irreducible operators and T<i has no reducing subspace M. such that T2I.M £ ( # / ) . Applying Lemma 4.1.17, we have

Thus

V(^'(W/*(T(fc))))^V(-4'(iy*(:ri)))®\/(-4'(W'*(T2)))

(Isomorphism Theorem).

Let R be a projection in A'(W*(T2 )) for which [i?] is a free generator of \J(A'(W*(T2))). By Lemma 4.1.16, T^m)\ranRe(RI). By the similar argument above, we find a nonzero projection Q iGA' (W* (T2 )) such that T3 := T2 |ranQi is the direct sum of irreducible operators and T4 := T;j; |ran(j-Qi) n a s n o reducing subspace .M with T4|^€(i?7). Using Lemma 4.1.17 again, we have

A'(W*(T^m))) = A'(W*(T3))®A'(W*(T4)).

Thus Q\ commutes with every operator in A'(W*(T2 )) and QiGA'(W*(T2 )) by von-Neumann double commutant theorem. Therefore Q = S(m\ where S is a nonzero projection in W*(T2), and

rp <7-t(m)| jrn | \(m)

Since T3 is the direct sum of irreducible operators, so is T2\TanS by Proposition 4.1.13. This contradicts the assumption on T2. Thus Q = 1^ and T(fc) is a direct sum of irreducible operators. By Proposition 4.1.13, T is also a direct sum of irreducible operators.

By Theorem 4.1.14 and its proof, we have the following theorem.

Theorem 4.1.18 Let A,B£C(H). If A is the direct sum of irreducible i , ,

operators, i.e., A = J2 (BA),n , 1 < / < O O , l<rij<oo. Then each two of {Ai\ i = l

are not similar if and only if there exists an isometric isomorphism (j) such that

(j>{\J{M{W*{A®B)))) = N^fcl)©(N+U{oo})(fc2\ Q<kx,k2<oo

i

and (f)(1) = 2^2,riiei, where {e,} are the generators of\J(A'{W*(A))). i=l

Note that for A£C(H), if A is a direct sum of infinitely many irreducible operators, then KQ(A'(W*(A))) = 0. Thus, in this case, i^o-group can not describe the unitary equivalent relation between operators. But we have the following theorem.

Theorem 4.1.19 Let A,Be£.(7i). Assume that A is the direct sum of i , ,

finitely many irreducible operators: A = ^Z ©Aj™ , 1<Z < oo, l<7ii such that

cj>(K0(A!(W*(A®B)))) = N^ }

and <j>(I) = 2 ^ n^ei, where {e,} are the generators of Ko(A'(W*(A))). i = i

Note that for A€Bn(Q.). Since A is a direct sum of finitely many irreducible operators, we can get the following theorem by Theorem 4.1.19.

Theorem 4.1.20 Let A,B£Bn(fl), then the following statement hold. (i) A has a unique (SI) decomposition up to unitary equivalence.

k (ii) IfA= J2®A\ni>,Aie{RI) and{Ai}k

i=l are pairwise unitarily in-2 = 1

equivalent. Then A=B if and only if there exists an isomorphic map <f> such that

<f>(K0(A'(W*(A®B)))) = Z<fc)

and 4>(IA'(W(A®B))) = ^Ylniei' where {e;} are the generators of \j(A!(W*(A))).

4.2 Strongly Irreducible Decomposition of Operators and Similarity Invariant: K0-Group

It is more convenient in terms of iô-group to describe the unitary invariant of operators which are direct sum of irreducible operators. The proof of it depends strongly on the tools of C*-algebra theory. We give, in terms of semigroup theory, a necessary and sufficient condition to that an operator can be expressed as a direct sum of irreducible operators. In this section, we first prove the following theorem.

Theorem 4.2.1 Given Te£(H), the following are equivalent:

(i) T is similar to ^2 ®A\n under the decomposition of the space »=i

k . .

H=J2 ®KT , where k,ni < oo, Ai£(SI), AiÂj for i ^ j and each T(n> i=l

has a unique (SI) decomposition up to similarity;

(ii) \J(A'(T))= N(k>, and this isomorphism 4> sends

[I]—miei + n2e2 H h n ^ ,

where {ej}f=1 are the generators of N^ and N = {0,1,2, • • • }, n, ^ 0. From Theorem 4.2.1, we obtain the following corollary.

Corollary 4.2.2 Let T i ,T 2 e (5 / ) ,T = Ti®T2. If \f(A'(T))*N, then Ti~T2 . Moreover, if T^ has a unique (SI) decomposition for all natural numbers up to similarity, then Ti~T2 if and only if Ko(A'(T))=Z, where Z := {0, ±1 , ±2, • • • }.

Before we prove Theorem 4.2.1, the following lemmas are needed.

Lemma 4.2.3 Given A, BGC(H). Assume that ip is an isomorphism from A'(A) onto A'(B). {Pi}^ i is an (SI) decomposition of A if and only if {y>(P;)}"=1 is an (SI) decomposition of B. In particular, if A~B, then A!(A)Â!(B). Proof Since <p is an isomorphism, 0 = (p(PiPj) = ip(Pi)tp(Pj) for i ^ j and

n

J2 tp(Pi) = I- We need only to show that B\v^Pi)H£(SI), i = 1,2, • • • , n.

Otherwise, there exist two nonzero idempotents Qi and Q2&A'(B) such that Q2Q1 = Q1Q2 = 0 and Qi + Q2 = <f(Pi) for some i. Note that lP~1(Qi),<P~1(Q2) are two nonzero idempotents and P; = ip~1(Qi) + 9~l(Q2)- This contradicts A\piu^(SI), and we proved the first part of the theorem.


If A~B, there exists an invertible operator X such that XAX 1 = B. Define Tt-^XTX'1 for all T^A'(A). This mapping is an isomorphism from A'(A) onto A'(B).

Lemma 4.2.4 Let Te£(H) and Pi,P2£A'(T) be idempotents. If Pi~A'(T)P2> then T |P 1 -H~T|P 2 7^ , where ~A'(T) means similarity in A'(T). Proof Since Pi~A'(T)P2, there is an invertible operator X£A'(T) such that XPiX-1 = P2. Thus XranPi = ranP2, Xran(I - Pi) = ran(I - P2) . Set

X\ = X\ranp1, X2 = X\ran(i_pxy

Then X = X\+X2, the topological sum of X\ and X2, and

XiCGLiCiPtH, P2H)), X2GGL(C((I - P{)H, (I - P2)H)),

where GL(A) is the set of invertible elements in the algebra A. Note that

T = Ti 0 0 T2

PiH (I-Pi)H

T[ 0

o n p2n {1-P2W

where Tx = T\PlH,T2 = T\{l_Pl)H,T[ = T\P2H and T2' = T\(I_P2)n. A simple calculation shows that

T[ 0 0 Ti

T h a t i s T | P l W ~ T | p 2 > i .

Xx 0 0 X2

Xx 0 0 X2

Ti 0 0 T2

Lemma 4.2.5 Let TeC(H) and let {Pi}?=1 and {Qi}?=1 be two (SI) decompositions ofT. If there exist Xi&GL(C(PiH,QiH)) such that

then

Xi(T\PiH)X-1=T\Qin,i = l,2,.-

X = X!+X2+ • • • +XneGL(A'(T)).

Proof Since

H = ranP\+ranP2-\ \-ranPn = ranQ\+ranQ2+ • • • +ranQn,

QlH

; •>

QiH

where Tt = T\PiH,T[ = T\QiH,i = 1,2, • • • ,n. Clearly, XT = TX and X is invertible.

Lemma 4.2.6 Let

\ - * l i ' ' ' ) * m > * 771+1•>' ' ' i -* n /

and

• ,Qn}

6e two spectral families in A'(T), where TGC(H). If there are X,YGA'(T) and a permutation n of Sn = {1,2, • • • , n) satisfying

(i) XPiX'1 =Qi,l<i<n; (ii) YPiY-1 = QWi,l<i<n. Then for each Qr, m < r<n, there exist a Pri with m < r'<n and Zr, a

product of finitely many Y 's and X 's, such that ZrQTZ~x = PT>. Moreover, {Pr'} is exactly a rearrangement o / { P r } " = m + 1 . Proof Given Qr,m < r<n, it follows from (ii) that exists Pj1,l<ji<n, such that YQrY~l = Pjx. If m < ji<n, then we set

Zr = Y

and

P, = P

If l< i i< r a , then by (ii) there exists an operator Qjx,ji ^ r, such that

XYQrY^X^^Qj,.

By (ii), YQjxY~l = PJ2 for some j 2 . If m < J2<n, then set

Zr = xXY, Pri = Pj2.

If 1 < 7 2 < T I , it is obvious that j i ^ j 2 . Otherwise,

Qh = Y~lPhY = Y~lPhY = Qr,

which contradicts m < r<n. Using (ii) again, we can find Pj3 such that

YQhY-'=Ph.

Ti

J-n

pxn

PxH

T{

0 Ti

Similarly, j 3 g {31,32}- If m < j3<n, then set Zr = YXYXY,Pr, = PJ3. Or we can continue the procedure above. Since n is finite, after s<m + 1 steps we will force Pse{Pm+1, • • • , Pn}. Set Pr = Pjs, Zr = YXY- • -XY, where X appears S times. Then ZrQrZ~x = Pja. We claim that if ri =/= r2, where ri,r2&{m + 1, • • • , n}, then j S l ^ j S 2 . Otherwise, there exist Zri — YXY- • -YXY (X appears Sl times) and ZT2 = YXY- • YXY (X appears «2 times) such that

^riVri^n = ^/r2^ir2^T2 •

Without loss of generality, we may assume that Si>S2- If s\ > S2, then

^r2 ZriQnZri ZT2 = Qr2^\Qm+li ' ' • ,Qn}-

Note that Z~2Zri = XF- • -XF (X appears j S l — j S 2 times). Set

R = YXY- • -XY,

where X appears j S l — j S 2 — 1 times. By the procedure of the choice, we have RQrxR-1e{Pi,P2, • • • ,Pm}. Thus

XRQriR-1X-1e{Q1,Q2,--- ,Qm}-

But XRQriR~ X~ = Z~2 ZriQriZ~i ZT2 = <2r26{Qm+l> • • • ,Qn}- A

contradiction. Thus sx = S2- But if s\ = s2l we can easily prove that Qrx = Qr2, which is also a contradiction. This completes the proof of our Claim and the lemma.

By the similar argument of Lemma 4.2.6, we can prove the following result.

L e m m a 4.2.7 Letre£(H) and let

i - ' l ) ' * ' I M H I I ' " j P-m.k-i — 1) ' ' ' i-• mjt; •• mfc + 1) ' " ' i*nf

and

i V l ) ' * ' i V m n ' " iVmib_i- l )" ' j Wmk > Wmk+li " " " j VnJ

be two spectral families in A'{T). If there exist

X1,X2,---,Xk:YeGL(A'(T))

and a permutation IT of sn such that

XiPjX~l — Qj,mi + l<j<mi+i, i = 0,1, • • • , k - 1, mi = 0

and Y~1PjY = Q-K(J), l < j < « - Then forVr,mk < r<n, there exists Zr, a product of finitely many {Y, X\,--- ,Xk), such that {ZrQrZ~1}™=mk+1 is a rearrangement of {Pr}™= +1.

Lemma 4.2.8 Let TeC(H) and let {Pi, • • • , Pm, Pm+i, ••• , Pn) and {Qi> •' • > Qm, Qm+i, ••• , Qn} be two spectral families in A'(T). If the following conditions are satisfied:

(i) For each Pi, there is an Xi&GL(PiH,QiH) such that

XiT\PinX-1 = T\QiH, l<i<m;

(ii) there exist a Y&GL(A'(T)) and a permutation ir of Sn such that

Y~~ PtY = <2„-(,),

then given Qr, r£(m + 1, • • • , n), we can find r'G{m + 1, • • • , n} and

Zr€GL(QrH,Pr<H)

such that

Zr(T\Qrn)Zr =T\piH-

Furthermore, if r\ ^ r2, then r{ ^ r'2. Proof Given r£{m + l,--- ,n}, by (ii) of the lemma there exists an operator Pj1€{Pi}?=1 such that YQrY~x = Pj1. If m < ji<n, set Zr = Y\Qr-)-i. Otherwise it follows from T\(YQrY-i)H = T\pj n and (i) that XJ1T\PJIUX~I

1 = T I Q ^ ^ . Using condition (ii) again, we can find J2S{1,2, • • • ,n} such that YQj1Y~1 = Qj2. Clearly, j \ ^ J2-If j2e{m + 1 , . . . , n} , set Zr = Y\Qj7iXhY\Qrn,Pr, = Ph. Thus Zr(T\QrT-i)Z~1 = T\prli-c. Otherwise, by the similar arguments used in the proof of Lemma 4.2.6, after finite steps, we can find Pr'£{Pk}k=m+i s u c n

that Zr(T\Qr-n)Z~1 = T\PT,-H- Similarly, we can prove that r[ ^ r'2 if r\ ¥= r2-

Lemma 4.2.9 Let TG£(H). If T has a unique (SI) decomposition up to similarity, then for each idempotent PGA'(T), T\P-H has a unique (SI) decomposition up to similarity. Proof Since T has a unique (SI) decomposition up to similarity, each spectral family of T|p-^ consists only finite elements, and the (SI) decomposition spectral families consist of same number, say m, of elements.

Let {-Pi}£Lx and {Qi}^ be two (57) decomposition of T\PH, and let {Pi}"=TO+1 be an (SI) decomposition of T\^_P^n. Then

{Pi •• l<i<n}

and

{Qi • l<i<m; Pk:m + l<k<n}

are two (57) decomposition of T. By the uniqueness of the decomposition, we can find an operator YGGL(A'(T)) such that

{YPiY-1} = {Ql,.-- ,Qm,Pm+u... ,P„} .

By Lemma 4.2.6, we can find Zi&GL(C(QiH,PiH)) and a permutation TT of Sn, such that

Zi{T\QiH)Z^ = T\p^n, l<i<m.

Set Z r = I\pkn,k>m + 1 and Z = Zi-i i-Zn. By Lemma 4.2.5,

ZGGL(^'(T)) and Z P W e G L ( ^ ' ( r ) | p W ) . Note that

( Z | P W ) Q i ( Z | p „ ) - 1 = P T ( i ) , l<»<m.

The proof of the lemma is complete.

Lemma 4.2.10 Let TG£(H). IfT has a unique (SI) decomposition up to similarity. If P and Q are two idempotents in A'(T), then the following are equivalent:

(i) P~A>(T)Q; (ii) T\PH~T\QH-

Proof (i)=^(ii) is a consequence of Lemma 4.2.4. (ii)=Ki). By Lemma 4.2.9, T\pH,T\QH,T\{I_P)n and T\{I_Q)n all

have a unique (57) decomposition up to similarity. Since T\PK~T\Q-H,

there exists an operator X £GL(C(PH, QH)) such that

X(T\pU)X~ =T\QH.

Thus if {Px,P2, • • • ,Pm} is an (57) decomposition of T\PH, then (XP1X-\XP2X-1,--- ^PmX-1} is an (57) decomposition of T\QH.

Assume that {Pm+i, • • • , Pn} and {Qm+i, • • • , Qn} are (57) decomposition of T|(/_p)W and T | ( / _ Q ) ^ respectively. Then {Pj}™=1 and

{XPiX~l : \<i<m; Qk:m + l<k<n}

are two (SI) decompositions of T. By the uniqueness of the decomposition, there exists an operator YeGL(A'(T)) such that {YPiY~l}f=1 is a rearrangement of {XPjX"1 : l<i<m; Qk : m-\- l<fc<n}. By Lemma 4.2.8, for each r£{m + 1, • • • , n} , we can find Pr< with r'&{m + 1, • • • , n} and Zr€.GL{£(QTH,Pr<'H)) such that

and r[ = r'2 if and only if n = r%. Set Z = Zi+ • • • +ZneGL(A'(T)). Since ZPZ""1 = Q, by Lemma 4.2.5,

P~A'{T)Q-

Lemma 4.2.11 Let T££(H) and let P and Q be idempotents in A'(T). IfT\pn is not similar to T\Q-H, then P©0W<» is not similar to <5©0W(n> in A' (T^n+1^) for each natural number n. Proof If there is an XeGL(A'(T(-n+1'>)) satisfying

X(P®OnM)X^ = Q®0Hin)

for some nGN, then by Lemma 4.2.4,

T(n+1)l(Peow(„,)«(»+» ~ T ( n + 1 ) |(QeoK(n) )W(»+D •

But we have

r ("+ 1 ) l(p©ow („ ))w("+1)- r |pw

and

T(n+1)\{Q®On{n))H<-"-+V-T\Q-H-

Therefore T\PK~T\Q-H- A contradiction.

Lemma 4.2.12 Given Te£(H). IfT^n) has a unique (SI) decomposition up to similarity for each natural number n, then for two idempotents P, Q in A'(T), P~A>{T)Q if and only if [P] = [Q] in \/(A'(T)). Proof The "if" part is obvious.

Assume that [P] = [Q], then there is a natural number k such that

P©0W(io~_4,(T(k+i))<3(B0H(fc).

By Lemma 4.2.4, T\P-H~T\Q-H- By Lemma 4.2.10 we conclude that

P~A'(T)Q-

Proof of Theorem 4.2.1 (i)=S-(ii). Let P be the orthogonal projection from H onto Hi, and let E be an idempotent in A'(T^). Since T<") has a unique (SI) decomposition up to similarity, by Lemma 4.2.9 T^\E-H(„) and T(n}\,I_g)-H(n) have unique (SI) decompositions up to similarity. If {Qi, ••• , Qa} is an (SI) decomposition of T^\EnM and {Qa+i, ••• , Qb) is an (SI) decomposition of T^\^I_E-)H(n), then {Qi, • • • ,Qb} is an (SI) decomposition of T K Since {P 1

( n n i ) , . - - ,Pfc(rmfc)} is also an (SI) de

composition of T^n\ by the uniqueness of the decomposition, there is an XeGL(A'(T^)) such that

XQjX-1 = Pt.

Since E = Qi + Q2 + • • • + Qa, XEX'1 = X P}mi) • Define a mapping

\J(A'(T)) -> N<fc)

by h([E\) = (mi,--- ,m*). Then /i is well defined. In fact, if [E] = [F], k k

then by Lemma 4.2.12, F~E~ E P ^ in Moo(A'(T)). U F~ E P ( m i ) in i = l i = l

Moo(^'(T)), then h([F]) = h([E}) and F ~ E . Thus /i is one to one. For a &-tuple (mi, • • • , mfc) of nonnegative integers, we can find a number n

m i i such that mi<nrii, i = 1, 2, • • • , k, which implies that h maps E pr} to i = l

(mi, • • • , mfc) and is onto. Thus

\J(A'(T))^N^\

By the construction of h, we know that h([I]) = (n\, • • • , n^).

(ii)=*-(i). Suppose that V(-^'(T '))=N(fc) a n d h is the isomorphism. Then there exist a natural number r and k idempotents Qi,--- ,Qk in A'(T^) satisfying

h([Qi]) = eu\<i<k. Since \J(A'(T^))^\J(A'(T)), we need only to prove that T has a unique finite (SI) decomposition up to similarity. First we verify the following assertions:

(a). For any idempotent P in A'(T), if T\PH£(SI), then there exists a natural number i, l<i<k, such that h([P]) = e .

fc fc Assume that h([P}) = £ A*ei = £ A;/i([<3i]), AêN. Set w = r J2 <**,

i = l i = l find a natural number n > w such that

fc t " , , - - ( n - « 0 . P©ow(„-i)~i4,{r(n)) ] T QlA,)©ow

1=1

By Lemma 4.2.4,

T<n>|,p«n . „ W W ~ T W | (E Q l ' ®°„(«-»))w(n)

l(P©oM(„_„)«(»)r - . - (Ai,

Thus

( E Qî^nî

Note that T | P - H € ( S 7 ) , thus only one Aj equals 1 and the others are zero, i.e.,

h([P}) = e*

for some i. (b). For arbitrary idempotents P and Q in „4'(T<">), if h([P]) = h([Q]),

then T\PH~T\Qn. Repeating the arguments in (a), we get (b). Let {Pi, Pi, • • • , Pm} be a spectral projections of T and assume that

fc

K[Pi\) = J2XiJeJîJ^N-. 7 = 1

Then

m m k

i = i i = i j = i

fc ra fc fc fc

From h(I) — Y2 nê*, we have J ] J3 ^»j — 12 n»- Thus m< J^ m. This i= l i = l j = l *=1 i = l

implies that the number of elements in each set of spectral projections of T is finite and T is the direct sum of only finitely number of (SI) operators.

Furthermore, let {P1 ; P2, • • • ,Pi} be an (SI) decomposition of T, then fc

E i= l i = l

h(j2iPi}) = H[i]) = j2n^-

k By (a), t = ^2 ni a Qd for each i, l<i<k, there exist

i= l

Pil,---,Pinie{P1,P2,-.-,Pt}

such that

h([Pil]) = -- = h([Pitli]) = ei.

By (b), T\PijU~T\pikH, l<j,k<Ui. Denote A, = T\Pi.H, l<j<rii, we have

»=i

Assume that {P{, P2,-- • i P's) ls another (SI) decomposition of T. Then k

repeating the above arguments, we have r = J2ni a n d f° r e a c n h l<i<&,

there are n, idempotents in {P[, P^ • • • , P^} such that ft maps each of them k

to <*. By (b), if h{[Pi\) = h([P$),l<i,j< £ ni: then T\Ptn~T\p.n. By j = i

Lemma 4.2.5, T has a unique (SI) decomposition up to similarity. The proof of the theorem now is complete.

Proof of Corollary 4.2.2 Note that if TW has a unique (SI) decomposition up to similarity, then by Theorem 4.2.1 Ti~T2 if and only if \J(A'(TX®T2))^N. Therefore, if Ti~T2, then KQ(A'(T1®T2))^Z, since \J(A'(T1®T2))^N. Conversely, if K0(A'(Ti®T2))^Z, by Theorem 3.2.1, V(-4'(Ti©T2))SN(fc\fc<2. Since K0(A'(T1eT2))^Z, VM'( T i©?2))=N. Thus Ti~T2 . This completes the proof of the corollary.

The following proposition tells us that besides normal operators, a class of analytic Toeplitz operators whose unitary invariant and similarity invariant are same. In fact, we have stronger result.

Proposition 4.2.13 Let ipi and tp2 be two univalent analytic functions on the unit disk D. Then the following are equivalent:

(ii) kerTTvl<Tie2 # {0} and kerrTv2tTvl ¥= {0}. Proof (i)=>(ii) is obvious.

oo

(ii)=>(i). Assume that ifi(z) = £} AJzJ',zeZ?,i = 1,2. Since </?, is j=0

univalent, X\ ^ 0 for i = 1,2. If there are X, Y£C(H2) such that

Denote fli = <î(D),J72 = ^(D). Clearly, fii and Q2 are simply connected and TêB^flt^TêBiiQ^). If fi2 ^ fi2, by Lemma 4.2.3

kerrTipi)Tv2 = kerrT^Tify = {0},

this contradicts our assumption. Thus we may assume that Q.i = Q2 = ^ and o^T^) = cr(TV£,2) = fi. Without loss of generality, we assume that Oefi and v?i(0) = 0, (^2(20) = 0,zo£D. Then there exists a Mobius transformation

X :£>->£>

satisfying x(0) = z0. Thus y2(x(0)) = 0. This implies that TVtM*TV2UW). Therefore we may assume that ^2(0) = 0. Note that T*VT*2 have the following matrix representations with respect to the ONB {1,2, z2, z3, • • • }:

0 X1 A2 A3 • • •

0 A; A2 ••.

0 Ai '••

.0

rp#

0 X1 A2 A3

0 Af Al

0 A?

.0

Prom T*XY* = Y*T*2, computation indicates

y* __ 2/11 2/12 2/13 • '

2/22 2/23 • '

0

A2

and t/nn = [-n"]n-12/ii, n A i

2,3,

We claim that | ^ | < 1 . Otherwise, since Y* is bounded, yn = 0 and ynn = 0 for n = 2,3, • • • . Similarly, j/y = 0 for i, j>l. This contradicts Y* j^0.

Similarly, | j t | < l . Thus |Aj| = |A?| = A. Denote

6/j = a^ff-f-

and

f/j = diag(l,e >,e îdj • ).*' = !. 2-

Obviously Uj is unitary. Denote

0 A * " 0 A

0 '•• '

0

Since UjT*U* = e~ie'T*, RjSA'(T*),j = 1,2. Thus, there exists a function gjGH00 such that Rj = T*.. Since T*.^T*.,gj(D) = <pj(D) = ft and T*.£Bi(Cl*). Clearly, gj is a univalent analytic function on D and

9j(0) = 0,^(0) = A > 0 , j = 1,2.

It follows from Riemann mapping theorem that gi = gi and therefore

•L<fil—-L<fi2-

In the following we will compute the iô-groups of some Banach algebras in terms of Theorem 4.2.1.

Theorem 4.2.14 K0(H°°)^Z, V(#°° )=N. Proof Consider the analytic Toeplitz operator Tz. It is well known that

A'(TZ)^H°°.

By Theorem 4.2.1, we need only to show that T = TJ has a unique (SI) decomposition up to similarity for each natural number n. Since T*£Bn(D), there are only finitely number of elements in each spectral family of T*. Thus we need only to prove that if P is a minimal idempotent in A' (T), then T\P(H2)M~TZ. Note that T is an isometric isomorphism and P(H2)^ is an invariant subspace of T, it follows from the famous von-Neumann-Wold theorem T\P^H2^)—TZ. This completes the proof of the theorem.

Corollary 4.2.15 K0(H°°(n))^Z and V(#°°(fi))=N, where ft is a nonempty bounded simply connected domain. Proof Since ft is nonempty bounded and simply connected, there exists a univalent analytic function <p such that <p(D) = ft, and A'(TV)=H°°. The corollary follows Theorem 4.2.14.

Proposition 4.2.16 Let <pi,<f2,--- , fn&H00 be univalent analytic functions, then there is a natural number k,k<n, such that

n n

\/M'(5Z®T^))-N(fc) and Ko(A'(J2®T^-zik)-»=i i = i

RÛjTÛ^

Furthermore, T — ^ ®TVi has a unique (SI) decomposition up to similar-

ity.

Proof By Proposition 4.2.13, A'(Y, ®TVi)^ £ ®(H2)n*. Thus it follows i = i j = i

from the isomorphism of if-groups and Theorem 4.2.14 that

VCA'(E ®^j)- £ © V((# 2))K)=N(fc)

and

n

i = l

To summarize the facts discussed above we have the following corollary.

Corollary 4.2.17 Let <pi,tp2£H°° be univalent analytic, then the following are equivalent:

(i) TVl~TV2; (ii) lVl=llfi2;

(Hi) K0(A!(TVl®TV2))^Z.

4.3 (SI) Decompositions of Some Classes of Operators

Using Lemma 2.2.4 and Theorem 4.2.1, we get the following theorem.

Theorem 4.3.1 Let Ai,A2,--- , Ak€(SI)n£(H) satisfying

A'(Ai)/radA'(Ai)^C, i = 1,2, • • • , k.

Then the following statements hold: (i) Ai~Aj if and only if K0(A'(Ai®Aj))=Z;

(ii) LetT = J2 ®A<f'i\ where Ai^Aj,i ^ j , then i = l

\J(A'(T))^N(-k\K0(A'(T))^Z{kl

Furthermore, T has a unique (SI) decomposition up to similarity.

A unilateral weighted shift T on Hilbert space Ti. is an operator that maps each vector in some ONB {e„}^L0 into a scalar multiple of the next

vector, i.e.,

Ten = anen+i,aneC

for all n. If an ^ 0 for all n, T is injective.

Lemma 4.3.2 Let A££(H) be an injective unilateral weighted shift and power bounded, i.e., | |An | |<M < oo for a constant M > 0 and all n>l. Then there is an injective unilateral weighted shift B with ||J3||<1 such that A~B. Proof Suppose that

eo ei

e2*

It is well known that A is unitarily equivalent to the unilateral weighted shift with weighted sequence {|an |}. Thus we can assume that a „ > 0 for all n.

If an<l, n = 1,2, • • • , then set B = A. Otherwise, denote

m = min{j : ctj > 1}.

k If Y\ aj>l for all k>ni, then set mi = oo. Otherwise, set

j=ni

k

mi = min{k>ni : 1 [ ctj < 1}, j=n1

then mi < oo and ami < 1. If mi < oo, consider the set {k > mi : a^ > 1}. If it is empty, let n^ = oo. Otherwise let ri2 = min{k > mi : a^ > 1}. If

k

Y[ Oj>l for all k>ri2, set m<z = oo.. Otherwise set m,2 = min{k>ri2 :

k J2 ctj < 1}. Similarly, we define finitely many or countably many rij's and

mi's inductively satisfying (1) m0 = 0 < ni < mi < n2 < m2 < • • • < rik < mk < rifc+i < • • • (the

last one is oo if the sequence is finite); ( 2 )oy< l i f j < m ;

3

(3) [ ] ai if nk<j<mk,k = 1,2, ••• ; i=nk

A =

0 ai 0

a2 0

0

0

(4) atj<l i£mk<j<nk+i,k = 1,2,

Define

Xj

0<j < n\

fl ai nk<j < mk

mk<j < nk+1.

Denote M = sup{\\Ak\\ : k>l}, then l<Xj<M,j = 1,2,---. Define Xe£(H) by

OO OO

3=0 j=0

for all J2 oijejCH. It is easily seen that X is invertible. Set B = X~1AX, j=o

then B is still an injective unilateral weighted shift. If the weighted sequence of B is {bj}^1, then bj — xJ1&jXj-i,J>l-

Note that 0 < bj = <x,<l if j <ni; bj = a~lankXnk-i = Xnk-i if j j-i

j = nk, k>l, since mfc_i<nfc -I <nk\ bj = (\\ ai)~1aj{ \[ at) = 1 if

nk < j < mk,k>l. If j = mfc, fc>l, X j - i> l , Oij < 1, thus

0 < bj = l-ajX-lx<l.

If TO* < j < rafc,fc>l,0 < fej- = oy<l . Therefore 0 < 6j-<l and | | £ | |<1 .

L e m m a 4.3.3 Let A be an injective unilateral weighted shift with ||^4||<1. Then there exists an algebraic homomorphism 4> : H°°—>A'(A) satisfying

(1) If h is a polynomial, then </>(/i) = h(A); (2) If 4(h) =0, thenh = 0; (3) Given h£H°°, there is a sequence {pn}^=i of polynomials such that

{Pni-^)}^! converge in strong operator topology. Denote

4>{h) = lim pn(A). sot n~+oo

(4) For each h£H°°, ||ft(i4)||<||h||oo, where ||, is the norm of h in

H°


Proof Denote R = H{oo) and

T =

A 0 {I-A*A)i 0

/ 0

0

Ho

n2-

where Hi = H,i = 1,2, • • • . Clearly, T*T = IR, thus T is an isometry. Since

T* =

'A* (I-A* A)* 0 0 /

0 I

0

Wo

0 « i 0

a2 0

0 " eo e i

e2

and since {ker(A*)k : k>l} = H0,H0c\J{ker(T*)k : k>l}. If

A =

For xfc = 0©0 © • • • © 0 © efc©0© •• •€# , obviously (T*)fc+2a:fc = 0 for all Ar>0. Hence HiC\/{ker(T*)k : k>l}. Similarly, Hjc\J{ker{T*)k : k>l} for all j>0. Therefore T is a pure isometry and for each heH°°,h(T) is well-defined,

ii^cmi = iwioo and there is a sequence {pn}^Li of polynomials such that

h(T)=SOT- lim p„(T). n—>oo

Let Po be the orthogonal projection from i? onto Ho. Define

$(h) = PoHT)\no.

Then it is not difficult to see that $ is a homomorphism from H°° to .4'(A) satisfying (l)-(4).

Lemma 4.3.4 Lei A be a power bounded injective unilateral shift with o~e{A) = {z : \z\ = 1}, then given T£A'(A), there exists an h£H°° such that T = h(A).

Proof By Lemma 4.3.2 and Lemma 4.3.3, h(A) is well-defined for each h£H°°. Since 0^ae(A) and kerA — {0}, there exists an injective unilateral weighted shift B such that A*B = I. Since

ae(A) = {z : \z\ = 1},

ae(B) = {z : \z\ = 1}.

Since B has no eigenvalue and is not invertible, o~(B) = {z : |,z|<l}. Denote

eo a unit vector in kerA*. For |A| < 1, denote f\ = J2 XnBneo, then 71=0

A* f\ = A/A and < eo,f\ >= 1. It is obvious that /^ is a vector-valued analytic function for X&D. Given TGA'(A), define

/ i ( A ) = < T e 0 , / x > = < e 0 , T 7 x > .

Thus h is analytic in D. From fj ^ 0, (A*-X)fj = 0 and dimker(A*-\) = 1 for X&D, there exists an a€a(T*) such that T*fj = afj because T* commutes with A*. Thus

h(X) =< Te0, fx >=< e0, T*fj > = < eo, afj >= a.

Since \h(X)\<\\T\\,h€H°°. Note that h(A)* fj = M*)/x = T * / A for all Ae£> and V{/x = 1*1 < 1} = W, we have T = h(A).

Using Lemma 4.3.2, Lemma 4.3.3 and Lemma 4.3.4, we have the following proposition and theorem.

Proposition 4.3.5 If A is a power bounded injective unilateral weighted shift with ae(A) = {z : \z\ = 1}, then A'(A) is isomorphic to H°°.

Theorem 4.3.6 If A is a power bounded injective unilateral weighted shift with o-e(A) = {z : \z\ = 1}, then A ( n ) has a unique (SI) decomposition up to similarity for each natural number n. Furthermore, \/(A'(A))=N and K0(A'(A))^Z.

Example 4.3.7 Let A be an injective unilateral weighted shift with increasing weight sequence {wn}^=1 and let r = lim |iun|. Then A/r is a

n—>oo

power bounded injective unilateral weighted shift and ae(A/r) = {z : \z\ = 1}. By Theorem 4.3.6, \J (A'{A))*?N, K0(A'(A))^Z and A™ has a unique (SI) decomposition up to similarity for each natural number n. Example 4.3.8 Let T be an injective unilateral weighted shift with decreasing weight sequence {wn}^=1 and lim wn = 0. A simple compu-

tation shows that A'(T)/radA'(T)^C. By Theorem 4.3.1, \/(A'(T))^N, KQ{A'(T))=7i and T^"' has a unique (SI) decomposition up to similarity for each natural number n.

In the following we will discuss the class of bilateral weighted shift. An operator AGC(H) is called a bilateral weighted shift if there is an ONB {en}^L_00 of the space H and a sequence {a„}^ = _ 0 0 of complex numbers such that Aen = anen+i. If an ^ 0 for all n, then A is injective. It is well-known that A is unitarily equivalent to the bilateral weighted shift with weight sequence {|«n|}^L_oo- ^n w hat follows we always assume A is an injective bilateral weighted shift with (real) monotone weight sequence and A is not invertible. Denote

A =

0 ax

0 a0

0 a-i

0 a_2

0 Q!_3

0

e2

e i

eo

e_i

e-2

e-3

S R 0 T

\fiej : j>0} = H+ \Z{ej :j<0} = H--

We assume that a,j>aj+\ > 0 and lim an = 1. Since A is not invertible,

lim an = 0. n—* + oo

For polynomial p(.z) = £) OjZJ', denote j=0

PW = p(S) i?p

0 p(T)

Then MT^WpW^ and |b(S)||<|W|oo- Let Q and P* be the orthogonal

projections from H onto H- and , respectively,

V w:= \ A A e * : A e C > ' k = ° - ± 1 > ± 2 ' • • • • Note that when /c>0,

71 &—1

A;-Rp = ^ Q j P f c ^ Q = E (otj J J ai)ek®ek-j. j=0 fc<j<n i=k—j

Thus

t. 1 „ L. 1 fa ^

\\pkRPh<{ n «o( E «?)*<( n «onpii2<(n aoiHioo, i= —1 j=fc+l i= —1 i= — \

where ||Pfc7?p||2 is the Hilbert-Schmidt norm of PkRp, \\p\U is the norm of p in H2(dD). Therefore,

+oo +oo fc—1

PPII2< E n*wi2< E ( n ai)iipn- < +°°-fc=0 fc=0 i = - l

If {pn}n°=i a r e uniformly bounded in H°° and converges to some h£H°° uniformly in every closed subset of D, then {pn(S)}^=i and {pn(T)}n°=1

converge to h(S) and h(T), respectively, in strong operator topology, and {Rp^nî converges to some operator Rh in the Hilbert-Schmidt norm. Thus for each h£H°°, we get an operator

'h(S) Rh ' 0 h{T)\ '

denoted by h{A). Obviously, H/iÛÛ/iH^ and h(A)eA'(A). Furthermore,

(/ii + h2){A) = hi(A) + h2(A), hxh2{A) = fti(i4)fea(4)

for all h1,h2£H°°.

Theorem 4.3.9 A'(A)/radA'(A^H00, where A is an injective bilateral weighted shift with decreasing weight sequence {oin}t^-oo' ^m an = 1

n—> — oo

and A is not invertible. Proof By the analysis above, we need only to show that for each XeA'(A), there exists an h£H°° such that X = h(A) + Q, where

QGradA'(A). Assume that

X =

Xn -Xio

^oi -Xoo

X_21 X_20

-X^l-1 -X l-2

Xo-1 Xo-2

X _ i _ i X _ i _ 2

A"_2-l -X"-2-2

ei

eo

e - i

e-2

For each integer k, denote Ak(X) = (£ijXzj)i,j, where e^ 1 i — j = k

. That is, Afc(X) is the operator, which reserves the fc-th 0 i-jj£k

diagonal in the matrix representation of X and let the other entries be zero. It is obvious that X commutes with A if and only if A commutes with every Ak(X). Moreover, computations indicate that Ao(X) = xool, ^k(X)A^ commutes with A and A0(Ak(X)A^) = Ak(X)A^ for k < 0. Thus there is a constant a such that Ak{X)A^ = al. If a ^ 0, we have

{a~lAk{X)A\k^l)A = A{a~1^k(X)A^-1) = I.

This contradicts that A is not invertible. Therefore a = 0 and Afc(X) = 0 for all k < 0. Thus

X = Xi X2

0 X3

and X commutes with T. A simple computation shows that

eradA'{A). Xi X2 0 0

Since ae{T) = {z : \z\ = 1}U{0} and ||T|| = 1, we can find heH°° and QeradA'(A) such that X2 = h(T) + Q by Lemma 4.3.4. This completes the proof of the theorem.

Theorem 4.3.10 Let AECCH) be an injective bilateral weighted shift with monotone weight sequence. If A is not invertible, then

\f(A'(A))^, K0(A'(A))^Z


and A^n> has a unique (SI) decomposition up to similarity for each natural number n.

Given two power bounded injective unilateral weighted shifts A~{ak}'j?Li and B~{/3fc}^=1, by the arguments used in the proofs of Lemma 4.3.3 and Proposition 4.2.13, we have the following proposition.

Proposition 4.3.11 A~B if and only if kerTA B ¥" {0} and kerrg A ¥" {0}.

Note, if A is not similar to B, then kerTA,B = {0} or kerrs,A = {0}. Assume that kerTA,B = {0}. Define

B 0

. ° A

It follows from kerTA,B = {0} that

A'(T) = { TB TBA

0 TA :TBeA'(B),TAeA'(A) and TBA&kerTB,A}.

By Proposition 4.3.5, A'(T)/radA'(T)^H00®Hco. Thus we have the following proposition.

Proposition 4.3.12 Let A, BeA'(T) be two power bounded injective unilateral weighted shifts, then the following are equivalent:

(i) A~B if and only if A'(A®B)/radA'(A®B)^M2(H°°);

(ii) A is not similar to B if and only if

A'(A®B)/radA'(A®B)^H°°®H'x;

(Hi) A~B if and only if K0(A'(A®B))^Z; (iv) A is not similar to B if and only if KQ(A'(A®B))=Z®Z.

Let S be the unilateral shift on H2 given by Sf = zf(z), feH2. Let 9GH°° be a nonconstant inner function and Pg denote the projection of H2 onto H(6) = H2eOH2. The Jordan block S(9) is defined by S(9) = PBS\H(6) [Bercovici, H. (1988)]. In the following, we will prove that for the singular inner function 9 with S(6)e(SI),S(6)(-n^ has a unique (SI) decomposition up to similarity. Applying this result we get that V^ has a unique (SI) decomposition up to similarity.

Theorem 4.3.13 Let 9GH°° be a singular inner function such that S(9)£(SI), then for each n>l,S(9)^ has a unique (SI) decomposition up to similarity.

Proof Applying the six-term exact sequence of if-theory [cf. [Taylor, J. (1975)]] and the short exact sequence of Banach algebra

O-^0H°° - U H°° -^ Hoo/0H°° —• 0,

we obtain the following exact sequence of groups

Q^K0(H°°) - ^ Ko(H°°/0H00) -£+ KiiOH00) -±> Ki[H°°), (4.3.1)

where 7r* (resp. i») is the induced homomorphism of -K (resp. i) on KQ{H°°)

(resp. Ki(9H°°)) and d is the connected homomorphism. It is proved [Tolokommokov, V. (1993)] that Bar((9H°°)+) = 1, where

Bar((0H°°)+)=min{n:(alr-- ,am+1)T<=Lgm+1((0H°°)+)

such that

(aj + b\, am+i, • • • , am + bmam+i )TeL<7m((0tf°°)+)

for some {6 i}^1c(6'ii '0 0)+ ,V m>n}, and

n

Lgn((0H°°)+) = {(ai,---,anf : aie(0Hoo)+, (i = 1,2, • • • , n ) , ^ a ^ = 1 2 = 1

for some (&!,••• ,bn)T,bi£(6H°°)+ (i = 1,2,-•• ,n)} . From this i(0H°°)+

is an isomorphism [Wang, Z.Y. and Xue, Y.F. (2000)]. Thus for any a£Ki(9H°°) with n(a) = 0 in Ki(H°°), there is an / = 0g + leGLrd&H00)-*-) for some geH°° such that [a] == [/] in Ki(0H°°) and i»([/]) = 0 in ^ ( t f 0 0 ) . Thus we can find heH°° such that f = eh

so that e ' W = TT(/) = 1. Noting that A'(S(0))S*H°°/0H°° and since S(9)G(SI), we have that H^/OH00 contains no non-zero idempotent. Therefore ir(h) = 2km for some integer k [Taylor, J. (1975)] and hence he(eH°°)+. So / = eh£(6H°°)+. This means that d = 0 by (4.3.1). Finally, we conclude from (4.3.1) and [Wang, Z.Y. and Xue, Y.F. (2000)] that Ko(H°°/9H00) = {n[l] : neZ} .

Let P be a nontrivial idempotent in A'(S{0)W) = Mn(A'(S(0))). Since g<r(H°°/9H°°)<1 + Bar(H°°/0H°°)<l + Bar(H°°) = 2 by [Tolokommokov, V. (1993)] and since

A'{S(0)) = {u(S(9)) : ueHco}^H°°/9H°°,

there are X&GLi(A'(S(0))) and k with l < f c < n - 1 such that

P = Xdiag(Ik,Q)X-1.

Set Ti = X\ diag(ikfi)H<-n) a n d T2 — X 1\PH(n). It is easy to verify that

T1T2 = IpHW

and

^ 2 ^ 1 = Idiag(Ik,0)HM

and

T a S(6 l )< n W.or i = diafl(S(0)<fc>,O)

on #(fc>e0 which is similar to S{6)W. Thus S(<9)(TI) has a finite decomposition.

Suppose that {7Ji}£I1C.4'(S'(#)(")) a r e idempotents,

m

i=l

and

s(e)^\PiHin)e(Si)(i = i,2,---,m).

By the above arguments, S(9)^\P.H(n)£(SI) implies that there exists an operator Xi€GLi(A'{S(6))) such that

Pi = Xidiag{IH2,0)Xr1, (i = 1,2, • • • , m).

m m Thus n[/#a] = [ £ P.] = £ [ # ] = m[IH2] in 7fo(,4'(S(0)))=Z. Since [J*,]

i = l i = l

is the only generator of Ko(A'(S(6))), we get that n = m. Therefore we can choose YieGLi(A'(S{0)^)) such that

Pi = Yidiag(0,0,--- , 0 , 7 ^ , 0 , •• • .O)^" 1 = YieiYr\ (i = 1,2,- • • ,n) .

n Set V = J ] PiYi. Then it is easy to check that

FeGL 1(^ ' (5(0)(" )))

with F - 1 = £) y . " 1 ^ and y - ^ - y = e*. That is 5(0)<n>) has a unique i = l

finite (SI) decomposition up to similarity.

Let V be the Volterra operator on H = L2([0,1]) denned by

Vf(t)= [ f(s)dsjen,te[0,i}. Jo

Then ||V|| = ^,a(V) = {0} and A'(V) is the weak closure of the algebra generated by IH and V. Put Tv = (IH - V){IH + V)'1. Then A'(TV) = A'{Tv) and Ty is unitarily equivalent to 5(e) by [Bercovici, H. (1987)] and hence A'(V)^H°°/eH°°, where e(z) = exp(^).

Corollary 4.3.14 Ve(SI) and 0 n > has a unique finite (SI) decomposition up to similarity for each n>\. Proof We have supp(e) = {1} and every nontrivial divisor of e has the form et(z) = Aezp(*f±i),0 < t < 1, |A| = l,zeD. Take

zn = 1 , n = 1,2, ••• . n

Then lim et(z„) = lim (e/e t)(z„) = 0, i.e., ( e t , e / e t )T £ Lg2(H^). So

n—»oo n—>oo

5(e) G (57), since it is not difficult to see that S(9) 0 (57) if and only if there exists a nonconstant division B\ of 0 such that (6i,6/9i)T£Lg2(H(-00^).

Now suppose that {P^^dA'(V^) is an idempotents, PjP, = 0(i ^ j ) m

and ^ Pi= I. Let Q be an idempotent in .4'(Ty |p.W(„)). Then

gPiG^'(2v n )) = ^ ' ( ^ ( n ) )

is an idempotent and moreover Q^A'(V^\PiH(n)). Thus lA™)|p.-H(„)€(57) means that Q = 0 or Pu i.e., T^")|PiH(«)e(57), so that {Pi}^ is an (57) decomposition spectral family of Ty . By Theorem 4.3.13, n = m and there is YzGLi(A'(V^)) such that P< = YeiY~l,(i = 1,2,--- ,n) . The corollary follows.

4.4 The Commutant of Cowen-Douglas Operators

As indicated in Theorem JW2 of Chapter 1, Cowen-Douglas operators have very rich contents, including many of hypernormal operators and subnormal operators. In order to characterize the similarity invariant of Cowen-Douglas operators, we will discuss the commutant of Cowen-Douglas operators in this section, which is the preparation for characterizing the similarity invariant of Cowen-Douglas operators in Chapter 6.

In this section, we always assume that TeSn(f i ) , thus V ker(T — z) =

H, where H is a complex, separable, infinite dimensional Hilbert space. Note that B1(Q,)c(SI) and for every TeBi(fi) we can easily prove the following result.

Proposition CD [Cowen, M.J. and Douglas, R. (1977)] A'(T) is

isomorphic to a subalgebra of H°°(Q).

Proposition CD indicates that if Tei3i(f)), then A'(T) is commutative. In the finite dimensional space C", J n (A)s£(C n ) , and A'(Jn{X)) is commutative. For Volterra operator V, A'(V) is also commutative. In Section 4.5, we will see that for multiplication operator Mf on the Sobolev disk algebra, MjeBi(D) if and only if A'(Mf) is commutative. In Chapter 5, we will show that the set of strongly irreducible operators, whose commutant is commutative, is dense in the set of all Cowen-Douglas operators in the norm topology.

Example 4.4.1 Let Xk = \ and Wk = Xk + J-i Afc 1 0 Afc

Define

T =

0WX 0 0 W2

C2

C2 °° C 2 < E £ ( J > C 2

l

Proposition 4.4.2 Suppose that TG.C(T(.) is given in Example 4-4-1> where H = 8 C 2 . Then

(i) Te(SI); (ii) A'{T) is not commutative; (Hi) A'(T)/radA'(T) is commutative.

Proof (i) Suppose that PGA'(T), then

P i i A2P13 • • •

P22 P23 • •

P33 ••

0

c2

c2

C 2 (cf. [Jiang, C.L. and Li, J.X. (2000)])

and Pfc+i fc+1 = Ek 1PuEk for all k > 1, where Ek = W1W2- • -Wk. Compu-

tation indicates that Ek =

Then

1 fe+i fc! 2(fc-l)!

0 h [_ fc!

. Assume that P n an a12

0-21 0,22

Pk+lk+l=Ek PnEk

0 A;! an a12

0-21 0,22

\_ fc+i fc! 2(fc-l)!

an - <»2i2^\y.k l (an - a 2 2 ) # I J T * ! + °ia ~ « 2 I ( 2 ^ I ) T ) 2 ( ^ ) 2

0.21 a^2U^vk] + 0,22

„ fc(fc+l) / \k(k+l) . fc2(fc+l)2

Oil ~ Q21 2 ( a n - a 2 2 j V 2 + Q12 ~ a 2 1 4

fc(fc+l) . A21 «21 2 + «22

Since k(k + 1) —> oo as k —> oo, a,2i = 0 and a n = 022-It is easy to see that if P 2 = P , then P = 0 or 7, i.e., Te(SI).

° * Since WkA0 = A0Wk for all k>l, (ii) Denote 4Q = 0 0

A := diag(i40, A), • • • )zA'(T).

Denote B02 10 0 0

Set B fcfc+2 = Ek-1B02W3Wi- • -Wk+2 for fc>l. Then

-Bfcfc+2 = fc!_M^i)fc,

0 jfc! 10 0 0

1 3+4+5+--- + (fc+2) 3-4-5 (fc+2) 3-4-5 (fc+2)

0 1 U 3-4-5 (fc+2)

/•I i(*+B)-fc 3-4-5 (fc+2) "" 3-4-5 (fc+2)

0 0

fc(fc+5) (fc+l)(fc+2) 2(fc+l)(fc+2)

0 0

Thus ||5fcfc+2|| < 2 for all k>l and

B:=

0 0 B o 2 0 0 0 0 -0 0 0 B13 0 0 0 • • • 0 0 0 0 5 2 4 0 0 • • €C(H).

Computation implies B£A'(T). But since A0B02 ^ BQ2AO,AB ^ BA, and therefore A'(T) is not commutative.

(iii) Suppose that BGA'(A), then

5 =

Bu B\2 5 i 3 • • •

B22 B23 • • •

B33 • • •

0

Similar to the proof of (ii), we have

Bij = 0 4

L e t {ek}kLi be an ONB of H = ®C^ such that

'OWi 0

0 0 W2 0 •••

0 0 0 W3 ' •.

respect to {efc}^=1, where Wk =

We may rearrangement {ek}kxL1 denoted by {fk}k

x'=i such that

* 1

4-" y l i _ S "

0 Ai W2'


oo oo

where Hi = V {f2k},H2 = V {/2/t-i} and A ; = l fc=l

Ai =

0 1 0

0 0 ±

0 0 0

h h -k / a ' A l =

0 1 0 ••• 0 0 ± 0

0 0 0 1

/ l

/ 3

/ 5 "

S72fc-i = /2it-2 and 5 / i = 0. Note that

Bij 0 b£

Thus

5 5 i B12 0 5 2 W2'

Without loss of generality, we may assume that Hi —H^- Then

A: Ax S 0 Ai

H H'

where S us an injective back shift with power sequence {1}. By former argument, we have

B = B1B12

0 B2 , V BGA'(A),

where Bi&A'(Ai),i = 1,2. Suppose that B,TeA'(A). Then

B = •Bi B12

0 £ 2

ft

Furthermore,

BT - TB

T =

0*" 00

TiT 1 2" 0 T2

H H

H -; H'

it follows that A'(Ai) is commutative. It shows that A'(A)/radA'(A) is commutative.

In the chapter 6, we will show that the class of operators C := {TeCH)D(SI) : A'{T)/radA'(T) is commutative} is dense in (57) operators in norm topology. Therefore we have the following conjecture.

Conjecture Given TeC(H)n(SI), A'{T)/radA'(T) is commutative.

If Te6n(f i)n(S7), we can give a confirmative answer to the conjecture.

Theorem 4.4.3 Let AGBn(Q.)n(SI), then there exists a natural number m, 0 < m<n and a bounded connected open set $Ccr(A) such that A£Bm($) and for each T£A'(A), a(T\ker(X-A)) = {/(^)},^£3>i where /(A) is a function analytic in 4>. Furthermore, A'(A) jrad A' (A) is commutative.

Before we prove Theorem 4.4.3, the following definition and proposition are needed.

Definition 4.4.4(Minimal index) Let

TGBn(n),AeA'{T)

and

A(z) := A| fcer(T_z).

If a(A(z)) is disconnected at zo£Q, then there exists a positive number 6 such that cr(A(z)) is disconnected in each point in D(zo, 5) := {z : \z — ZQ\ < 5}. Thus we can find a positive number s such that a(A(z))r\D(\(zo),s) = \(ZQ),Z&D(ZO,5), where A(zo) is an eigenvalue of A(ZQ). Set

P(z)= f {A{z)-X)-1dX.

dD(\(z0),e)

P(z) is said to be a holomorphic idempotent defined on D(\(zo), e) induced by A'(T). If each holomorphic idempotent P(z) induced by A'(T) satisfies

dimker(T\ v r an(P(0)_Zo)) < n,

then n is called the minimal index of T or we say that T has the minimal index n.

Example 4.4.5 IfTGBi(Q), the minimal index ofT is 1.

Example 4.4.6 Assume that f(t) = z(z~ | ) , then Tj£B2(Q)n{SI),OeQ. 2 is not the minimal index ofTi. But there is a connected open set fii such thatT^B^VLx).

Example 4.4.7 For the analytic Toeplitz operator Tzi, T*2€B2(D). A simple computation shows that T*2 has no minimal index.

By the definition, we have the following proposition.

Proposition 4.4.8 Given A£Bn(Q,), n is the minimal index of A if and only if there is no B£Bm(Q),m < n, such that BGA'(A).

According to the literature of [Cowen, M.J. and Douglas, R. (1977)] (Chapter 3), we have the following proposition.

Proposition 4.4.9 T£Bn(Q)r\(SI), there is a natural number m,0 < m<n, and a connected open set fli such that T£Bm(Q) and m is the minimal index ofT.

Proposition 4.4.10 Given A£Bn(Q) with minimal index n, if P(z) is a holomorphic idempotent defined on a connected open set $ , induced by A'(A) and dimranP(z) = k < n,z&$. Denote Hi = V ranP(z), then

A\-Hi&Bm(Cl) for some m,k<m < n. Proof Let P be the orthogonal projection from H onto Hi and

A! = A\Hl, A2 = (A*\ni)

Then

Ai A12

0 A2

Hi

By Lemma 1.2 of [Herrero, D.A. (1990)], erp(A*) = 0. Thus dimH{ = oo. Since A — z is right invertible for zefl, let

B =

be a right inverse of A — z, i.e.,

(A - z)B

Bi B12

0 B2

Hi

Hi

~A\-z . 4l2 0 A2-z_

~IHI 0 "

. ° I H ± .

'B1B12 0 B2

Thus (A2 — z)B2 = Iftj-, and Ai — z is right invertible for zef2. It is not difficult to prove that B2&Bi{Q) for some I < n. Let ir be the

canonical mapping from £(H) to £(H)/JC(H), where £(H)/fC(H) is the

Calkin algebra. Then we have

n(B)n(A - z) = ir(IWl) 0

0 7T(/Wx)

This implies that ran(A\ — z) is closed. Denote (ei(?/),••• ,ek(y)) the holomorphic frame of ranP(y), then for l<j<k

{A! - z)ej(y) = (\{y) - z)ej(y).

Since Hi = \J ranP(y),ran(Ai — z) = H\ for each zG<&. Thus

Ai€Bn-i(Q). Let m = n — I, the proof of the theorem is complete.

Lemma 4.4.11 [[C owen, M.J. and Douglas, R. (1977)] Let f : ri—+Gr(n,H) be a holomorphic curve, where fi is a connected open set. Let Ef be a vector bundle defined on fi and let r i (z) , - -- ,rn(z)) be the holomorphic frame of Ef. If ri(zo),-- • ,rn(zo) form an ONB of f(z0), then there exists a holomorphic frame rj", • • • ,7^. of Ef, defined on some open set A containing ZQ, such that

ri(zo) =n{zQ),i = 1,2, ••• ,n

and

< ^k)(zo),rj(zo) >= 0, l<i,j<n,k= 1,2, • • • .

Proof of Theorem 4.4.3 Without loss of generality, we assume that n is the minimal index of A and D c O . If there is a B£A'(A) such that a(B(0)) = {Ai, A2}, Ai = A2, then since B(z) is holomorphic for z€.Q, there exists an £ > 0 such that

<x(B(z)) = {X1(z),\2(z)},zeD(0,e) = {z : \z\ < e},X1{z) ? X2(z),Xi(0) = Ai

and A2(0) = A2. Since B(z) is holomorphic on Q, we can find an £1 > 0 such that

D(Xue1)na(B(z))=D(X1,e1)na(B(z)) = {Ai(z)}

and

D(X2,e1)n<T(B(z)) = D(X2,e1)n<j(B(z)) = {X2(z)}.

Structure of Hilbert Space Operators

Set

Then

P(z)= J (B(z)-\y1dX,zeD(0,e1).

SD(Ai,ei)

I-P{z)= f {B(z)~\)-1d\,zeD{0,ei).

dD(X2,El)

Denote M = \J ranP(z). By Proposition 4.4.10, we may assume that «e£»(0,ei)

Ai := A\M&Bk{Cl).

Denote A2 = {A*\Mi.)*,Bl = B\M,B2 = (B*\Mx)*. Note that Me{LatA)n(LatB) and

A = Ai A12

0 A2

M M x . B

Bi B\2 0 B2

M

where LatA denotes the lattice of invariant subspaces of A. Let {e\(z), • • • ,ek(z)} be a holomorphic frame of EA1 and

{ek+i(z), • •• ,en(z)} be a holomorphic frame of

E2 = {(x, z) : x€(I - P(z))ker(A - z), ZGD(0, e)},

then {ei(z),--- , ek(z), ek+i{z), • • • ,en(z)} is a holomorphic frame of EA-I o n £ ( 0 , e ) .

Since AB = BA,B\A\ = A\B\. By Proposition 4.4.10 it follows from Ai€6fc(fi) that A2£Bn-k{Sl). Denote

Hi = kerA,H2 = kerA2GkerA, • • • ,Hm = kerAmekerAm'1,- • • .

By Theorem CD in Section 1.4, we have

Wi©W2® • • • ®Hm = V k 0 ^ 0 ) : !<*<», 0 < j < m - 1},

and

0 A l 2 A 1 3 • •

0 A 2 3 • •

0 '•

H2 B

Bn B\2 B\z

B22 B23

B-33

n2

Unitary Invariant and Similarity Invariant of Operators

Denote A = kerA1,L2 = kerA\ekerAi,- • • ,Cm = kerAfekerAm~1,-• • . By Theorem CD again, Cii

A, =

0 A12 A13

0 A'23

0

0

®Cm = V R 0 ^ 0 ) : l<i<k,0<j<m - 1} and

C2

rBn B12 B13

B22 B23

B'33

0

Since BGA'(A),B1€A'(A1). Therefore

Bk+i,k+i~Bk,k

and

Since

B'k+i,k+i~B'kk, k= 1,2,-•• .

B'n =-Bl\/{ei(0),-,e f c(0)},

°{B'n) = {Ai} and <r(£^) = {\l},k = 1,2,

Set

then

Since Bu

Set

{B\)m =

B'w ••• B'lm

. ° B'mm_

£1

r

a( (S! ) m ) = {Ai}.

= B| f c er^ff(5ii) = {Ai,Aa}. Thus

a(5fcfc) = {Ai,A2},fc = l ,2 , - - -

Bm —

B\\ • • • B\m

v -^mm _ Hm

A

c3-

then

a(Sm) = {Xl,X2}.

Define

Pm = f (Bm - A)-xdA

M>(Al,£l)

and

Pm = Pm®0 £ ®Uk • k>m.

Then Pm is an idempotent and PmAm = AmPm, where

Hi

"lm-1 fc>m

Tim

Denote Af = \J{ranPm : m = 0,1,2, • • • } , then N&{LatA)C\{LatB) and

J W { 0 } . OO

{Claim 1} Af = M = V f o C O . - " ,en(z) : z€D(0,ei)} = V An-

Since

V { e P ( 0 ) : l<i<k,0<j<m - l } c V/R^C 0 ) : 0<i<n,0<j<m - 1},

A © • • • ©AnCfti© • • • ©Wm, A © • • • ®Cm£LatB

and

Cx@---®Cm£LatBm.

Note that

fcer(S! - A)n = ker{Bx ~ Ai)* = kerAx = \ /{ei(0) , • • • , efc(0)}.

A simple computation indicates that

dimker(Bm - \i)mn = mk.

For each x£C\® • • • ©An, we have

(Bm - Ax)m"x = ((ST)m - X^x = 0.

0 A12 • • •

o ••

Air,

o

This implies that

A © • • • ®£mCker(Bm - Xi)mn = ranPm.

Since dim{Ci® • • • ®£m) = mk, C\® • • • ®Cm = ranPm and M = M.. Denote

K = \J{ek+l{z),--- ,e„(z) :ze£)(0,e 1 )} ,

then

JCe{LatA)r\(LatB).

Define

Qm= f (Bm - A)-^A and Qm = Qm®0 E @7ik. J k>m

SD(A2,ei)

Similarly, we can get K. = \J{ranQm : m = 1,2, • • • }. Since {e\(z),- • • ,ek{z),ek+i(z), • • • ,en(z)} is a holomorphic frame

of EA, we can obtain an ONB of kerA by Gram-Schmidt orthogonalization. Without loss of generality, we may assume that {ei(0),--- ,efc(0),efe+i(0),--- ,en(0)} is the ONB of kerA.

Since \\(z) ^ \2(z),z£D(0,ei), by Lemma 4.4.11, Theorem JW 1 in Chapter 1 and the argument of Claim 1, we have the following Claim 2. {Claim 2} There are two subspaces E\ and E\ of Wj, i = 1,2, • • • , satisfying

(i) Hi = E[+Ei,E\C\Ei1 = {0}-(ii) <T(Bu\Ei) = X1(Q),<j(Bii\E0 = A2(0); (iii) \J{E) : \<i < oo}eLatB,j = 1,2; (iv) V{^i : 1<* < oo} = M and \J{E\ : 1<* < oo} = K. Let {9l---,9i} and {<£+1\ • • • ,g'n

i) be ONB of E\ and E\ respectively. By Gram-Schmidt orthogonalization, we can get an ONB {ffi>--- ,5fc>fffc+i,--- ,9n} olHi,i = 1,2,--- . By (i), (iii) of the claim and a simple calculation, we have

n2 w3 ' A =

0 A12 Au

0 A23

0

W2

W3 '

Bn B\2 Bis

B22 B23 B =

B. 33

where

Aij — "11 "12

0 a\\ Bij = 0 4

{Claim 3} K + M = H and K,C\M = {0}.

By Claim 2, we can assume that V{ej (0) : l ^ J < n } = 'Hm+i- Set

X(d)

0 A 1 2 0 •••

0 A 2 3 • • •

0 "-.

Wi B n 0 0 B22 0

B 33

Wl

n2 Hz-

then ,4(d)€#„.(fii), where OeftiCft. Since AB = BA,

A{d)B{d) = B{d)A{d).

Since a(Bn) = {Ai,A2} and since Sfc+1:fc+1~Bfc,fc,

a(B(d)) = {X1,X2}

and

(B(d) - Ai)*(5(d) - A2)"-fc = 0.

Set P = J (B(d) — X)~1dX. Then P i s an idempotent. To complete 3£>(AI ,EI)

the proof of Claim 3, we need the following Claim 4. {Claim 4} r a n P = M.

Set

Bm(d)

> n • • •

0

0

•Bn

: , Pm(d) = J (Bm(d) - X)^dX Hm aD(All£l)

and Pm(d) = Pm(d)®0 £ &-yik. Then {Pm(^)}m=i a r e uniformly bounded. fc>m

By Banach-Alaoglu Theorem, Pm{d) converge to P in weak operator topology. It is easy to see that

ranPl{d)=ranP1 = {ei(0),--- ,efc(0)}.

Note that \/{ef(p) : l<j<k,0<i<m\£LatB, V{ejm)(0) : l < j < n } = Hm+i and (iv) of Claim 2, we have ranP2CranP2(d). Since

dimranP2 = Ik — dimranP2(d),

ranP2 = ranP2(d).

Inductively, we can prove that ranPm(d) = ranPm for all m. Thus ranP = M and Claim 4 is proved. Similarly, we can prove that ran(I — P) — K. Thus Claim 3 is proved. Since n is the minimal index of A, M. and K. must be nontrivial subspaces of "H and M. + K, = H,Mr\K. = {0}. This contradicts Ae(SI). Thus cr(B(z)) = {f(z)}. Since B(z) is an £(Cm)-valued holomorphic function, f(z) is an analytic function on Q. This completes the proof of the first part of Theorem 4.4.3.

For the proof of the second part of Theorem 4.4.3, we assume that Ai,A2€A'(A) and denote B = A\A2 - A2Ai. Suppose that a{B(z)) = {/(z)}. Since

(B(z) - f(z))ker(A - z)cker(A - z)

and

dimker(A — z) = n < +oo,

(B(z) - f{z))nker{A - z) = 0, zGQ..

Since B(z) = Ax{z)A2{z) - A2{z)Al(z), f{z) = 0. Thus B(z)nker(A-z) = 0. It follows from \J{ker(A - z) : zefl} = H that Bn = 0. Therefore BEradA'(A) and the proof of second part of Theorem 4.4.3 is now complete.

By Theorem 4.4.3 and its arguments, we have the following corollary.

Corollary 4.4.12 Let AeBn(Q,)n(SI), then

radA'{A) = {BeA'(A) :Bn=0}.

Corollary 4.4.13 Let AeBn(£l), then AG(SI) if and only if A'(A)/r ad A'(A) is isomorphic to a subalgebra of H°°. Proof Since H°° contains no nontrivial idempotent, A'(A)/r ad A'(A) has no nontrivial idempotent, thus nor does A'(A). This proves the sufficiency. The necessity follows from Theorem 4.4.3.

By Factorization Theorem / = \Q f° r e a c n function f£H°°, where x is an inner function and Q is an outer function. If \ is a finite Blaschke product, then by the knowledge of Toeplitz operator Tf£Bn(Q). Thus we have the following corollary.

Corollary 4.4.14 Let feH°°. If there is a XQ&D such that inner part °f f — /(-^o) is a finite Blaschke product, then Tf^(SI) if and only if A'(Tf)^H°°. Proof We first prove that if Tfe(SI), then radA'(Tf) = {0}. Without loss of generality, we assume that / = \Qi where x is a Blaschke product of order n and Q is a outer function. It follows from [Cowen, C.C. (1978)], that A'{Tf)cA'(Tx). But Tx^

n). Thus A'(Tx)^Mn(A'{Tz)). Since

radMn(A'(Tz)) = {0},

radA'(Tf) = {0}.

By Corollary 4.4.13, A'(Tf) is commutative. Since H°°<lA'(Tf), A'{Tf)^H°°.

4.5 The Sobolev Disk Algebra

Let Q be an analytic Cauchy domain in the complex plane and let W22(Q) denote the Sobolev space W22(ti) = {f&L2(Q,,dm) : the distributional partial derivatives of first and second order of / belong to L2(Cl,dm)}, where dm denotes the planar Lebesque measure. For f,g€W22(Q,), we define

<f,g>:= Y, IDafD^dm, | a | < 2 7

then W22(fi) is a Hilbert space and a Banach algebra with identity under an equivalent norm. By Soblov embedding theorem, f&W22(Q) implies that feC(Ti) and ||/ | |Cm)<M||/| |n/22 (n) for some M.

For f£W22(Cl), the multiplication operator Mf on W22(fl) is denned as

Mfg = fg, geW22(n).

Let

W(n) := {Mf : f£W22(Q)},

then W(Q.) is a strictly cyclic operator algebra with strictly cyclic vector e(s,t) = 1. It has been proved that a(Mz) = aire{Mz) = Q. and A'(MZ) = W(Q). Let Aa(Mz) be the algebra generated by the rational functions of Mz with poles outside Q and R(Q) := Aa(Mz)e. Since W(Q) is strictly cyclic, there exist positive numbers N and K such that for each / G W / 2 2 ( 0 ) ,

N\\f\\w>Hn)<\\Mf\\<K\\f\\W22{n). [Lambert, A (1971)]

Thus R(Q.) is the closure in W22(Q) of all rational functions with poles outside il.

Denote

MZ(Q) := Mz\R{n).

Proposition 4.5.1 (i) a(Mz{ft)) = U, oe(Mz{$l)) = dSl, nul(Mz{Q) -ZQ) = 0 and ind(Mz(Q) — ZQ) = — 1 for z0efi;

(ii) The maximal ideal space of A'(MZ{Q)) is equivalent to Q. The homomorphism k*Q corresponding to zo€fl is

k*Z0(Mf(n)) = f(zQ)=<f,kZ0>

for some kZo€R(Q), and kZo£ker(Mz(Q.) — z0)*; (Hi) A'(MZ(Q)) = Aa(Mz{Q)) = {M/(fl) : feR(Q)}, where M,(fi) :=

Mf\mn). Therefore A'(MZ(U)) is strictly cyclic and Mz(£l) is rational strictly cyclic. Proof (i) If 20 £ Q, then

(*o - z^eRiCl),

M(*o-*)-i = (ZQ - M , ) " 1

and

M (^_ x )- i ( f i ) = ( z 0 - M , ( f i ) ) - 1 .

Therefore, a(Mz(Cl))(Zil. Assume that z0 6 (T and for all g£R(£l), the function (z0 - Mz(Q,))g = (zo - ^)s(z) vanishes at z0. Thus zo - M2(fi) is not onto and a{Mz(Q))cCl.

Let z0efi, then for /ei?(fi) there exist a number (5 > 0 and a function h analytic in T such that

f(z)-f(z0) = (z-z0)h(z),zeY,

where T := {zeC : \z - z0\ < 6}cTcCl. Denote

and define

1 A

E := Q. - - F = n - {zeC : |z - z 0 | < - }

«.)-{*<*> /(z)(z-z0)-1 -*e£

Then f(z) — /(zo) = (z — zo)<7(z) for all zefi and g is analytic in fi. Note that

f Q f 9 y \9\2dm<—J | ( z -Zo)0(z) | 2 dm<-^ | | / - /o | | 2

y 2 2 ( f 1 ) ,

i.e.,

<?eL2(£)

and

d [ / ( ^ / ( 2 ° ) ] = / ' (*) = g(z) + (z- z0)g'(z)eL\E).

We have (z - z0)s'(*)€L2(£;). Thus

j f \g'{z)\2dm< 1 ^ |(z - z0)5 ' (z) |2dm< J ^ l / - / o | | U ( n ) -

That is g'eL2(E). Since

2 5 ' + (z - z0) f l"eL2(£) , a 2 [ / (z ) - / (z 0 ) ]_o„, , ,_ , „ - ,

ds2

we have (z — zo)g"&L2{E) and

^|5"W|2rfm<|||/-/0 | |2v22(n) .

Therefore, 5 e W 2 2 ( £ ) . For k = 0,1,2 and z e f r = {zeC : |z - z0 | < §<J}, by Cauchy formula,

I^WI-ISi/K^*1^!*1"01-«esr

Therefore, (?,(?', 5"eZ,2(§r) and #eW 2 2 (§ r ) , Since geW22(E) at the same time, we have g£W22(Q,). It follows from /(z) = /(ZQ)+(Z—zo)g(z) that the

codimension of ran(Mz(tt) - ZQ) is 1. It is obvious that ker(Mz(Q) — ZQ) = {0}, thus z0Gps-F(Mz(n)), ps-F(Mz(n)) = fi and ind(Mz(Q) - zQ) = - 1 .

(ii) and (hi). Denote A := {M/(fl) : feR(Q.)}. It is easy to see that A is a commutative algebra with identity closed in weak operator topology and Ae = R(fl), that is A is strictly cyclic. By [Lambert, A (1971)], the adjoint space

A* = {g* : there exists a function g£R(Q) such that

g"(Mf{Sl))=<Mf{Sl)e,g>}.

Thus, for homomorphism k*0 : k*ZQ{Mf(Q)) = f(zo),f£R(Cl), there exists a kZa£R(il) such that

K0(Mf(n)) = f(z0) =< Mf{Q)e,kZ0 >=< f,kZ0 > .

On the other hand, if p : A —> C is a homomorphism, then

p(M,(n)) = z 0 Gff(M / (n) )=n .

For a rational function r{z) with poles outside fi, p(r(Mz(Q))) = r(zo). Since {r(M0(fi))} is dense in A, p(Mz(Q)) = f(z0) for each /G.R(fi).

Let s€fl(n), then < g,(Mz(n) - z0)*kZo >=< (z - z0)g,kZo >= 0. Therefore, kzoeker(Mz(tt) - z0)*.

Assume that T€.A'{Mz(Q)), then for z0ett,

Mz(n)*T*kZ0 = T*Mz(nykZ0 = ztr*kzo.

Since nul(Mz(Q) — ZQ>)* = 1, there is a complex number t(zo) such that

T*kZ0 =t(z^)kZ0.

Thus

( r / ) ( z ) = < Tf,kz > = < /,T*/ez > = t(z)f(z), f£R(Q), ZGQ.

Because t(z) = (Te)(z)eR(Sl),T = Mt(il). This proved that

A'(MZ(Q)) = A = Aa(Mz(Q)).

Proposition 4.5.2 Let Vi be a bounded simply connected Cauchy domain, then the set of polynomials is dense in R(£l). Proof Given a positive number e and an f&R(Q), let r(z) be a rational function with poles outside fl such that | | / —7i|iy22(fi) < £• For this r, there

is a bounded Cauchy domain fii such that QiDfiiDfiiDfi and the poles of r(z) are outside fii. By Mergelyan theorem, there exist polynomials pn such that pn converge to p uniformly on fii. If follows from Cauchy formula that p'n—>r' and p'^—>r" uniformly on fi. In conclusion, pn^

>r in W22(Q,) and there is an integer N such that ||pjv — ^||iy22(n) < £- Therefore, WPN - f\\w™(si) < 2e.

Because of the special definition of the inner product in Sobolov space for general fi, it is very difficult to discuss the properties of the space R(fl) and the multiplication operators further. But if we choose a better fl, many results can be obtained.

When fi = D, the unit disc, we call R(D) Sobolev disk algebra. For simplicity of symbols, in what follows we will denote Mf(D), the multiplication operator with symbol f&R(D), by Mf.

Proposition 4.5.3 e n = Hnz i Pn

(i) Hilbert space R(D) has an ONB {en}^L0, where

[(3n<'-7r i+2n+l)7r]2>n = °> *> 2> ' ' ' ! (ii) R(D) is a functional Hilbert space, the reproducing kernel of which

oo

is given by k(u,v) = J2 / ^ u " ^ n - For zoSfi, n=0

K = f>^)*n;

(Hi) If f(z) = Yl fnzn is analytic in D, then fsR(D) if and only if

n=0

E l/n -" 32

=o Hn < +oo.

If fGR(D), then Sn = J2 fkzk converge to f uniformly in D; fc=o

(iv) Mz is an essentially normal unilateral weighted shift, Mzen =

Qn^n+l) ®n —

(v) Assume that f(z)

,n = 0,1,2, ••• , and \\MZ\\ =

52fnzneR{D), then

/o 0

Mf = /iff f° h%f\% /o / 3 / 3 3

j203 Jl03 j 0

/ 9 8 . 15'

eo e i e 2 .

e3

Proof (i) Computation shows that zn = (s + it)n,n = 0,1,2,-•• ,is n2+2i n+1

an orthogonal system and | |zn | |2 = (,a»*-n*+2n+i)* ^ B y p r o p o s i t i o n 4 5 2 ,

{e„}£°=0 forms an ONB of R(D). (ii) For z0£D,\f{z0)\<\\f\\cl5<M\\f\\w*2{D) for each f&R{D). Then

R(D) is a functional Hilbert space with reproducing kernel

00 00

k(u,v) == £en(«)i>) = £/£«nl/n. n=0 n=0

Since < f,J2(3Wlzn >= f(z0) for feR(D), kZa = E ( ^ o n ) ^ n . (iii) Assume that f(z) is a function analytic in D with the Taylor ex

pansion / = ^2fnzn, then / = £) J£en£R(D) if and only if X) laH2 < +°°-

If feR(D), then «„(«) = £ fkzk = f ) £e f c converge to / in W22(£>).

fe=0 fc=0

By Sobolev imbedding theorem, sn(z) converges to f(z) uniformly on D. (iv) Since

Mzen = zf3nzn — — — e n + 1 = a n e n + 1 , n = 0,1,2, • • • ,

Pn+l

MZ*M2 - M 2 M ; = diag(al, a\ - a2,, a2, - a?, • • • ) .

Since a2n+1 - a 2 - ^ 0 , M.GC1.

(v) For / = £ / „ * " = Zfcen&R(D),

^ Ji/f ^ V ^ t Pn ^ I fm-nTS2- m>n < Mfen,em >=< yfk-3 en + f c ,em > = < _ ^

~ Pn+fc [ 0 m <n. Thus M/ can be represented as a matrix in (v).

Proposition 4.5.4 Let f be analytic in D, then f£R(D) if and only if f'eH2,f"eL2(D). Proof If f€R{D) and f(z) = *£fnz

n. By Proposition 4.5.3, E 1 ^ <

+00. Thus ]>>2 | /„ |2 < +00 and f'&H2. From the definition of R(D),f"€L2(D).

Conversely, if f'eH2,f"eL2(D). A theorem in [Duren, P.L. (1970)](Theorem 3.11) asserts that function / is analytic in D, continuous to C = dD and absolutely continuous on C if and only if f'£Hx. Thus f£C(D) and feR(D).

For f&R(D), denote fr := f(rz),z£~D,0 < r<l.

Proposition 4.5.5 (i) fr converge to f in R(D) as i—>1; (ii) fr(Mz)—>Mfasr->l.

Proof (i) Since f(z) is uniformly continuous in D, f(rz)—>f(z) uniformly in D. Thus

/ \f(rz)-f(z)\2dm-,0 ( r—,1). JD

Given a positive e, since / ' and f"€L2(D), we can choose ro such that 0 < r0 < 1 and / \f'(z)\2dm < e and / \f"(z)\2dm < s.

D\(2r0-1)D D\(2r0-1)D Thus

fD\rf'(rz)-f(z)\2dm

= IDVOD k / ' ( « ) - f'(*)\2dm + froD \rf'(rz) - f(z)\2dm

< WDVOD \rf'{rz)\2dm + j D V < j D \f'(z)\2dm} + JroD \rf'(rz) - f(z)\2dm.

Similarly,

JD\r2f"(rz)-f"(z)\2dm

< WDVOD \r2!"{rz)\Hm + fD^D \f"(z)\2dm] + JroD \r2f"(rz) - f"(z)\2dm.

Since rf'(rz) and r2f"(rz) uniformly converge to f'(z) and f"(z) respectively in any closed subset of D, we can find a r% such that when r > r j ,

1. If |2|<r0, then \rf'(rz) - f'(z)\ < e and \r2f"(rz) - f"(z) < e; 2. If z£D\r0D, then rz > 2r0 - 1. Thus, if r > n, we have

I Vf'irz) - f'(z)\2dm < 4e + ns2

JD

and

' \r2f"(rz) - f"(z)\2dm < 4e + ne2. I ID

Therefore, fr—>/ (r—>1) in R(D).

(ii) Given g£R(D),

fr(Mz)g=^ f f(rm-Mzrlg{z)d£

= ^l I /(rOte-*)-1^)^ ICI=J

= f{rz)g{z), zED.

Assume that f(z) = ^2fnzn, then fr(z) — Y^fnfnzn. Since

By Proposition 4.5.3, fr&R(D). Thus fr(Mz) = Mfr. Since A'(MZ) is strictly cyclic, fr—>f in -R(D) is equivalent to Mfr = fr(Mz)—>Mf.

Theorem 4.5.6 Assume that f£R(D), then (i) a(Mf) = f(D); (ii) ae(Mf) = aire(Mf) = f(C). If z0eD and f{z0)<?f(C), then

ind(Mf - /(*„)) = -nul(Mf - f(z0))* = - » ,

where n is the number of zeros of f(z) — f(zo) in D, including multiplicity. Proof (i) Let ZQ^LD, then the value of the functions in the range of Mf — f(zo) is zero at z0 and Mf — f(zo) is not onto. This implies that a(Mf)Df(D). On the other hand, if u>o^f(D), it is easy to see that

[f{z)-w0)-^R{D)

and

[Mf - t«o]M(/_10o)-i = M ( / _ W o ) - i [M / - w0] = I.

Thus a{Mf) = f(D). (ii) By Proposition 4.5.1, ae(Mz) = crire(Mz) = C. By [Conway, J.B.,

Herrero, D.A. and Morrel, B.B. (1989)],

ae(Mfr) = ulre(MfT) = o-e(fr(Mz)) = fr(C) = {/(*) : |«| = r} .

If z 0eC, f(rz0)eaire(Mfr). Since

Mfr-f(rzo)—+Mf-f(z0) as r —» 1,

f(z0)&aire(Mf) and f(C)Caire{Mf).


Conversely, if there is a z0GD such that / (zo)^ / (C) , then f(z) — f(z0) has only finitely many zeros in D, denoted by {zo,zi,--- , zn}c.D. From the proof of Proposition 4.5.1(i), we know that

f(z) - f(z0) = {z- z0)k°(z - Zl)

k> • • • (z - zn)k»g(z), geR(D)

and g(z) ^ 0 for z&D. By Proposition 4.5.1,

Mf - f(z0) = (M,_Zo)fco • • • (M 2 _ ,J f c "M s

n is a Fredholm operator and ind(Mf — f(z0)) = - ^ ^ = — n. It is easily

i=0 seen that nul(Mf — f(zo)) = 0 and therefore nul(Mf — f(zo))* = n.

Proposition 4.5.7 Let feR(D), then (i) IJZQ&D andf(zo) & f{C), thenMj£Bn(Q,), where Q. is a component

of pa-p(Mf) containing /(zo) andn is the zeros of f(z) — /(zo) in D; (ii) Mf is an essentially normal operator.

Proof (i) It needs only to prove

\J{ker[{Mf - f{zQ)Y]k : k>l} = R(D).

In fact, if

yeR(D)e\J{ker[(Mf - /(z0))*]fc : fc>l},

then for each A;,

ye[ker[(Mf - / ( z 0 ) )1 Y = ran{Mf - /(z0)) fe.

Thus we can find a function h€R(D) such that

y=[Mf~ f(z0)]kh = (z - z0)

kok • • • (z - zn)k"kgkh.

Note that ZQ is the zero of y of order kko. Since k>l can be any natural number, ZQ is an essentially singular point of y. It is a contradiction.

(ii) By Proposition 4.5.3, there is a sequence of polynomials pn converging to / in R(D). Since A'(MZ) = {Mg : g£R(D)} is a strictly cyclic algebra,

MPn—>Mf (n—KX>).

By Proposition 4.5.3, Mz is essentially normal, and n(Mz) is normal in the Calkin algebra £(R(D))/K{R{D)), where TT is the canonical mapping from

C(R(D)) to C(R{D))/K(R(D)). Therefore 7r(pn) are normal elements and so is

TT(M/) = lim 7r(MpJ, n—>oo

i.e., Mf is an essentially normal operator in C(R(D)).

Example 4.5.8 f(z) = z3 + z2, g(z) = ^ £ - £ j £ , M < 1, |/?| < 1.

/ (C) separates f(D) into three components: fii,^' an(^ ^ 3 , l eQi , —16^2

and i e 0 3 - M;GBi(ni)nB2(Q2)nS3(n3). M;eB2(D).

In the following, we will discuss the commutant of the multiplication operators.

Lemma 4.5.9 Let feR(D), M}eBn{fl), z0eDx := / _ 1 ( f i ) <""*

/(2) - / (z0) = (z - zo)'11 (2 - Zl)h> •••(z- Zl)

h^gZ0{z),

1+1 — where {zi}i=1<zDi are pairwise distinct, ^2 hi = n,gZo(z) ^ 0,z£D. Then

i=i there exist n linearly independent vectors

is — ru 1.1 . . , uhi-i i. kh2~l . . . hht+1~i\ n-za • — \H-zo,^Zot iKz0 i ^ u I'WSI i i^zi }

such that

kerM}_f{zo) = \jKZo.

Proof Choose kl0,k%0,--- , k ^ _ 1 £R(D) such that

M* 1.1 _ u A/f* 1.2 _ JL1 11/* j u h i - l _ i . h i - 2 J w z-z 0

K zo — "201 J W0-zoK2o — K2o> ' Mz-z0

Kz0 — Kz0

Then for every usR(D) and 0<j<hi — 1, here let /c°0 = fcZ0, we have

< u, M}_f(zo)kl0 > = < ( / - /(*«,)«, *£, >

= < (z - z0)hi-j{z - Zl)

h* •••(z- zi)h^gz0u,kZo >

= 0.

Thus k{oekerM*_f(zoy Similarly, choose k\.,--- ,kz*+1~2eR(D) such that

M*z_Ziki = ki;\o< i<i, i<j<h%\'-2.

Then keekerM}_f{zo).

If there is a sequence of complexes {c^ : 0<i<l,0<j<hi — 1} satisfying

that Y.4kii = 0. aPP!y o n b o t h s i d e s by M* ! „ , we get

c j 1 _ 1 = 0. Similarly, c{ = 0,(0<i<l,0<j<hi - 1). Thus kZQ is linearly independent and

kerM}_f{zo) = \ / ^ o -

Given feR(D),M*feBn(n),D1 = / _ 1 ( n ) . For z 0 eDi ,

/ ( 2 ) - /(^o) = (z - 2o)(^ - zi) • • • (z - z„_i)5Z0(z), #Zo(z) ^ 0 for ze£>. (4.5.1)

Let iV~z0 = {zi}2=Q, the numbers in Nzo can repeat. Denote

T = Li{NZo : ZQ£DI, there is at least one Zi(0<i<n — 1) such that

/ ' ( * ) = 0}.

Lemma 4.5.10 T is an at most countable subset of D\ and ZQGDI\T if and only if the numbers in DZo are pair-wise distinct. Proof If zkGNZo with f(zk) = 0 for some k, 0<k<n - 1. Then

f'(z) = (z - Zi) • • • (z - zn-i)gZ0(z) + •••

+ (z - z0) • • • (z - Zfc_i)(z - zk+1) • • • (z - zn-i)gZa(z) -\

+ ( z - z 0 ) - - - ( z - z n _ i ) ^ 0 ( z ) .

Let z = Zfc, we have

(z f c - Z0) • • • (Zfc - Zfc_l)(Zfc - Zfc+l) • • • (Zfc - Z n _ l ) g 2 0 ( Z f c ) = 0 .

Thus there exists at least an i ^ k with z; = z&. The converse is obvious.

Lemma 4.5.11 Let f€R(D), Mj?eZJ„(fi), then there exist an open subset Acf~1(Q) and analytic functions

ai{z),a2(z),--- ,an(z)

and

Zi(z) ,Z2(z) , . . - ,Z n _i(z)

on A such that for each A&A'(Mf),

(Ag)(z) = a1{z)g(z)+a2(z)g(Z1{z)) + - • •+an(z)g(Zn-1(z)),z€A,g£R(D).

Proof Denote Dx = / _ 1 ( 0 ) . If the set {z&D^ : f'(z) = 0} is finite, then T is finite. Let A = Di\T. If the set {zeD\ : f'(z) — 0} is a countably many set. Choose an open subset D2CD1, then D2C\T is finite. Let A = D2\T.

For zGA, by Lemma 4.5.9 and Lemma 4.5.10,

kerMf_f(zj = y\kz, kZl, • • • ,kZnlj

and the numbers in Nz are pairwise distinct. Since A&A'(Mf),

A*kz&kerM*f_f(z).

Thus there exist n complex numbers cei(z), 02(2), •• • , an(z) such that

A*kz = ai{z)kz + a2(z)kZl + \- ^ ( z ) ^ . ; .

Therefore,

(Ag)(z) = <Ag,kz >

= <g,A*kz >

= ai(z)g(z) + a2(z)g(zi) + ••• + an(z)g(zn_i). (4.5.2)

For Nz = {z, z\, • • • , zn-i}, we choose ui, u2, • • • , un-\&D\ such that (i) the n open balls B(z,e), • • • ,B(zk,£)(l<k<n — 1) are pairwise dis

joint in D\ for some e > 0; (ii) {wfc}fc~i are pairwise distinct and

n - l

{uk}nkzln[B(z,e)U(\jB(zk,e))}=<H.

k=i

Set

fk(Z, Zi, • • • Zn-x) = (uk-Z)(uk-Zi)... (uk-Zn-.i)gz(uk)-f(uk)+f(Z),

where (k = 1,2, • • • , n — 1) and ZeA.. Computations show that

lAl = Me*[|^|(z,21,...,zn_1)=(*,«i>-,*»-,)]!

= I[I1 («i - «i)][Il (* - ^)][nff(«i - z)][U gz(ui)]\ ± 0 iytj ijij t = l 1=1


By the implicit holomorphic function theorem [Griffiths, P. (1985)] (Theorem 9.6), there exists a 6 > 0 and analytic functions Zk = Zk(v), (k = 1,2, • • • , n — 1) in B(z, S) such that for v£B(z, S),

f(uk)-f(v) = {uk-v){uk-Z±{v)) • • • {uk-Zn-i{v))gv(uk), k = l,2,--- , n - l .

Thus for vGB(z,S), V,ZI(V),- •• , Zn_i(v) satisfying (4.5.1) and the analytic functions Z\(v), ••• , Zn-\{v) can be extended analytically to A satisfying

f(u) ~ f(z) = (u- z)(u - Zi{z)) • • • (u - Zn_1(z))gz(u).

From (4.5.2), we get

(Ag)(z) = a1(z)g(z)+a2(z)g(Z1(z))+---+an(z)g(Zn-i(z)), V g£R(D),z€A. (4.5.3)

Set g = e,z,z2,- • • ,zn~1 respectively and denote

h\ = Ae, h2 = Az, • • • , hn = Az 7 1 - 1

then

cti(z) + a2(z) -\ + an(z) = hi(z)

zoti(z) + Zi(z)a2(z) H + Zn-i(z)an(z) = h2(z)

zeA

{ zn-lai{z) + Z^-1(z)a2(z) + • • • + ZZll(z)an(z) = hn(z)

Since the coefficient determinant J(z) is a Vandermonde determinant,

J(z) =

1 1 z Z\

1

Zn-\

Zn-X Zn-l . . . Zr

^ 0 .

By Cramer's rule, a\(z), • • • , an(z) are analytic in A. Since Ag is analytic, (4.5.3) determines the vector Ag and describes AeA'(Mf).

Corollary 4.5.12 J /M;eBi(f i ) , then A'(Mf) = {Mg : gGR(D)}.

Let n>2 and w be the n-th root of 1, i.e., weC, wn = 1. Let A n denote the Vandermonde determinant of order n:

1 1 1 w w

1 n - l

1 wn-1 ••• w(n-1]

For l<i, j<n, the (z,jf)-cofactor will be denoted by A,.,-.

Proposition 4.5.13 An operator A&A'(Mz^) if and only if for each geR(D),

n

i=l

n

where cn(z) = £ ^ ( 2 A»J'57^T) and {hj}nz=1 are n functions in R(D).

Proof For f{z) = zn,M*neBn(D), A = {zeD : z + 0}. If

then

Let

denote

z, z0eA, zn -Z% = (Z- Z0)(Z - wz0) •••(z-wn 1zQ),

Zi{z0) = wl Lz0

AeA'{Mzn),

hk = Azk~1 (fc = l ,2 , - - - ,n).

Using Theorem 4.5.11, computations indicate that

A n z 2 ^ = 1 A n

L ^ •• '2J-3 = 1

On the other hand, for arbitrary hi, hi, • • • , hn€.R(D), set

For gGR(D), formally, set

n

i = l

{Claim} A is bounded. For each integer m, let m=m(modn) and 0<m<n — 1. Assume that

oo

g{z) = J2 9mZm' m=0

then

(Ag)(z)=Zai(z)g(wi^z)

n n . . oo

£ EKE A„M^)( E s^^'-1^*")] i = l j=\ m = 0

n oo n

s r E [ E ffn(E Afc-u/"*-1 V n _ J ' + 1 ) ] i = l m = 0 j = l

oo n n

= £ E <?4E(E A ^ ^ - D ^ " - ^ ) ] m = 0 i = l j = l

co n n

= £ E 5 m [ E ( E A y ^ C - ^ ^ z " - ^ 1 ) ] . TTI=O i = i * = i

It is easy to see that

i = i

Therefore,

gAy^^^l A n m = j - 1 0 m ^ ' - l .

(Aff)(z) = E (Qknz^hiz) + gkn+1zknh2(z) + ••• + gkn+n^zknhn)

k=0

n oo = E ( E Qkn+i-lZ^hi).

i = 0 fc=0

For each i, l<i<n,

£ gkn+i-izknhi\\a

k=0

fcn||2 < M\\hi\\2\\ £ 9kn+i-lZ fc=0

0 0 a

fc=0

= Af | |M2 E |f^±^i|2|%±i=i|2 fc=0

<N\\g\\2-

Thus ||j4<7||2<n./V||g'||2 and A is a bounded linear operator. It is easy to show that AGA'{MZ~).

For an operator TG£(H), there exists a Banach reducing decomposition to each idempotent P in A'(T), i.e., H = H1+H2, where

HiGLatT, Hi = ranP

and

H2 = ran{I - P).

The following proposition characterizes all of the Banach reducing decompositions of Mz2.

Proposi t ion 4.5.14 If PGA'(MZ2), the following statements are equivalent:

(i) P 2 = P ; (ii) (Pg)(z) = ax(z)g{z) + a2{z)g(-z) (z =/= 0) for g<ER(D) and one of

the following two must be true: (1) a2{z) = 0 and P = I or 0; 00

(2) ai(z) = ^ + 1 + E «2fc+i22fc+1 andai(z)ai(-z) = a2(z)a2(-z); fc=0

00 00

(Hi) If hi,h2&R(D),hi(z) = £ bnzn, h2(z) = £ cnz

n, then one of n=0 n=0

the following two must be true: (1) zhi(z) = h2(z) and P = I or 0; (2) b0 + ci = \,b2k = -c2k+i (k>l) and hi(z)h2(-z) = hi(-z)h2(z).

Proof (i)=>(ii). If 0:1(2;) and 0:2(2) correspond to P , then ^1(2:) = -z (z ± 0).

Since P 2 = P ,

a\(z)g{z) + ai(z)a2(z)g(~z) + a2{z)a1(~z)g(-z) + a2(z)a2(-z)g(z)

= <xi(z)g(z) + a2(z)g(-z) (z ^ 0)

for each g£R(D). When g = e,

a\{z) + ax(z)a2{z) + a2(z)cti(-z) + a2(z)a2(-z) = ax(z) + a2(z).

When g = z,

a\(z) - ax(z)a2(z) - a2(z)ai(-z) + a2(z)a2(-z) = 0:1(2:) - a2(z).

Simple computations show that

a21(z) + a2(z)a2(-z) = a1(z) (4.5.4)

[ai(z) + ai(-z)]a2(z) = a2(z) (4.5.5)

Since a i and a2 are analytic functions, a2(z)=0 or 0:1(2:) + a\{—z)=l. If a2(z)=0, then ai(z)=l or 0 and P = I or P = 0. If ai(z) + a i ( - 2 ) = l , if follows from Proposition 4.5.13 that 0:1(2:) is analytic in -D\{0} with a pole of order 1 at z = 0. Therefore, 0:1(2:) can be expressed as 0:1(2:) =

^ + \ + E a2k+1z2k+K By (4.5.4),

fc=0

a2(z)a2(-z) = ai(z)[l - 0:1(2:)]

OO

fc=0

= a i ( z )« i ( - z ) .

(ii)=Ki).

{P2g){z) = a\(z)g{z) + a2{z)a2{-z)g{z) + a2{z)g(-z)[a1(z) + ax{-z)}

= a\{z)g(z) + a1(z)a1(-z)g(z) + a2(z)g(-z)

= cti(z)g(z)[ai(z) + <*i(-2)] + (*2{z)g{-z)

= ai(z)g(z) + a2(z)g(~z) = (Pg)(z), g€R(D).

Similar computation shows that (ii)<£=>(iii).

Example 4.5.15 Operator P is defined by

{Pg)(z) = (-+ sinz)g{z) -{- + sinz)g(-z), z ^ 0,geR(D).

Then PeA'{Mz2) and P2 = P.

In the following we will discuss the commutants of the multiplication operators on R(D) with the symbol of the form f(z) — znh(z).

Proposition 4.5.16 Let f£R(D),f'(0) ^ 0,T£A'(Mf), then T = Mv

for some (psR(D) if and only if T admits a lower triangular matrix representation with respect to the ONB {en}^Lo-Proof If T&A'(Mf) admits a lower triangular matrix representation

T = (tijhj^ij =0(j >i).

Let f{z) = J2 Uzn with fx ^ 0. Compare the (2.1) entries of T M / = MfT, n=0

ft A> _ f + ^° Pi Pi

Since f\ ^ 0, t n = t22. Compare the (3.2) entries, we get t22 = £3. In general, tu = t22 = • • • = CQ. Compare the (3.1) entries, we get

/ / P° J. f t @° - t * ^° _L f + &1 i\Hi-z- -r JiHz— — ]2t\x— + Jit2i—.

Pi Pi Pi P2

Since £33 = *n, £32 §7 =t21@±. Thus t32 = cx@±, where cx = jf^*2i- Suppose

that £(+1,; = ci^f^ (l<k — 1). Compare the (fc + 1,k — 1) entries, we get

t . Pfc-2 . t . Pk-2 j. . Pk~2 . t . Pk-1 Jl£fc+l,fc"3 r /2tfc+l,fe+l—;5— = 72lfc-l,fc-l—5 1" Jltk,k-1—5—•

Pfe-l Pk Pk Pk

It follows from tk+i,k+i = *fc-i,fc-i that

0k-2 . Pk-X Pk-2 Pk-X tk+l,k-Z - t fc ,k- l—5— = C\- —

Pk-X Pk Pk-X Pk

and tk+i,k = c i ^ r 1 - By the mathematical deduction, tk+i,k = c i^ip 1 f° r

all k>l. Similarly, by the same arguments we can prove that if tk x = Ck-x S2—,

' Pk — 1 t h e n tk+i,i+i = Cfc_i- r, i = 0 , l ,2 ,

Co

Therefore,

0

C l ^ Co

Since </? = Te0eR(D), by Proposition 4.5.3, y> = ]T c„z™ and T = Mv. n=0

Conversely, if T = Mv, by Proposition 4.5.3, T admits a lower triangular matrix representation with respect to the ONB {en}^L0.

Lemma 4.5.17[Deddens, J.A. and Wong, T.K. (1973), Lemma 2] Let N be a nil-potent on H, X0 = A + JV,0 ^ AeC. If B,A0,Ai,-••££(?{) satisfying ||,4fc||<M and AkX0 = X0Ak-i + B (k = 1,2, • • •), then A0 = Ai = A2 = • • • .

Lemma 4.5.18 Let T£A'(Mz~)nA'{Mf), f&R(D), f = zrg, l<r < n and g(0) ^ 0. Then T(EA'(Mzs), where s = (n,r) denotes the maximal common divisor of n and r. Proof By Proposition 4.5.3,

Ma

Gu 0 G21 G22 and M,T =

0 Wx 0 0 W20

0

with respect to the ONB {efc}j?L0, where Gij,Wk are rxr matrices. Wk is invertible and since g(0) 7 0, Gu is also invertible (i,j, k = 1, 2, • • •). Since

TMZ- = MZ~T,


'Tn 0 T21 T22

with respect to the ONB {ek}kX3

=o, where Tij is nxn matrix (i, j = 1,2, • • •). {Case 1} If r is a divisor of n, n = pr, r > 1 and s = r. Suppose that

Tkk

'Vft Vifc 12 nv

^ J j 2 - *£

Vk Vk • • • Vk

L- pi p2 PP

where Vk- is a r x r matrix (k,i,j = 1,2,- • •). Denote m = (k — l)p,k = 1,2, • • • . It follows from TMzrg = MzrgT that TkkFkk = FkkTkk and Fkk

equals to

0 W m + l ^ m + l . m + l

Wm+2Gm+2,m+l

0 W / m+2G m +2 ,m+2 0

. *^m+p— l^m+p— l ,m+l **m+p— l^m+p— l ,m+2 ' * ' "m-l-p—l^m-f-p—l,m-f-p—1 " m

Compare the (1, p — 1) entries of TkkFkk = FkkTkk, we get

* / l p ^ m + r - l G m - ) - p _ i i m - ( - p _ l = 0.

Since Wm+r-i and G m + P _i ) m + P _i are invertible, V^ = 0. Similarly, Vjj — Oif j > i,k = 1,2, ••• . Thus, Tfcfc admits a block lower triangular matrix representation and

T =

Vu 0

^22

where Vkk is a r x r matrix. By the arguments used in the proof of Proposition 4.5.18, T&A'{MZT).

{Case 2} If r is not a divisor of n. Find positive integers p and q such that qr — pn = s. Since TMzqrgq = MzqrgqT,TM2.+»tJ1 = Mzs+ni,gqT. Because

TMzP» = Mz P-T,

TMzsgq = MzsgqT.

Note that s is a divisor of n and g9(0) ^ 0. Repeating the proof of Case 1, we get T£A'(Mzs) and complete the proof.

Using Lemma 4.5.17 and Lemma 4.5.18, we get the following proposi

tion.

Proposition 4.5.19 Let f(z) = znh{z)&R(D),h(z) ^ 0 for z&D and n > l . Then A'(Mf) = A'{Mzn)C\A'{Mh) = A'(MZ*), where

s = (n,ni,ri2,- ••),/i(.z) = ao + a\znx + a,2Zn2 -\ .

Proof Assume that f(z) = hnzn + hn+izn+1 H .Set

Bk

Pnk — n

fink- n+1

0

Pnk-

and

Hk

hnk ftnfc-1 - - ' fonfc-(n-l)

hnk+1 hnk • • • / l n f c - ( n - 2 )

.hnk+n-l hnk+n-2 • • " /ijifc

where hi = 0 if i < n. Then

Mf =

0 0 i=2i 0

-?31 -F32 0

where Fk+iik = B^HiB^ Since h(z) ^ 0 for z£D,

n/c—1

ker(M})k = \ / {e,}-i = l

Suppose that TeA'(Mf). It follows from T*M} = M*fT* that nk—1 nfc—1

T*( V {ei})cker(Mf)k - V {ei} a n ^ hence T* admits a block upper i = 0 i=0


triangular matrix representation with respect to the ONB {e j}^ 0 , i.e.

T n 0

T21 T22

Compare the (2, 1) entries of TMf = MfT, we get F2xTu = T22F2i. It follows from F2\ = B^H^x and B^lHiBxTx = TnBÊxBx that

HiBiTuBi = B2T22B2 Hi.

Compare the (i,i — 1) entries, we get

i.e.,

Thus

By Lemma 4.5.17,

i.e.

Thus

TuBi HiBi-i = Bi H\Bi-.\Ti-\,i-\.

HiBiîTi-itiîBi_1 = BiTuB~ Hi.

Bi-iTi-i^-iBi^ = BiTaB~ ,

Bi Bi-\Ti-iti-i = TaBt -Bj_i.

Ti+1,i+iWi = WiTH,

where Wj = B^Bi. Similarly, by mathematical deduction we can prove that

Bk+i-iTk+i+i,i+iBi+1 = Bk+iTk+iiiBi ,

i.e.,

Tk+i+iti+iWi = Wk+iTk+i,i,k = 0,1,2,-•• ,t = 1,2,3, ••

This implies that

TMZ-=M^T.

Since

TMz~Mh = Mz~MhT,

MznTMh = Mz„MhT.

This means that

zn(TMhg)(z) = zn(MhTg)(z)

for g£R{D) and z&D. Thus

TMh = MhT

and

T&A'(Mzn)nA'{Mh).

Conversely, A'(Mzn)nA'(Mh)cA'(Mf) is obvious. Thus,

A'(Mf)=A'(Mzn)nA'(Mh).

II TeA'(Mf) = A'(Mzn)nA'(Mh). Suppose that

h(z) = a0 + alZni + a2z

n2 + ••• ,

where at ^ 0(z = 0,1, 2, • • •). Then h-h(0) = znklhlt where nj = nki+ri for ki,riGN and n < n and /ix = z r i gi . Since TeA'(Mh),

TMznklMhl =Mx^lMhlT.

Since T&A'(Mzn),

MznklTMhl =MznklMhlT,

i.e.,

for geR(D). Thus

TMfcl = MhlT.

By Lemma 4.5.18, Te ,4 ' (M z n) , where si = (n,r i) = (n,ni). Similarly, if n2 = n/e2 + r2 for /c2, r 2 e N and r2 < n, then

a 2 z " 2 + a £ 3 + --- = znfc2/i2,

where /12 = ^Qi- By the same argument Te^4'(Mz»2) for si = (n, ni,ri2). In general, T€LA'{MZ,), where s = (n,ni,ri2, • • •).

Conversely, if r e ^ ' ( M z . ) , then it is obvious that TeA'(Mf). Thus

^ ' ( M / ) = >4'(Mzn)a4'(Mfc) = A'(Mz.),a = (n,n1 ,n2 , •• •)•

Corollary 4.5.20 Suppose t/iat feR(D),f(z) = znh(z),h{z) =£ 0 /or zG-D and for each k > 0, /i(z) is not a function of zk, then

A'(Mf) = A'{MZ) = {M9 : g&R(D)}.

In the following, we will discuss the strong irreducibility of multiplication operator M/ on R(D).

Theorem 4.5.21 Given fGR(D), the following are equivalent. (i) M;eBi (n ) ; (ii) A'(Mf) = {Mg : g£R(D)}; (Hi) A'(Mf) is commutative; (iv) If M*f € S„(fii), then for each AeA'(Mf),

(Az)(z) = zhxiz), {Az2){z) = z^z), • • • , (Azn'l)(z) = z^h^z),

where hi = Ae\; (v) Mf£{SI).

Proof (i)=Kii) Corollary 4.5.12. (ii)=^(iii). Obvious. (iii)=>(iv). If there an operator AeA'(Mf) such that (Azk)(z0) =£

Zohi(zo) for some k, 1 < k<n — 1 and z0eDi = f~1(£l),z0 ^ 0. Then

(MzkAe)(z0) = Z^ZQ) £ (Azk)(z0) = (AMzke)(z0).

Thus MzkA ^ AMzk. But A and M2t are in A'(Mf). A contradiction. (iv)=^(v). For z€A, since ai(z) + a2(z) + • • • + an(z) = h\{z) and

hk+i = zkh\ for l<fc<n — 1, by Theorem 4.5.11 and (iv)

f (Zi(z) - z)a2(z) + (Z2(z) - z)a3(z) + ••• + (Zn^(z) - z)an{z) = 0

(Zf(z) - z2)a2(z) + (ZUz) - z*)a3(z) + ••• + {Z*_x{z) - z*)an(z) = 0

(Z?-\z) - z"-i)a2(z) + (Zr\z) - zn~l)a3(z) + ... + (ZZll(z)-z"-l)an(z)=0

Computations indicate that the coefficient determinant is still a Van-dermonde determinant:

V =

Zx-z Zl - z2

Z2-z Z2

2-z* Zn-\ —

zLi - •

7TI — 1 _ n — 1 yn—l z?~l - ~n—1 yn-i _ n - 1

(-1) n - 1

1 1 ••• 1 1 Z\ Z2 • •• Zn-\ z zl zl ... zu z*

yn—l yn—l yn—l -n—1 Z l L1 " • Z n - 1 Z

^0.

an(z) = 0 and A'{Mf) = {Mg : Therefore a2(z) = a${z) = • • geR(D)}.

Since A'(Mf) does not contain nontrivial idempotent, Mf€(SI). (v)=^(i). Suppose that the minimal index of MJ is n and n>2. By

Theorem 4.5.11, Ag = ot\{z)g{z) H + an(z)g(Zn-i(z)) for A£A'(Mf), geR(D) and zeA. Since M*eA'{M*f),M*ku - uku and M*zkzx{u) = Zi(u)fc21(u) for uGA. By Theorem 4.4.3, the spectrum of M*\ker(M}-f(u))

is connected. Thus u = Z\{u). This contradicts z ^ ^1(2) when zeA. Therefore n = 1 and M;&Bi(fi).

Proposition 4.5.22 Mz™e(S7) j / and on/?/ i / n = 1. Proof If n>2, define an operator P by

Pf = \\J{?) + / M + • • • + /K"1*)] / e / W

By Proposition 4.5.13, PeA'(Mzn). For each i, l<i<n - 1,

Pf(w'z) = £[/(w<z) + f(*i+lz) + ••• + fi^-^z)}

= £[/(*)+ /("*) + "- + /("n-1*)]-Thus

p2f = £ [ " ( / ( * ) + / ( "* ) + ••• + tt"n-lz))\

= ![ / (*) + / H + • • • + /(a;""1*)] = P / .

It is obvious that P is nontrivial and Mzn$.{SI).

Proposi t ion 4.5.23 Let feR(D) with f(z) = znh(zm), h(z) ± 0, zeD and n>\. If for each k > m, h(zm) is not a function of zk, then Mf£(SI) if and only if (n, m) = 1. Proof By proposition 4.5.19 A'(Mf) = A'(Mzs), where s = (m,n). By Proposition 4.5.22, A'{Mzs) contains no nontrivial idempotent if and only if s = 1. Thus Mfe(SI) if and only if s = 1.

Proposition 4.5.23 requires n>l. One may asks the question: Is the conclusion of Proposition 4.5.23 true when n = 0? In fact the answer in general is negative unless / is an integral function.

oo

Example 4.5.24 Let f(z) = 2z - £ ^zn, then Mfg(SI). n=l

Proof It is easy to see that f(z) is analytic in 2D = {zEC : \z\ < 2}. Thus f&R(D). Define an operator P as follows:

(Pg)(z) = \g{z) + \g{2 + - A _ ) , g£R(D).

Then

Simple computations show that P2 = P and f(z) = / (2 + j r j ) . Thus

PMfg(z) = \f{z)g{z) + 1/(2 + ^)g(2 + ^ )

= lf(z)g(z) + y(z)g(2+^)

= MfPg(z), for all geR{D).

Set g(z) = 10 + f(z), then g satisfies the requirement of Proposition 4.5.23 with n = 0. But Mg£(SI).

Proposition 4.5.25 A'{Mz2)/radA'{Mz2) is noncommutative. Proof By Theorem 4.5.21 A'(MZ2) is not commutative. Assume that R is a left ideal of A'(MZ2). Set £1 = {Ae : AeR},R2 = {Az : A&R}. {Claim(i)} Ri,R2 are ideals of R(D), and if hi£Ri,h2€R2, then

/n(-*)efli, hl{z)~hl{'z)eR1,h2(-z)eR2

and

h2{z) - h2(-z) eR2.

Let A&R and hi = Ae. Denote h2 = Az£R2. For arbitrary fi,f2&R(D), denote B^A'(MZ2) determined by / i , / 2 - For g£R(D),

(BA)g(z)

= r/i(z)hl(3!)+h '(^l+/2(z)'' l(^-^(-) + / l ( z ) h2(«Hha(-.) + h { z ) /.2(»)-h2(-»)

r / 1 ( 2 ) h t ( ^ + " ' ( ^ + / 2 ( Z ) , ' l ( j ) - ^ ( - ^ _ /1(Z)h a ( j )+' ,8 (- )+/2(Z)h i» ( ,>-^ (- ' )1 , ,

+ I 2 22 J V Z>-

Choose f2 = zfi, i.e., B + M/j , then

(BA)g(z) = [ * < f ! ^ + * ( ^

which means /i / i i€i?i and Ri is an ideal of R(D). Set A = e, / 2 = 0, i.e., (B<?)(z) = §[g(z) + g(-z)], then

(BA)»(z) = [M*)+M-») + h2(z)+£(-z)]g{z)

_|_ rfei(z) + >ti(-z) _ h2(z)+h2(-z)-i / ,N

which means /ii(z) + /ii(—z)Gi?j and /ii(—z)Gi?i. Set A - 0, / 2 = e, i.e., (5ff)(z) = £[<?(z) - g(-z)], then

>»i(«)-hi(-») ^ ( ' l - ^ C - * )

(BA)<?(z) = [ 1 + £ ]fl(*)

h l ( « ) - h l ( - « ) fc2(')-''2(-»)

+ [—^ &—]$(-*),


which means z €-Ri- Similarly, we can prove that R2 has the same properties. {Claim(ii)} For arbitrary z\, z2£D, z\ ^=Q,z2^ 0, Set

Ri(zi) = (z2 - zf)R(D), R2(z2) = (z2 - z2)R(D),

then Ri(zi) is a "maximal" ideal which has the properties: if /ii£i?i(zi) then hi(-z)eR1(z1) and hl(z)~^l(~z)eRi(zi). Similar properties hold for #2(22)- For each hxeR1(z1) and h2GR2(z2), ^ n d -^-A'C^z2) with

(A9)(z) = (*£> + *£>),(*) + (*£> - ^ ) 5 ( - z ) , z ^ 0.

Let R(z\, Z2) be the set of all such operator A. Simple computation shows that R(zi,Z2) is an ideal of A'(MZ2). If there is an ideal S of A'(MZ2) such that R(zi,Z2)CS, then Si :— {Ae : AGS}DRI(Z\) is an ideal and by (i) and the "maximality" of Ri(zi), Si = R\{z{). Similarly, S2 := {Az : AGS} = i?2(^2)- Hence S = i?(zi,Z2), i-e-, R{zi,z2) is a maximal ideal of A'(MZ2). {Claim(iii)} radA'(Mz2) = {0}.

Let

then

i2 = n{i2(zi, z2) : 21, Z2G£>, ZX ^ 0, z2 ^ 0},

.Ri = {Ae : AeR} = n{Ri(zi) : zi&D,zx ^ 0}

= {heR{D) : h(z) = h(~z) = 0,zeD} = {0},

R2 = {Az : AGR} = n{R2(z2) : z2e£>,22 ^ 0}

= {heR(D) : h(z) = h(-z) = 0,zeD} = {0}.

Thus R = {0} and radA'(Mz2) = {0}. Since A'(MZ2) = {0} is noncommu-tative, A'(Mz2)/radA'(Mz2) is noncommutative.

In the last part of this section we will consider in invariant subspace of Mz on R(D).

Proposition 4.5.26 Let feR(D). (i) IfBa(z) = f^,a£D, then MfoBa~Mf; (ii) Mf~Mz if and only if f(z) = \fE=£, |A| = 1 and aeD.

Proof (i) It is obvious that foBa€R(D). Define an operator Sa as follows: Saf = foBa, /G-R(D). Computation indicates that there are positive numbers Mi, M2,M$ and M4 such that

/ \foBa\2dm<Mi I \f\2dm, [ \(foBa)'\

2dm<M2 [ \f'\2dm JD JD JD JD

and

f \(foBa)"\2dm<M3 [ |/'|2dm + M4 f \f"\2dm. JD JD JD

Thus | | / o5 a | | <M| | / | | for some number M and all feR(D), i.e., Sa is bounded. Note that for each g£R(D),

SaMfg = Sa(fg) = f(Ba)g(Ba)

and

MfoBaSag = MfoBag(Ba) = f(Ba)g(Ba).

Thus SaMf = Mf0BaSa. Since Ba is an invertible analytic function, Sa is one to one and onto. Thus Sa is invertible and Mf0sa = SaMfS*1.

(ii) If f(z) = \f5£, then Mf~Mz by (i). On the other hand, if Mf~Mz, by Theorem 3.5.6,

f(D) = a(Mf) = a(Mz) = D.

Thus f(D)cD. Since / is continuous on D, by maximal module theorem f(D)cD. For arbitrary XeD, since \&a(Mz)\ae(Mz) = f(D)\f(C), Xef(D) and Dcf(D). Therefore / maps D onto D. Note that for each AeD,

nul(\ - M}) = nul{\ - Mz*) = 1,

this implies that f(z) — A has only one zero. Thus / maps D one to one onto D. It must be a Mobius transformation up to a coefficient of module one.

Lemma 4.5.27 Assume that / i , /2, • • • , fn are n functions in R(D) without common zeroes in D. Then there are functions gi,g2, • • • ,gn in R{D)

n such that ]T] fkgk = e.

fc=i Proof Denote J = {gifi + g2f2 + • • • + 9nfn • GR(D)}.

It is obvious that J is an ideal of R(D). If J is a nontrivial ideal, it must be contained in a maximal ideal. But this is impossible, because

the maximal ideal space of R(D) is D and / i , /b, • • • , fn have no common n

zeroes in D. Thus eej and ^ /£<% = e for some 51,52, • • • ,5nSi?(J5). fc=i

Theorem 4.5.28 (i) Subspace M with finite common zeroes {zi, ^2, • • • , zn} in D, multiplicity included, is an invariant subspace of Mz

if and only if

M = (z-z1)---{z-zn)R{D);

(ii) If M is an invariant subspace in (i), then the projection onto M.

is PM = Mx{M^Mx)-lM^, where X=f[ f5^-

i= l

Proof (i) If M = (z — z\) • • • (z — z„)R(D), it is obvious M is an invariant subspace of Mz with common zeroes {xj}f=1 in D.

Conversely, if MeLatMz with common zeroes {z,}"= 1 . Denote

N := {geR(D) : (z - Zl) • • • (z - zn)g£M}.

Then it is not difficult to see that N is an invariant subspace of Mz. For each weD, since {zi}?=1cD, there is a function fw£N such that fw(w) ^ 0. Since fw is continuous in D, choose a neighborhood U(w,e) such that fw does not equal zero in U{w,e). Thus we can find a finite open cover U(wi,ei),--- ,U(wk,ek) of D and functions fWl,fW2,--- ,fwk in N such that fWi has no zero in U(iVi,Ei) (l<i<k). Thus {fWi}i-i have no common zero in D.

By Lemma 4.5.27, there are 51,32, • • • ,9k^R(D) such that

k

Y^fwigi = e.

Note that since polynomials are dense in R(D), each invariant subspace of Mz in fact is an ideal. Thus eeN and N = R(D). Thus

{z-z1)---{z-zn)R{D)dM.

Since M c ( z - z i ) • • • (z-zn)R(D) is obvious, M = ( z - z i ) • • • (z-2„)i?(D). (ii) Since M* is a Fredholm and kerMx = {0},M*Mx is invertible. It

is obvious that Mx{MxMx)~lM* is a self-adjoint idempotent. Thus it is

an orthogonal projection. By Proposition 4.5.7, M* is a Cowen-Douglas operator of index n, and ranM* = R(D). Thus

ran\Mx{MlMx)-x Mx] = ranMx = xR(D) = M,


i.e., PM is the projection onto M.

It is well-known that in Hardy space H2 Mz similar to the restriction of it on any nontrivial invariant subspace M. The following result indicates that this statement is valid in Sobolev disk algebra if and only if M has only finitely many common zeroes in D.

Proposition 4.5.29 Let M&LatMz, then MZ~MZ\M if and only if

M=(z-Zl)---(z-zn)R(D),

where {zi}f=1cD. Proof Assume that M = (z — z\) • • • (z — zn)R(D). Denote

P= {z-zi)---(z-z„).

Define Tp : R(D)—>M by Tpf = pf,feR(D). It is easy to see that Tp

maps R(D) one to one onto M. Since

T;^Mz\MTpf = zf = Mzf, feR(D),

MZ = T~1MZ\MTP.

On the other hand, if there is an W : R(D)—>M such that

MZ = W-1MZ\MW,

then

WMZ = MZ\MW.

Denote h = We. Computations indicate that Wzn — znh. Thus Wp = ph = Mhp for each polynomial p. By Proposition 4.5.2, Wf = Mhf for all feR(D). This implies that W = M/, and Mh has a closed range. By Theorem 4.5.6, O0/i(C). Since Mh is essentially normal, h has finitely many zeroes in D. By Theorem 4.5.28, M = (z — z\) • • • (z — z„)R(D) for

Theorem 4.5.28 and Proposition 4.5.29 describe the structure of invariant subspaces of Mz with finitely many common zeroes in D. The following example is an invariant subspace of infinitely many common zeroes.

oo

Example 4.5.30 Let f(z) = ( J ] f$£)(z - l ) 5 , where an = 1 - 4j-.

Then feR{D).

Proof By a result in [Gardner, B.J. (1989)], f(z) is continuous in D. oo

Let B(z)= n bak, where bak = f ^ . fc=i

{Claim} B'(z) = £ (^^ ft bak(z)).

Let G = {z£C : \z\ < r < 1}, choose r so that zeG. Assume that N oo

an > r when n > N. B(z) = ( fj 6„J( f] 60J. Thus fc=l fc=N+l

JV oo N oo

B'(z) = (Ubak)'( I] &«*) + ( II *«J( n &«J' fc=l fc=JV+l fe=l k=N+l

N N oo

= ( I l * . J ' B ^ ) + (n i . JB j»W. where BN(z)= ft bak. fc=l fc=l k=N+l

Note that 5jv(z) 7 0, z£G. Thus OO

In(Bjv(z))= ]T M f ^ ) -k=N+l

But

^ ( 2 ) £ 5iv(^) k£^+l K - 2)(1 ~ akz)

converges uniformly in G. Thus

k=N+l v ~ akZ' n^k,n>N+l

Note that

|S'(z)(z - 1)5| = I £ ia\:ll{k7f n *«„(*)!

fe=l V n^fc

<16E( l -a f c ) l l ^ f l l 2

fe=i

< 64 £ (1 - at) < 00, fc=i

thus B'(z)(z — l ) 5 is bounded on Z). It is easy to see that B(z)(z — l ) 4 is bounded on D. Therefore (B(z)(z — l ) 5 ) ' is bounded.


Similarly, (B(z)(z - l ) 5 ) " is bounded on D, these imply f(z)&R(D). Denote M{an} := {g£R(D) : g(an) — 0,n = 1,2,•••}. It is obvious

that M{an} is an invariant subspace of Mz with infinitely many common zeroes and f£M{an}.

For a given set M, [M] denotes the minimal invariant subspace of Mz

containing M. If M ^ {0} is an invariant subspace of Mz on Hardy space H2, Beurlig Theorem asserts that dim(MQzM) = 1 and [MQzM] — M. Comparing to this, the invariant subspaces of Mz on Bergman space L\ are more complicated. [Apostal, C , Bercobici, H., Foias, C. and Pearcy, C. (1985)] proved that if n is any positive integer or +oo, there is an invariant subspace M in L\ such that

dim(MQzM) — n.

For Sobolev disk algebra R(D), we have the following result.

Proposition 4.5.31 Let M ^ {0} is an invariant subspace of Mz in R{D), then dim(MQzM) = 1. Proof [Richter, S. (1987)] studied a Banach space B of analytic functions, which satisfies the following conditions:

(i) For each XQD, the point evaluation functional is continuous; (ii) If feB, then zfeB; (iii) If f£B and /(A) = 0, then f = (z - X)g for some g&B. S. Richter also proved that if the above Banach space B is an algebra

and M is its closed ideal, then dim(MQzM) = 1. It is obvious that Sobolev disk algebra and each nonzero invariant sub-

space M of Mz satisfy these conditions. Thus dim(MezM) = 1.

4.6 The Operator Weighted Shift

Let {WAJ^LI D e a sequence of uniformly bounded operator on C". An oo

operator S in £( 0 C n) is called a unilateral operator weighted shift with fc=o

weighted sequence {Wk}™=1, denoted by 5'~{W/fc}^=1, if

S(x0,xi,--- ,xk,---) = ( O . W i Z o , - - - ,Wk+1xk,---)

+00 +00 for all (xk)eH+ = © C". Similarly, an operator S on H = © Cn

fc=0 k~—00

is called a bilateral operator weighted shift with weighted sequence

W}fcT-oo. d e ™ t e d by S~{Wb}£T_oo. if

for all (xk)£H In general, unilateral and bilateral operator weighted shifts are both called operator weighted shift, denoted by S~{Wk}, and n is called the multiplicity of S.

When n = 1, S is the scalar-valued weighted shift, from which the operator weighted shift is naturally generalized. We must point out here that it is not just a formal generalization. For example, using operator weighted shift, [Pearcy, C. and Petrovic, S. (1994)] proved that an n-normal operator is power bounded if and only if it is similar to a contraction operator.

In this section, we will discuss injective operator weighted shift, that is, each Wk is invertible. Given an operator weighted shift S~{Wk}, since S and el9S are unitarily equivalent for each #£[0,27r] [Lambert, A (1971)], cr(S),ae(S) and oy(S) have circular symmetry. It is easily seen that

r(S)= Km (sup\\Wi+k...Wi+1\\)K k—>oo i

n(S)= l\m(mf\\Wi+k---Wi+1\\)l, k—>oo 1

where r(S) is the spectral radius of S and

ri(S)= lim (m(Sk))i, fc—>00

m(S) := inf{\\Sx\\ : ||z|| = 1}

is the lower bound of S. Since dirnker(S — \)<n < 00 and dimker(S — \)*<n < 00 for each

AeC, ps_F(S) = PF{S)- For unilateral operator weighted shift S^{Wk}kxL1-l

we have

<r,(S) = <r*(S) = <re(S)c{\ : r i (S)<|A|<r(S)}.

Clearly, <rp(S) = 0, thus ar{S) = a(S) = {A : |A|<r(S)} [Lambert, A (1971)], where av(5) is the approximate point spectrum of S.

An operator weighted shift 5~{Wfc} is said to be upper triangular, if there is an ONB {ei}™=1 of C™ for each k such that Wk admits an upper

triangular matrix representation with respect to this ONB. By the basic matrix theory, we have the following proposition.

Proposi t ion 4.6.1 Every operator weighted shift S~{Wk} is unitarily equivalent to an upper triangular operator weighted shift S~{Wk}-

Theorem 4.6.2 Let S~{Wk}'£L1 be a unilateral operator weighted shift, if cre(S) is disconnected then S^(SI).

To prove this theorem we need the following lemmas.

Lemma 4.6.3[Herrero, D.A. (1990)] Let A,BG£(H), then

an(AB) = <JI{TAB) = ai{A) - ar{B).

Lemma 4.6.4 Let A££(Hi), B&£(H2) and

dirnHi = dimH-z-

If oi(A)C\ar{B) = 0, then G := {YGranTAB '• ||V"||<1} is closed in weak operator topology (WOT). Proof Without loss of generality, we assume that H = Hi = Hi. Let {ya}a £A be a net in G and Y = (WOT) — l imi^ . Then there exists

a

Xa££(H) such that Ya = TAB{XO). By Lemma 4.6.3, 0 ^ O-I(JAB)- Thus there exists TGC(C(H)), such that TTAB = I- Therefore,

||xa|| = ||T(yQ)||<||T||||rQ||<||T||

for each a. Since each bounded closed set in £(H) is compact in weak operator topology [Conway, J.B. (1990)], we can find an operator XG£(H)

such that X = WOT — l imX a . Therefore, for arbitrary re, y£H, a

lim < (TABXO)X, y > = lim < AXax, y > — lim < XaBx, y >

= < AXx, y > - < XBx, y >=< (TABX)X, y > .

Thus Y = WOT - lim TABXa = TABX. Clearly | |F | |<1, therefore Y<=G.

L e m m a 4.6.5 Let ^4~{^4fc}j£Li and B~{Bk)V=i ^e unilateral operator weighted shifts with multiplicities n andm respectively. If ai(A)(lo-r(B) = 0

and if C is an operator of the form

C

then CEranTA,B-Proof Let M = sup{||cfc||},

k

0 ci 0

c2 0

0 cj 0

C2 0

0 c3

c™ c n

0

c

0 "

n

cm

cm

cm

Cm ,

Fk = cfc 0

0 0

and Ek =

0 0 0

0 0

Cfc 0

0 0

Then \\Fk\\<M,Fk = £ Ej and C = WOT - limjFfc. By Lemma 4.6.4 we

need only to prove that each Ek&anTA,B- Set

Xk-i = A^ Ck,

Xk-2 = (AkAk-i)~1CkBk-i,

Define

X0

(Ak- • •A2)~1ckBk-v • -B2,

(Ak---A1)-1ckBk-V--Bi.

X

X0

Xk-i

r~in /~m /"m r~*n

cr

Cr' Cr'

A straightforward computation shows that AX — XB = Ek-

Let Hk be the k-th subspace in the orthogonal direct sum H+ = 0 C". k=0

Then for each unilateral operator weighted shift S~{Wk}kxLi, ^k can be

regarded as an invertible operator from Hk-i to Hk- In what follows we will always identify x&Hk with (0, • • • ,0 ,x,0, • • • )GH+, the k-th of which is x.

Lemma 4.6.6 There exists an ONB {ef'}f=1 of Hk {k = 0,1,2,---) , such that the weighted Wk of S is of the form

Wk =

wi (fc)

„(*) wkk)

Jfe-i)

Sk~i) (4.6.1)

and di>di+i (i = 1, 2, • • • , n — 1), where

di=]im(f[\WiU)\)i.

3 = 1

(4.6.2)

Proof For convenience denote Mk = WkWk-i- • -W\ and MQ = / . Choose an ONB {e^0), ef\ ••• , e{°}} of H0- Set

dx =max{TInr||M fcef ) | |* : \<i<n}. k—*oo

M\ Without loss of generality, we may assume tha td i = lim | |M f cef J | |s. Set k—>-oo

=(°) l|Mfce<°>||

Then

Wke\' (fc-i) ||Mfcel' (0),

„(*) . = ,mjk)

\\Mk-A (0)| = u)K 'e (4.6.3)

and S i := {e[ f c ) }^ 0 is ONB of H+. Let P[k) be the projection from Hk

onto the subspace [\j{e{k)}]x and Pj(0) = / . By (4.6.3),

P^WkPf-V = P[k)Wk (4.6.4)

Suppose that we have found an orthogonal set Bj of H+,

Bj = {ef] : l<j<i,k>0}

and dj = '^^\\PJt\Mkef)\\i, where P,W (l<j<i, fc>0) is the projection fc—>oo J J J

from H+ onto (\/{e[k\ ••• , e<fc)})x and PJ0) = I. They satisfy the following properties:

(i) d\>d2> • • • >df,

(h) PJk_\Wke?-V = WJk)ef\ (j = l,2,.--,i). (4.6.5) (hi) Pf ]WkPf ~l) = P$k)Wk, 0 = 1,2,-.- ,i). (4.6.6) Without loss of generality, we may assume that

- H - l - ^ ^ W » ' - ( 0 )

k—*oo

Thus di>di+i. Set

di+1= l imll^Mfee^l fc—>oo

(k) _ ^i Mkei+i

f^ef -eftfc.

i + l l

By (4.6.6),

Fi Mk&i^ - WP^M.-^w

J"°Mfce^11 . c(fc) (4-6-7) - | |if-1,M,k-ieS"'iir+1

(fc) (fc)

Besides, Si+i = {e^ : l < j 0} is still an orthogonal set of H+. Let

P i+{ be the projection from Hk onto the subspace (V{ei > • • • i ei+i})"1 a n d

i^+i = / . It follows from (4.6.6) and (4.6.7) that P^WkPfc1* = P$\Wk.

Repeating the procedure above, we can find {rfi}?=1 and ONB {e[k)}?=1

oiHk such that d i>d 2 > • • • >dn and {e(k) : l<i<n, k>0} is an ONB of H+.

From the choices of e\ and w\ ', each wk has the form (4.6.1) and

fc ~ n 5 " ( i n « W =TS"||i'i(*)iMtei

0)||* =di(» = l)2,.-- ,n). fc—»oo -*•-*• fc—>oo

i= i

By Proposition 4.6.1, the following lemma can be easily proved.

Lemma 4.6.7 Let S~{Wk}kLo be an operator weighted shift, then

S\ S\2 • • • Sin

s2 '•• !

o ' *~*n — l , n

Hi

n2

'•'rin

whereH+ = W ^ ' . ^ ^ K ^ l f c l i is an injective unilateral weighted shift and

S^ is a scalar-valued weighted shift (not necessarily injective). Without loss of generality, we may assume that Tii = I2. Lemma 4.6.8 Let S~{Wk} be an operator weighted shift of the form given

n in Lemma 4-6.7. Then cre(S) = \J ae(Si). Proof We only prove the lemma when n = 2, and let

S = Si S\2 0 S2

Let n be the canonical map from C{H+) onto C{H+)/K{Ji+). Then

n(S) w i ( S i ) 7 r i ( 5 i 2 )

0 7Ti(52)

where TTI : C{Hi)—+C{Hi)/)C{Hi). We need only to show that (T(TT(S)) = O-(-ITI(SI))UO-(ITI(S2)).

By Gelfand-Naimark-Segal Theorem, we need only to prove that

a(S)=a{S1)Ua(S2).

If \€p(S), then (S - A) - 1 = (Aij)2x2 satisfying

S\ — A Sl2

0 S2-\ An AX2

A2i A22

Thus (S2 - A)^22 = I- Since dimker(S2 - A)<1, by Atkinson Theorem, (S2 — X) has a left inverse and \£p(S2). Since (52 — X)A2i = 0,^21 = 0. Therefore

(Si-\)Au=Au(Si-\)=I.

This implies that Aep(5i). Thus \Gp(S1)np(S2).

Conversely, if AGp(5i)np(52), set

X ( 5 1 - A ) - 1 - ( 5 1 - A ) - 1 5 ( 5 2 - A ) - 1

0 (S2 - A)-1

Then X{S - A) = (5 - X)X <r(51)Ua(52).

I. This implies XGp(S) and a(S) =

Now we are in a position to prove Theorem 4.6.2. n

Proof of Theorem 4.6.2 By Lemma 4.6.8, ae{S) = \J ae(Si). Since

ae(S) = an(S)c{\GC : r!(S)<|A|<r(S)} [Lambert, A (1971)],

there exist e and S, 0<e < <5, such that fl := {A : e < |A| < 5} is a bounded component of PF{S). Therefore, for each Si, at least one of the following holds:

(a) <r(Si)cUc := {A : \\\<e}-(b) ae(Si)cV5 := {A : |A|>«5}. Denote io = max{i : cre(Si)cVg}, then by the definitions of r(S) and

r i (5) and Lemma 4.6.6, ri(Si0)<di0<di<r(Si), if i<io- Therefore Si does not satisfy (a) and cre(Si)cVs if i<io! but if i > io, Si does not satisfy (b), thus a(Si)cUe. Note that {A : |A| = e or ^}ccre(5). We have l < i 0 < n.

io n Now denote Mi = © Hi and M2 = M^ = 0 Hi. Then

i=l i—io-\-l

s = AC 0 B M2'

where A and B are operator weighted shifts with multiplicities io and n — io respectively, and C is of the form given in Lemma 4.6.5. By Lemma 4.6.8,

io

<re(A) = \J<Te(Si)cV5.

n By Lemma 4.6.8 and its proof ar(A) = a(B) = \J a(Si)cUe. Therefore

oe{A)nar(B)=$.

By Lemma 4.6.5, there exists an operator X such that

AX - XB = C.

Set

Y = IX' 0 /

Then

and Sg(SI).

YSY'1 = A 0 0 B

Next we will discuss when the adjoint of a unilateral operator weighted shift is a Cowen-Douglas operator.

Proposition 4.6.9 Let <S~{Wfc}j£Li be a unilateral operator weighted shift. IfO€ae(S), then S* can not be a Cowen-Douglas operator. Proof If ae(S) = a(S), then S* is not a Cowen-Douglas operator. Thus we may assume that ae(S) ^ <r(S). Since 0£ae(S),ae(S) is disconnected. Let ft be a connected component of pp(S)na(S). Repeating the proof of Theorem 4.6.2, we can choose ft satisfying ft = {A : e < |A| < 5}, where 0<e < S. Therefore, there exists an invertible operator Y such that

So D* = (Y*)~1S*Y*

YSY'1 =

A* 0 0 B*

A 0 0 B

Mi

M2

= D.

M M

1 . Since a(B*) = a{B)c{\ : \\\<e}

(see the proof of Theorem 4.6.2), B* — X is invertible for all Asft. Thus

A ;e r (Zr -A)oMi .

This implies that \J{ker(S* - A) : Aeft} ^ H+.

Theorem 4.6.10 Let S^{Wk}^=1 be a unilateral operator weighted shift. Then the following are equivalent:

(i) M := sup ||Wfc-1|| < oo;

(ii) n ( 5 ) > 0 ; (Hi) 0£pF{S); (iv) There exists a connected open set ft containing 0 such that

S*eBn{n). Proof (i)=»(ii) and (ii)=^(iii) are obvious from the definitions.

(iii)=4-(iv). It is easy to see that there exists a connected open set fi containing 0 and riCpF(S). Since ap(S) = 0,5* — A is surjective and

dimker(S* - A) = ind(S* - A) = indS* = n

for all Aeft. Note that \/{ker(S*)k : fc>l} = H+, thus S*eBn(Q). (iv)=>(i). Since 5* is surjective, it has a right inverse T =

(Tk,i)%l=0eC{H+). Thus W^Tk>k-i = / and H ^ H = Wn^W^TW for each k.

By Proposition 4.6.9 and Theorem 4.6.10, the following corollary is obtained.

Corollary 4.6.11 Let S~{Wk}kLi be a unilateral operator weighted shift. Then B* is a Cowen-Douglas operator if and only if sup || VK^-1 j | < +oo.

fc

In what follows we consider the backward operator weighted shift S~{Wk} with the form

C"

Cn on H = 0 C n , n = l

and denote Ao(S) = A'(S)\kers-

Theorem 4.6.12 Let S~{Wk} be a backward operator weighted shift, then the following are equivalent:

(i) There exists AoGAo(S) such that card(a(Ao)) = m, (m<n); (ii) S = Si+S^-i \-Sm and for each i, l<i<m, there exists a uni

tary operator Ui such that UiSiU* = Si is an operator weighted shift with m

multiplicity n-i and ^2 rii = n; i=l

(Hi) S has a spectral family. Proof (i)=>(ii). For convenience, we assume that m — 2. Since A0eAo(S), there is A£A'(S) such that A\kers = A0. Let Aa = diag(A0,Ai,- • • ,Ak,---) be the diagonal part of A and let p(z) be the characteristic polynomial of AQ. Since Ak — Ek~

1AoEk,p(Aa) — 0, where Ek = Wv • Wk (fc>l). Thus p(a(Aa)) = {0}. This implies that a(Ao) = {Ai,A2}. Thus there are Jordan domains S7i and 0,2 such that Aiefi i ,A2en2 and 0[r^h = 0- Denote T* = dtli (i = 1,2). Clearly,

OWi 0

0 W2

0 ' • .

0


Ad&A'(S). Thus the Riesz idempotents

Pi = 2^ / (A " Ad)~ldX£-A'{S) (i = 1,2).

Note that

(A - Ad)'1 = diag((X - A0)-\- • • , E^(X - A0)-lEk, •••).

Thus Pi = diag(P£\ • • • ,P{ki] ,• • •), where

Pk] = Ek1^ I (A - A0)-1d\)Ek = E^P^Ek, (i = 1,2)

and

P f ' p f =0, P^ + P™ =/c»,(* = 0,l,2,...).

Denote Mk = P^C" , Nk = P?)Cn, Hi = PJi, (* = 1,2). Then OO

W = ®(Mk+Afk) = H1+H2, fc=0

and 5 = S1+S2, where 5i = S\Hl, 5 2 = S|Wa. Since

dimAik = dimMk-i = ni

and

dimMk = dimAfk-i — ^2 (&>!•)•

Therefore

Wk = W ^ - i - W ^ ,

where ^ e A . M f c . M f c - i ) , ^ 2 ^ / : ^ , ^ - ! ) . 00

Define the unitary operator Ui£C(Hi, ® Alt) as follows fc=0

^ 1 = ( ( j / o , 0 ) , ( y i , 0 ) , - - - , ( y f c ) 0 ) , - - - ) i - > ( y 0 , 2 / i , - - - ,Vk,---)-

Then U\S\Ul = Si is an operator weighted shift with weight sequence {Wjfc }£ii a n ( i multiplicity n i . Similarly, we can define a unitary operator

U2£C(H2, 0 A4) such that U2S2U2 = S2 satisfying the requirements of fc=o

the theorem. (ii)=>(iii). Obvious. (iii)=>(i). We still assume that m = 2. Since {Qi,Q2} is a spectral

family of 5 , Qi,Q2eA'(S). Let

Qf=diag{Q(i\..-,Qf,...)

be the diagonal of Qi{i = 1,2). Clearly, {Qd ,Qd } is also a spectral family of 5. Set Si = S\ „(i) (i = 1,2). Then

5 = S\+S2- Repeating the argument of (i)=>(ii), we can prove that Wk = W^+WP, where W^]eC(Mk,Mk-i), W™eC(NkMk-i) and dimMk = dimMk-i,dimAfk = dimNk-i for all k>l. Choose i±\,ii2&C such that \ni\ < |/n2|-

Set AQ = HIIMO+MI/SO and Afe = E^xA0Ek = fiilMk+^I^, k>l. Then

||Afc||<|H(IIQi|| + IIQ2||).

Thus

A:=diag(A0,Ai,--- ,Ak,---)eA'{S)

and

A0 = ^|fcers,cr(J4o) = { / i i , ^} ,

i.e., card(a(Ao)) = 2.

Theorem 4.6.13 Let S~{Wk} be a backward operator weighted shift and r

let r = ma,x{card(a(A0)) : A0&Ao{S)}. Then S~S = 0 Si, where each r

Si^(SI) is an operator weighted shift with multiplicity rii and £^ n» = n. i = l

Corollary 4.6.14 Let S~{Wk} be a backward operator weighted shift and A0eAo{S), then a(A0) = aAo{s)(A0). Proof Clearly, a(A0)CaAo(S)(Ao).

Assume that Ad = diag(Ao,Ai,---,Ak,---), where Ak = Ek~

1AoEk (k>l). By the preceding discussion, cr(Ao) = o-(Ad) and AdeA'(S). If A £ a(A0), AdB = BAd = I for some Be£(H) and BGA'(S).

This implies that B0 = B\kers€Ao(S) and A0B0 = B0A0 = J o - Therefore

< M O ( S ) ( 4 > ) C < T ( J 4 O ) .

Theorem 4.6.15 Let S~{Wt-} be an operator weighted shift, then the following are equivalent:

(i) Se(SI); (ii) cr(Ao) contains only one point for each AoGAo(S); (Hi) Ao(S)/radA0(S)^C; (iv) There is no nontrivial idempotent in Ao(S),

Proof (i)=>(ii). It is a straightforward conclusion of Theorem 4.6.12. (ii)=Kiii). Let J = {A0eAo(S) • <r(Ao) = {0}}. Clearly J =

radAo(S). Define a linear mapping C by

tp[Ao] = XAo, [Ao]eAo{S)/radAo(S), where XAoea(A0).

Note that a{Ao) consists of one point and for each Bo£[-<4o], Ao — B0€radAo(S). Thus a(A0 - B0) = {0}. Assume that a(A0) = {A}, a(B0) = {/j,}, then

0 = tr(Ao — B0) = trAo — trBo = nX — nfi

and A = v. This implies that <p is well defined.

If [Ao] ? [Bo], then a(A0 - B0) = {7},7 ^ 0, and

w) = tr(Ao — B0) = trAo — trBo = n-y 0.

This means that <p is injective. Clearly tp is surjective. Since

[A][B] - XAXB = [AB-XAB + XAB-XAXB] = [A- XA\[B\ + XA[B-XB],

it follows from a((A - XA)B) = a{XA{B - XB)) = {0} that <p([AB] -XAXB) = 0. This implies that y>([AB]) = </?( [A]) ¥>([£]) = A^AB and ip is an isomorphism.

(iii)=>(iv). Let ip be an isomorphism from Ao(S)/radAo(S) to C. Then

<p([Po]) = Ao

for all Po£Ao{S), where Ao = 0 or Ao = 1. Since

ip([PQ - AJc-]) = 0,

<r(Po - A) = {0}.

Therefore, 0 = tr(Po-X) = trPo—nX. This implies that P0 = 0 or Po = Ic-


(iv)=>(i). Assume that PeA'(S) is an idempotent, then Po = P\kers£Ao(S) is also an idempotent. Thus Po = 0 or Po = i o - Without loss of generality, we assume that PQ = 0. From PS = SP,

P =

OPoi * 0 Pl2

0 '••

cn

cn

C"'

Since P 2 = P, P = 0. Thus Se{SI).

Theorem 4.6.16 Let S~{Wfc} be an operator weighted shift, then SG(SI)

if and only if J := {A£A'(S) : c(A\kers) = {0}} is a maximal two sided ideal of A'(S) and A'(S)/J is abelian. Proof Suppose that Se(SI). {Claim a} J is a linear space. For A, B&J,

(A + B)\kerS = A\kerS + B\kerS = A0 + Bo.

By Theorem 4.6.15, a{A0 + B0) = {A}. Thus

nX = tr(A0 + Bo) = trA0 + trB0 = 0.

This implies that A = 0 and A + B e J. {Claim b} J is a closed two-sided ideal. Note that for A£j and BeA'(S), AB\kers = (A\kers)(B\kers)&MS)- Thus a(AB\kers) = {0} and Claim b holds. {Claim c} J is a maximal ideal. Suppose that J' is a two-sided ideal of A'(S) satisfying J' D J and J' ± J. Choose AeJ'\J, then Ad,

oo

the diagonal of A with respect to H = 0 C n , is in A'(S). Since A^J, fc=0

o-(Aa) = o-{A\kers) = {A}, A 7 0. This means that Ad is invertible. Denote Ar = A- Ad. Then AreJ. Thus Ad = A - ArSj'. This implies that J' = A'(S). Therefore J is maximal. {Claim d} A'{S)/J is commutative.

For A, BeA'(S), Set C = AB - BA. Then

Co = C\kerS = (AB — BA)\kerS = A0B0 — Po^O-

By Theorem 4.6.15, a(C0) = {fi}. Thus

n/i = trC0 = tr(A0B0 - B0A0) = tr{A0B0) - tr(B0A0) = 0

and fi = 0. Therefore C£j and A'(S)/J is commutative. Conversely, suppose that J is a maximal two-sided ideal of A'(S) and

A'{S)/J is abelian. If P is a nontrivial idempotent in A'(S). Set J' = A'(S)PA'(S) + J. It is not difficult to see that J' is a two-

sided ideal of A'(S) and J'DJ. Since P0 = P\kerS ¥= 0 ,P £ J and J ' ^ J ' . We assert that the identity 7 ^ J ' . Otherwise I = APB + C, where A,B<=A'{S) and C e J . Thus 7C" = liters = A0P0B0 + C0, here 4o = A\kerS,B0 = B\kerS and C0 = C|fcers. Since C0Sk7|fcerS,o'(Co) = {0}.

This means that

J c „ - C0 = A)P0So

is invertible. But the determinant detPo of P0 is zero (P0 7 0). Thus

deL40P0B0 = (detA0)(detP0)(detB0) = 0.

A contradiction. Therefore J'Z)J (but J' ^ J7) is also a maximal proper ideal. This contradicts the assumption that J is a maximal two-sided ideal.

Corollary 4.6.17 Let S~{Wk} be an operator weighted shift satisfying that Wk = W for all fc>l. Then SG(SI) if and only if We(SI). Proof By Jordan Theorem, there is an invertible matrix XGMn(C) such that

J = XWX~l = Jnj(^l) 0

Jnt(h)

and ^2 rii = n. i=l _

Set Y = diag(X, X, • • •), then Y is invertible, and S = YSY^^Wk} is an operator weighted shift, where Wk = J. By Theorem 4.6.15, S£(SI) if and only if I = 1 or if and only if W£(SI).

The following example indicates that the condition Wk = W (k>l) can not be omitted.

Example 4.6.18 Let

W2k+i 1 1 0 1 , w2k =

1 - 1 0 1

then

Wk-1 1 0 1

e(Sl).

Let S be the operator weighted shift with weighted {Wk}. Set

2k+l = 10 0 0 p2fc

1 1 00

and P — diag(Pi,P2, P3, • • •). A simple computation shows that PS = SP andP2 = P, i.e., S#(SI).

Now we will discuss the similarity of two operator weighted shifts.

Proposition 4.6.19 Let S~{Wk} andT~{Vk\ be two operator weighted shifts. Then 5 ~ T if and only if there is a sequence of invertihle matrices {Xt} such that sup{\\Xi\\, | |Xf i } < 00 and Wt = XiKiXi+

11, (*>1).

i

Proof If there are invertihle operators {Xi}^ on C n such that

supdl^llJX - l l < 00

and

Wi = XiViX-+\,(i>l).

Set X = diag(Xi,X2, •••). Then X is invertihle and SX = XT, i.e., S~T. On the other hand, if 5 ~ T or SX = XT and X^S = TX~l for some operator X. Note that X can be expressed as

X =

Xi X\i X\z X2 -^23

x3

where WiXi+i = XiVi and sup{||Xj||} < 00. Since X x has the form

X~l =

^1 ^12 ^13

X2 Y23

where Y,Wi = ViYi+\ and sup{||Yj||} < oo. Since i

xx-1 = x~1x = i,

XiYi = YiXi = ICn, (i = l ,2 , - - - ) .

Corollary 4.6.20 Let S^{Wk}f=1,T^{Wk}^=2 and S€Bn(Sl), then

Proof By Corollary 4.6.11, SeBn(tt) implies that

sup{| |W i | | , | |W i-1 | |}<oo.

i

Set Xi = Wi, then Wi = XiWi+iXr^v By Proposition 4.6.19, S~T.

Proposition 4.6.21 Let S~{Wk} and S~{Vk} be two operator weighted shifts, then S~S if and only if S~q.sS, where "~q . s" means quasisimilar. Proof We need only to verify that S~q.sS implies S~S. By the quasisim-ilarity, we can assume that S and S have the same multiplicity n and there exist X, Ye£(H) with trivial kernels and dense ranges such that SX = XS and YS = ~SY. Then X and Y have the form

X

Xx *

x2 0

Y2

0

Since kerXxckerX = {0},Xi is invertible. If follows from SXY = XSY + XYS that XYeA'(S) and

XY =

XiYx * X2Y2

0

Thus d%ag{XxYi,X2Y2, • • • )£A'(A). From SX = XS, we have

X2 = W^XM,.• • ,Xk+1 = W^XkVk, (fc>l).

This implies that Xk (k>l) is invertible. Similarly, Yk (k>l) is invertible. Thus XkYk (k>l) and diag(X\Yi,X2Y2, • • •) are invertible, and

[diagiX^XiYi, • • • J]"1 = diag{Y^ X^\Y2~l X^\ •••).

Therefore, there exists M > 0 such that

s u P { | | y f c - % - 1 | | } < M . k

Since diag(X\,X2, • • • )£kerrs•§, there exists N > 0 such that sup{||Xfc||} < k

N. Furthermore,

supdin-^D^supdin-^fc-^i.HXfciD^supdiy^Xfc-^D-supfHXfciD^Miv.

This implies that diag(Yi,Y2,- ••) is invertible and diag(Yi,Y2, • • • )€kerTgS. Thus S ~ 5 .

Proposition 4.6.22 Let S~{Wk} and S~{Vk} be two operators weighted shifts with multiplicity n. S,S£(SI) and S is not similar to S. T = S®S. Then Ao(T)/radA0(T)^C ® C. Proof Note that

XS<=A'{S) Xg£A'(S) 1 Xi2£kerrSg X2i€herrg s J '

For arbitrary Xi2&kerrSg and X21 £kerTg s, we have

'Xx * '

x2

. 0

, X21 =

'Yx *• Y2

. 0

Computations indicate that

X12X2i&A'(S),X2iX12€A'(S)

and

-^12-^2l|fcerS = XlYi, •^21-^12|fcer5 = YiX\.

Since S,'Se(SI), it follows from Theorem 4.6.15 that

<r{X{Y{) = o^Xx) = {A}.

{Claim} A = 0. We assume that Xi and X2 are invertible. Set X' = diag(X\,X2, •••),

then kerX' = {0} and ranX' = H. Similarly, if Y' denotes Y' = diag(Yi,Y2,- ••) then kerY' = {0} and ranY' = H. Since X'ekerrs^ and Y'£kerTg s, S~S. A contradiction.

-4'(T) = J -^21 -X"s


For each

and XGkerTs-g, \&

D = En E21

E12

E22

En E12

E21 E22 eMT)

0 0 = '0EnX\kers .0 E2lX\kerg_

By the claim above, cr(E2iX\kerg) = {0}. Thus a(D) = {0}. This implies that

0X1 \kerS 0 0

GradA0(T).

Repeating the argument above, if YGkerr-g s , then

GradA0(T). 0 0 y\kerS 0

Thus

Ao{T)/radA o(T) = { Xs + radAo{S) 0 0 Xg+radAo(S)

XS&A Xs€A' '(S)j-

By Theorem 4.6.15, Ao(T)/radAo(T)^C © C.

Proposition 4.6.23 Let S and S be two operator weighted shifts, S,Se(SI) and S~S. T = S®S, then

MT)/radA0(T)^M2(C).

Proof Without loss of generality, we assume that S = S. Then

MT) = ! \ X n

U * 2 1

IfcerS X\2\kerS

\kerS X22I kerS XijeA'(S),i,j = l , 2 . | .

Since S£(SI), by Theorem 4.6.15, A0(T)/radAo(T)^C. Thus we have Ao(T)/radA0{T)^M2{C).

Proposition 4.6.24 Let S~{Wk} and 5~{Vfc} be two operators weighted shifts. S,SG(SI) and S is not similar to S. T = S(BS. Then

Ao{T)/radA0{T)^C © C.

Proof Assume that S and S have multiplicities m and n respectively. By Theorem 4.6.15, we need only to prove the proposition in the case of m <n.

For

since E12E21 GAQ(S) is not invertible, by Theorem 4.6.15, a(Ei2E2i) = {0}. Denote

EuEkerTgg 1 E2iGkerT-gS J '

By the arguments used in the proof of Proposition 4.6.21 we have J = radAo{T). Thus Ao(T)/radAo{T)^C 0 C.

Summarizing the discussion above, we have the following theorem.

Theorem 4.6.25 Let S and S be two operator weighted shifts. S, S€(SI) and T = S®S. Then

1. S~S if and only if Ao(T)/radAo(T)^M2(C)] 2. S is not similar to 5 if and only if Ao(T)/radAo(T)=C © C.

As a matter of fact, we have a more general result by Proposition 4.6.21, Proposition 4.6.22 and Proposition 4.6.23.

Theorem 4.6.26 Let S and S be two operator weighted shifts and m

S~@Sj , where Si£(SI) is an operator weighted shift and S^Sj for i = i _ _

i ^ j (Corollary 4.6.13). Let T = S®S, then S~S if and only if m

Ao(T)/radA0(T)^ 0 M2fci(C). i = i

Proposition 4.6.27 Let S~{Wk} be an operator weighted shift and

k ~ |_0 1 J '

fc then A'(S) is commutative if and only i /sup | ^ Aj| = +00.

fc i= i

-E11 E12

E21 E22

J = radA'(S) Ei2_

E21 radA'(S)

Proof If TGA'(S), then T is of the form

T i i T\i T13 • • •

2~22 ?23 • • '

0 T 3 3 ' - -

and Tk+1>k+1 = Elk1TnE1,k+i-i (A;>1, Z>1), where

Elk=WlWl+v--Wk =

k

l E A i i=l

0 1

Set

Tu = *21 *22

then

Tk+i,k+i — Elk T\iE^k+i^i = Hi Hi i' t' '21 r 22

where

k

t'n = hi — ( E Ai)*2i i = l

k+l-1 k+l-1 t'12=t12 + ( E Wn - (Z W22 - (Z \i)( E AOfei

i = i t = l i = l i = i fc i - 1 fc+i-1 fc k+l-1

= *i2 + (EAi)(tn-*22) + ( - E A i + E A 0 * n - ( E A i ) ( E Ai)t2i i = l i = l i=fc+l i = l i=l

k+l-1

t'22=t22 + ( E A<)*21. i=l

If sup | E Ai| = +oo, since t'n are uniformly bounded, £21 = 0. Since fc »=i

sup{ | |W f c | | }<+oo, |A i |<M fc

l-\ fc+J-1 for some number M. Thus | - E A* + E A»| < 21M. Since t'l2 are

i = i »=fc+i uniformly bounded, tu = t\i. Thus

Tk+l,k+l = -Effc T'll-Bi.fc+i-l = *ll-f + *12«^2(0).

Therefore, .4 ' (5) is commutative. *

If s u p | ^ Aj| = N < +oo , set A = diag(A\,A2,---) and B = k »=i

diag(B\,B2, •••), where

-4fc+i

fc fe • E A i ( E A i ) 2

fc

EAi i = l

Bi = 0 1 0 0

. Then A, BGA'(S), but AB - BA ^ 0.

4.7 Open Problem

1. Is every operator in JC(H) is a direct integral of strongly irreducible operators? 2. What is the necessary and sufficient conditions for an operator T€£(H) to have only finitely many Banach reducing subspaces? 3. If Te£(H)fl(SI), is A'(T)/radA'(T) commutative? 4. Does the following statement holds for arbitrary Ti,T2&£(J~t)C\(SI)l

T!~Ta if and only if Ko(A'(T1®T2))=Ko(A'(T1))^Ko(A'(T2)). 5. Let T&£(H)r\(SI), then for each natural number n, does T^ have a unique (SI) decomposition up to similarity? 6. If TG£(H) is a direct sum of finitely many (SI) operators, does T have a unique (SI) decomposition up to similarity? 7. What is the "Beurling" Theorem for Mz in Sobolev disk algebra? 8. Given a necessary and sufficient condition for an injective unilateral operator weighted shift S~{Wk} to be strongly irreducible.

4.8 Remark

Theorems 4.1.3-4.1.20 are given by [Fang, J.S., Jiang, C.L. and Wu, P.Y. (2003)]. Theorem 4.2.1, Proposition 4.2.13 and Theorem 4.2.14 are due


to [Cao, Y., Fang, J.S. and Jiang, C.L.(2002)]. Theorem 4.3.1 belongs to [Fang, J.S. and Jiang, C.L. (1999)]. Proposition 4.3.5, Theorem 4.3.9, Proposition 4.3.11 and Proposition 4.3.12 are proved by [Ji, Y.Q. and Yang, Y.H. (2003)]. Example 4.3.8 is due to [Fang, J.S. and Jiang, C.L. (1999)]. Theorem 4.3.6 is given by [Jiang, C.L. (2004)]. Theorem 4.3.13 and Corollary 4.3.14 are proved by [Wang, Z.Y. and Xue, Y.F. (2000)]. Example 4.4.1 and Proposition 4.4.2 belong to [Jiang, C.L. and Li, J.X. (2000)]. C.L. Jiang also proved Theorem 4.4.3 [Jiang, C.L. (2004)]. The all results in Section 4.5 are proved by [Wang, Z.Y. (1993)], [Wang, Z.Y. and Liu, Y.Q. ], [Liu, Y.Q. and Wang, Z.Y. (2004)], [Liu, Y.Q. and Wang, Z.Y. ], [Jin, Y.F. and Wang, Z.Y.(l)]]. Theorem 4.6.2, Proposition 4.6.9 and Theorem 4.6.10 are given by [Ji, Y.Q., Li, J.X. and Sun, S.L. (2003)]. Theorem 4.6.12-Theorem 4.6.26 are proved by [Jiang, C.L. and Li, J.X. (2000)]. Example 4.6.18, Proposition 4.6.19-Proposition 4.6.27 are given by jia-jin-wan. The reader can refer to [Davidson, K.R. and Herrero, D.A. (1990)] about the (SI) decomposition of some special operator classes and [Behncke, H.], [Yan, C.Q. (1993)] about to the unique (SI) decomposition up to unitary equivalence.

Chapter 5

The Similarity Invariant of Cowen-Douglas Operators

5.1 The Cowen-Douglas Operators with Index 1

The backward unilateral shift is a typical Cowen-Douglas operator with index 1. Denote H = I2 — {(xi,x2,- • •) : X) W 2 < °°}- F° r («o,ai , - ••)&2, define T*(ao, a\, • • •) = (ai , «2, • • • )• For I'M < 1> w e have

r2*(l,A,A2 ,-. .) = A(l,A,A2 , . .-).

This implies that Dcap(T*). Clearly, T*eBi(D) and A'(TZ)^H°°. Note that Tz, the adjoint of Tj*, is an analytic Teolitz operator and also a pure isometry operator.

oo An operator SGC(H) is called a pure isometry operator if f] SnH =

n=l {0}.

von-Neumann-Wold Theorem Let S£C(Ti) be a pure isometry opera-i

tor. Then S^ ® Tz, where I = dimkerS*. fc=i

It is easily seen that Tz is also a pure isometry operator for each natural number n. The following is a well-known result.

Lemma 5.1.1 Let SGJC(H) be a pure isometry operator, then (i) 5 S T i ° and S*eBt(D), where I = dimkerS*; (ii) S€(SI)ifandonlyifS*eB1(D).

Lemma 5.1.2 Let P be an idempotent in A'(Tz), let S = TJ"^ |PW<n) and m = dimkerS*. Then there is a unitary operator U such that

149

(i) U(PH^)=H^®0^m\ i.e.,

UPU* = 0 0

(ii) LetV = U\ Pfi(rv), then VSV* = Tzm , i.e., S is unitarily equivalent

toZ (m)

Proof It is obvious that S is a pure isometry. By von-Neumann-Wold Theorem, S is unitarily equivalent to TJ . Thus there is a unitary operator

such that

V : PH{n) —• H{n)

VSV* = T ( m ) .

Note that if m < n, W ( n )©PW ( m ) is infinite dimensional. Therefore, there exists a unitary operator

W : H(n)ePH(n) —> Hin~m).

Set U = V®W, then U satisfies the requirements of the lemma.

Lemma 5.1.3 Let P be an idempotent in A' ((T*)^) and let S = (T*)(n^\p-H(n). If m = dimkerS, then there exists a unitary operator U such that:

(i) U(Pn^) = ft(m)©0(n-m\ i.e., UPU* = 0 0 y{n-m) •

(ii) Let V = C/|PH(n), then VSV* = (T*)^m\ i.e., S is unitarily equivalent to (r*)(m>. Proof Let Q = (IHM - P)*, then QGA'((T2)^) is an idempotent and

(Q(Tzr>Q)* = (/„(„, - P) (T; )W(/ H ( „ , - P ) . (n)/

Thus

Wl {{T*T%H^Y = {T;T\iH{n)-P)H^

Since PeA'((T*)^),

dimker((Tz)^\QnMr = dimker{{T*z)M)\(jH{n)_P)nM

= n- dimker((T*)^) =n-m.

The Similarity Invariant of Cowen-Douglas Operators 151

By Lemma 5.1.2, there exists a unitary operator XJ\ such that

0 0

and

U^QT^QUt = rp(n-m) ^

0 0

ft(n-m)

U(m).

Thus

and

UxPU* =

UiPiT^PU?

0 0 H(m)

0 0

* (T*)(m)

onto,

<^(n-m)

W ( m ) •

Define U2 : H{n) = H{n-m)®H{m) - ^ H{m)®H(m) by U2{x®y) = y®x for x€H{n~m) and y£H{m). Then t/2 is a unitary operator. Let U = U2UU

then U satisfies the requirements of the lemma.

For Te# n (Q) and zeQ, S(T - zl) = (T - zI)S for all SeA'(T). Thus

Sker(T - zI)cker{T - zl).

Define (TTS)(Z) = S\ker(T-zi)- In general, we substitute S(z) for (FyS1)^) (see Chapter 3). Clearly, TT is an injective contraction. But for T — (T*)(n^GBn(D), we can easily obtain the following lemma.

Lemma 5.1.4 Let T = {T*)^n\ then TT is an isometry isomorphism from A'{T) onto Mn(H°°).

Lemma 5.1.5 Let H = H2, .4e6i(fi)n£(W) and T = A<-nK For each idempotentPeA'(T), denoted =T\PnM. //Ti<=£m(ft), thenT^A^l Proof Without loss of generality, we can assume that DcQ. Then we can find W-valued holomorphic functions v(z) and e(z) on D such that

(A-z)v(z) = 0, (T; - z)e(z) = 0,

and v(z) and e(z) can be chosen to be the holomorphic frames of ker(A — z)

and ker(T* — z) respectively. Set

«fc(z) = ( 0 , " - , 0 ) v ( z ) , 0 , - . - , 0 ) )

k = 1,2, • • • ,n,z&D. efc(z) = (0 , - - - ,0 , e (z ) ,0 , . . . , 0 ) ,

Let P{z) = (TTP)(z), zeD, then P{z) = ( P ^ z ^ ^ e M ^ t f 0 0 ) is an idem-potent. By Lemma 5.1.4, P(z) is an idempotent in A'((T*)'")). Set

g = P(z)

and

s = (r;)(nW>. Since Ti£Bm(fl), dimkerS = ranP(0) = dimkerTi = m.

By Lemma 5.1.3, there exists a unitary operator U such that

U(PH{n)) = W ( m ) ©0 ( " _ m ) ,

i.e.,

UPU* = 0 0 <ft(n-m) (5.1.1)

and if V = *7|PW(n), then KSK* = (Tz*)(m>. Since l/*(V^m)©0("-m)) = l/*(ft(m)eO(n~m)) = QH™ and

VSV* = {T*)(m\ U*ei(z)£ker(S-z)cker((T*)(n'>-z),l<i<m. Note that (ei(z), • • • , en(z)) is a holomorphic frame of ker((T*)(n^ — z). Thus

U*ei(z) = Xn(z)ei(z) -\ h Xin(z)en(z), l<i<m,

where A^eC. Since < ei(z),ej(z) >= <% < e(z),e(z) >, l<i, j<n and U* is a unitary operator, we have

Aii(z)Aji(z)H h Ain(z)Ajn(z) = <5y, l<i, j<m,z<=D. (5.1.2)

From (5.1.1), UP(z)U*ei(z) = IH(m)ei(z) = e^z), l<i<m,z£D. Therefore,

P{z)U*ei{z) = U*ei(z),

i.e.,

(Pij(z))nx„(Aii(z),- • • ,\in(z)) = (\n(z),--- ,Aj„(z)), (5.1.3)

where Ki<m and zeD.

Set Wi{z) = Xn(z)vi(z) + • • • + \in(z)vn(z),l<i<m. Since

<vi(z),vj(z)>=Sij < v(z),v(z) >,l<i^ j<n (5.1.4)

by (5.1.2), < Wi(z),Wj(z) > = Sij < v(z),v(z) >,l<i ^ j<m. It follows from (5.1.3) that P(z)wi(z) = Wi(z), l<i<m and z£D. Note that

P(z)ker(T - z) — ker(T\ - z),

thus Wi(z)£ker(Ti — z), l<i<m and (wi(z), • • • ,wm(z)) forms a holomorphic frame of ker{T\ - z) for each z€D.

For zGD, define U{z) : ker(A^ - z)—>ker(T1 - z) as follows

U(z)vi(z) = Wi(z), l<i<m.

It follows from (5.1.4) and (5.1.5) that

< U(z)vi(z),U(z)vj(z) >=< Vi(z),Vj(z) > = S^ < v(z),v(z) >,l<i,j<m.

Since U(z) is a holomorphic isometric bundle map, using Rigidity Theorem wehaveTi^^™) .

Theorem 5.1.6 Let A&Bi{Q)nC{K), then V ( * 4 ' ( ^ ) ) - N and K0(A'(A))^Z. Proof By Theorem 4.2.1, for every natural number n and idempotent Pe(A'{A^)), if Ax = A(")|pW(„), then A^A. This is a straightforward corollary of Lemma 5.1.5.

Proposition 5.1.7 Let A,BGBI(£1), then the following are equivalent. (i) A~B; (ii) K0{A'(A®B))^Z.

Proof (i)=^(ii). It is a straightforward conclusion of Theorem 5.1.6. (ii)=Ki). We need only to show that if A^B, then K0(A'(A®B))^Z.

Otherwise, we assume that K0{A' {A®B)=Z. Since A^B, there exists a maximal ideal J in A'(A®B) such that A'(A@B)/J^C, where

I" J' kerrAiB [kerTBtA A'(B) _

and J' is a maximal ideal of .4'(.A) (see Section 5.2) following separating exact sequence:

n 0—>J -i-» A'{A®B)^A'{A@B)/J

A

Thus we have the


and the semi-exact sequence:

Ko(J) - ^ K0(A'(A®B))^K0(A/J). A*

Observing the six-term exact sequence:

K0(J) - ^ K0(A'{A®B)) - ^ K0(A'(A(BB)/J)

KX(A\A®B)IJ) *— K^A'iAQB)) — K^J)

It is easy to see that 3 = 0. Thus we get the exact sequence:

0^Ko(J) - ^ K0(A'(A®B)) - ^ tf0(„4/J)^o.

Note that K0{A'(A@B))^K0(A/J)=Z, therefore K0{J) = 0. This contradicts that K0 (J) ^ 0.

5.2 Cowen-Douglas Operators with Index n

L e m m a 5.2.1 Let AeBn{fl)n(SI), T = A{l)e£{H{l)) and let P be an idempotent in A'(T) satisfying T|pW(i)€(S'-r). Then A\ := T\PH(i) is similar to A. Proof Without loss of generality, we may assume that D. = D an n is the minimal index of A. For convenience we will prove the lemma only in the case of n = 2. Then T = A® A. Note that P is an idempotent in A'(T), by Theorem 4.4.3 we can find an idempotent P\€A'{T) and B£radA'(T) such that P(z) = P\(z) + B(z), where

and

B{z) =

fn(z) fn(z)

fn(z) /22O)

Bn(z)B12(z) B2i(z)B22(z)

ker(A — z) ker(A — z)

ker(A — z) ker(A — z)'

fijGH00 and Bij£radA'{A). Set G = -Inm + (2Pi + B). Since B€radA'(T), G is invertible in A'(T) and PG = GP^ This implies that G-1PG = P1eA'(T).


Without loss of generality, we may assume that P — P\, i.e.,

P{z) fn(z) h2{z) f2l(z) hl{z)

ker(A — z) ker(A — z)'

Set

P'(z) = fn(z) fi2(z)

. /2 l (z ) f22(z)

ker(Tz* — z) ker(Tz* — z)'

Since T\Pnm&{SI), tr(P'(z)) = 1 for all z£D. By the arguments similar to that used in the proof of Lemma 5.1.5, there exists an invertible element X(z) in M2(H°°) such that

and

X(z)P'(z)X-\z)

X(z)(I-P'(z))X-\z)

IcO 0 0

0 0 0 /c

Furthermore, X(z)\ranp>^ and X{z)\ran(j-pi(z)) are isomorphic bundle maps from ranP'(z) and ran(I — P'(z)) respectively onto ker(Tz* — z).

Set

X(z) = un(z) ui2(z) U2l(z) U22(z)

and

Then

X{z) = Ull(z)Iker(A-z) Ul2(z)her(A-z)

U2l{z)Iker(A-z) U22(z)Iker(A_z)

X(z)P(z)X(z) = her(A-z) 0 0 0

Note that X(z)ker(T - z) = ker(T - z). {Claim} G(z) = X(z)\ranp^ is an isomorphic bundle map from ranP(z) onto ker(A — z).

Note that G(z) = X(z)\ranpi^z-) is an isomorphic bundle map from ranP'(z) onto ker{Tz> — z).

Let e{z) be a holomorphic frame of ker{Tz* — z) and

ti(z) = e(z)®0, t2(z) = 0®e(z).

Then (£i(z), ^(z)) is a holomorphic frame of ker(T^,' — z). Set

Ai = Tz. |p/(z)W(2)

and Z(z) is a holomorphic frame of fcer(.Ai — z), then

i(z) = a(2)t1(z) + /8(z)t2(z),

where a(z),(3(z) are analytic functions in D. Since G(z) is a holomorphic isomorphic map, we can find a function

c(z) holomorphic in D such that

G(z)l(z) = c(z)e(z)

and

|K(z)||2 = (|a(z)|2 + |/3(z)|2)||e(z)|| = |c(z)|2||e(z)||

for zeD. Let (Si(z), • • • , Sn(z)) be a holomorphic frame of ker(A — z). Set

Vj(z) = Sj(z)eO, Uj{z) = OeSj(z), (j = 1, 2, • • • , n).

Then (fi(z), • • • ,vn(z),ui(z), • • • ,un(z)) is a holomorphic frame of ker(T— z).

Let fj(z) = a(z)vj(z) + /3(Z)UJ(Z), (j — 1,2,- • • ,n). Then (fx(z), • • • , fn(z)) is a holomorphic frame of ker(A\ — z). Set

V{z)fi(z) = c(z)Vj(z).

Let K\(z), • • • , Kn(z) be analytic functions in D and

g(z) = / f i(z)/ i(z) + • • • + Kn(z)fn(z)

= #i(z)(a(z)vi(z) + )8(z)ui(z)) + • • • + tfn(z)(a(z)t>n(z) + j8(z)un(z)).

Then G(z)<?(z) = c(z)(iT1(z)Vl(z) + • • • + tfn(z)un(z)) := g\z). Since < UJ(Z),WJ(Z) > = < Ui(z),Uj(z) >=< Si(z),Sj(z) > for z£D,

< g(z),g(z) >=t | ^ ( z ) | 2 ( | a ( z ) | 2 + |/?(z)|2)||Si(z)||2

t=i

+ t Ki{z)Kj{z){\a{z)\'> + \(3{z)\'>)<Si{z),Sj(z)>.

Furthermore,

<g'(z),g'{z) > = E |#,-(z)|2 |C(z)|2 | |Si(s)||a

+ E Ki(z)^-W|c(z) | 2 < ^ ( z ) ^ ) > .

This implies that ||G(.z)<7(.z)|| = ||ff(-z)|| and it verifies our claim. Similarly, we can prove that -^(z)|ran(/-P(z)) is also a holomorphic iso

morphism from ran{I — P(z)) onto ker(A — z). By Rigidity Theorem, we can find two isomorphisms

UitCiPH™,H®0) and U2££({I - P)t t ( 2 ) ,0©ft)

such that X = Ui + U2<=A'(T) and

X P X - l 0 0

This indicates that Ai~A and ends the proof of the lemma.

By Theorem 4.2.1 and Lemma 5.2.1, we get the following theorem.

Theorem 5.2.2 Let AeBn{n)r)(SI), then \J(A'{A))^N <™d

K0(A'(A))^Z.

Similar to the proof of Proposition 5.1.7, we have the following result.

Proposition 5.2.3 Let A,BeBn(fi)D(SI), then the following two statements are equivalent:

(i) A~B; (ii) K0(A'{A®B))^Z.

5.3 The Commutant of Cowen-Douglas Operators

n In this section, we always assume that T = 0 Tk, where TfcSBnfc (0fc)n(S7)

fc=i and \/ ker(Tk — z) = Tik- By the basic operator theory, we have the

following properties:

(5.3.1) A'(T) = {(Sij)nxn\Sij£kerTT T . , l<i, j<n} is a unital Banach algebra.


(5.3.2) kerrT, is a linear space and kerrT, T = A'iTi) is a unital Banach algebra.

(5.3.3) Denote eA,(T) the identity in A'(T). Then

eA'(T) eA'{T1)i

">t'(T„)-

(5.3.4) If Sij£kerrT. T. and Sjk&kerTT. T , then SijSjk£kerrT. T . (5.3.5) If (Sij^neA'iT), then

5(«,j)

o . . . o ••• o

0 • • • Sij ••• 0

o . . . o ••• o

£A'(T).

By (5.3.5), we can define a canonical map $ y : A'(T)—>kerrT. T. as follows:

(5.3.6) $„• is a linear map and $ i i(5)G^ /(T i) for all S G ,4'(T). In this section, J denotes a proper two-sided ideal. (5.3.7) Let J be an ideal of A'(T). Define

Uij — i*~^j • >Jij^.rCGTTrp_ rp_ clIlCl

o . . . o ••• o

0 • • • Sij ••• 0

0 • • • 0

eJ}

Then (5.3.7.1) Ju is an ideal of A'(Ti) or Ju = A'(Ti); (5.3.7.2) Jij is a subspace of kerrT. T.; (5 .3 .7 .3) S(i,j)&J for S = ( S y ) n x n G J .

By (5.3.7), we can define a canonical map from kerrT T. onto kerrT, T./$ij{J) as follows: 5jj—>[S'ij]l7, where kerrT, T./$ij(J) is the quotient of kerrT T. modulo Qij(J'). If i7 is closed, then

•A 'Cn/J - {([Sii]j)nxn •• SijGkerr^.}

is a unital Banach algebra. Thus we have a canonical map

$j : A'(T) —> A'(T)/J

as follows:

®j((Sij)nxn) = ([Sij]j)nxn-

(5.3.8) Let J be a closed ideal of A'(T). If

([Sii]j)nxn = *j(S)€A'(T)/J,

then

• Q . . . 0 • • • 0 _

0 • • • [Sii\j • • • 0

o . . . o ••• o

$j(S(i,mA'(T)/J.

Lemma 5.3.1(Lifting Lemma) Let T = © Tk and J\ he an ideal fc=i

of A'(Ti), then there exists an ideal J of A'{T) such that §i\(J) = J\. Furthermore, if there is another ideal J1 of A'{T) such that $n(»7') = J\, then JQ-J', where <J?n is given in (5.3.5). Proof Set

X = {

and

Rl R^D I : Ri£Ji,i = 1 , 2 , 3 , 4 ; B « , A i j € k e r T , i = 1,2}, A21K3 B21H4B12] " 3

J = {xi + X2 H h xn : \<n < 00, a^Gx}-

{Claim} J is an ideal of A'{T). Clearly, J is an additive group. Set

W =

Then

Wn W12

W21 W22 &A'{T) and X =

Ri R2A12 A21R3 B21R4B12 ex-

WnRi {WllR2)Al2 W21R1 W21R2A 12 +

(W12A2i)R3 (W12B21R4)B12

(W22A2i)R3 (W22B21)R4B12

Since

WuRi (WuR2)A12

W21R1 W21R2A 12

(W12A21)R3 (W12B21R4)B12

(W22A2l)R3 (W22B21)R4B12 ex,

WXeJ. Similarly, XWeJ. Thus J is an ideal of A'(T). Since $n(X)€j1 for all XGJ, eA, $ J. Therefore J" is a proper ideal of A'{T) and $ n ( J ) = J i . If $n(J') = J i for another ideal J' of A'(T), by (5.3.4) and (5.3.7), JcJ'.

Corollary 5.3.2 Let T = 0 Tk and J i be an ideal J of A'{Ti), then fc=i

there exists an ideal J of A'{T) such that $n(J") = J\. Furthermore, if $11 (J') = J i for another ideal J' of A'(T), then JcJ'.

Corollary 5.3.3 Let T = 0 Tk and J<=M(A'{T)), then $kk(J) =

.4'(Tfc) or $kk{J)£M{A'{Tk))7k = 1,2, • • • ,n.

Lemma 5.3.4 Let T = 0 Tk and S = (Sij)nxneA'(T). If for each fc=i

Rji£kerrT, T., RjiSij = 0, £/ien S(i, j)£radA' (T). Proof For each

R

Rn Rln

tlnl ' ' ' *Mi

RS(i,j) =

0 ••• RuSij ••• 0

0 • • • RjiSij • • • 0

0 • • • Knibij • • • 0

Since i^SV,- = 0, (RS(i,j))n = 0. This implies that 5(z, j)£radA'(T).

Corollary 5.3.5 LetT= 0 Tk, then

§kk{radA'{T)) = radA'(Tk), k = 1,2, • • • , n.

Corollary 5.3.6 Let T = ® Tk, JGM(A'(T)) and Si:i€kerTT, T./

®ij{J)- IfSijrji = 0 for all rêker-r^ T . / $ , j ( J ) , then Si:j = 0.

n Theorem 5.3.7 Let T = © Tk, then for each JeM(A'(T)), there is a

fc=i positive integer I j<n such that A'(T)/J=Mij(C). Furthermore, ifTk~Ti for k = 1,2, • • • ,n, then A'(T)/J^Mn(C) for all JGM(A'(T)).

Proof By Corollary 5.3.3,

$kk(J) = A'(Tk)

or

$kk(J)eM(A'(Tk)),k = 1,2,-•• ,n.

By Theorem 4.4.3, A'(Tk)/radA'(Tk) is commutative for k = 1,2, • • • ,n. Thus A'(Tk)/$kk(J)^C or A'(Tk)/$kk{J) = {0},fc = l ,2 , - - - ,n. With-out loss of generality, we may assume that there exists an integer lj<n such that

and

Thus,

and

A'(Tk)/$kk(J)^C,k = 1,2,-•• ,lj,

A\Tk)/$kk{J) = {0}, k = lj + l,---,n.

A'(T)/J = { ( [Sy] j ) n x „ : Sêkerr^.

[Skk] =0,h<k<n}.

By (5.3.4), 0 = [S^R^j = [S^jSA'^/îJ), where SijekerTT. T,, Rji£kerrT. Ti and / j - < i<n. By Corollary 5.3.6, [S'yjj = [Rijjj = 0 for Ij < i<n. Therefore / J - > 1 and

A'(T)/J={ ({Sij])ljxlj ° 0 0

I OijKzK&TTrp. rp_

{Claim 1} For l<i,j,k<lj, if

feerrTiiTi/$«(J)^{0}

and

kerTTjTJ$jk(J)^{0},

then

kerTTiTk/$ik(J) ^ {0}.

Note that > l ' (T i ) /$«(J )SC and ^ ' ( ^ / ^ - ( J ^ C , l<i,j<lj. If

fcerTTi>Ti/*o-(J)^{0},

by Corollary 5.3.6, there exist Sij&kerTTiTJ$ij(J) and Sji€kerTT. Ti/$ji{J) such that SySji = [e^>(Ti)]j = 1-

Similarly, there exist Sjk£kerrT. T /$jk(J) and Skj&kerrT iT./$kj(J) such that SjkSkj = [e '(Tfc)]>7 = 1- Thus, SijSjk ^ 0 and

{Claim 2} $U(J) ± kerrTiT, for l<i<lj. Otherwise we may assume that there is a jo '• l<io < lj such that

kerTTitTi/$u(J) ^ {0}, l<i<j0

and / c e r r T i | T i / * „ ( J ) = {0}, jo < j < / ^ .

By Claim 1, we have

kerTTj,Tj$ji(J) = {0}- l<«<7'o,jo < j< / j - .

By Corollary 5.3.6,

kerT^^./^jiiJ) = {°}» 1<*<J0, Jo < J < / j .

Thus

•A'CH/J-j ^og(([5y]j)J0xjo> ([sij}jhj-3oKh-Jo)) ° 0 0 '• Sij€'ier"rTi,Tj

This contradicts J G X ( ^ ' ( T ) ) .

{Claim 3} M{T)/J^Mls{C). For l<i<lj, denote en = [eA'(Ti)} = 1- By Claim 1 and Claim

2, there are eu£kerrT TJ$u(J) and en£kerTT.Ti/$ii(J) such that

eij-eji = en,l<i<lj. Since A'{Ti)/$u(J)=C,eiieu = en. Assume that Cy = e-ne-ij, then

&ij&ji ~ ei\BxjCjê\i — C 2'

and

ji ij — ej\C\i€-i\&\j — Gjj.

Since for each Sij£kerTT, T./$ij(J) (l<i,j<lj) there exists A^GC such that

ij = = îjîj

and since

— Xîj&ji "ijîijîj

— \ ij îj Jîiîj = *-M

Sij = Xijeij. Thus the first part of the proof of Theorem 5.3.7 is now complete.

lfTk~Tuk = 1,2, • • • ,n, then A'(Tk)/$kk(J)^C for all J e X CA'(T)). This completes the proof of the theorem.

Theorem 5.3.7 implies the following properties.

(5.3.9) If A'{TM$ii(J) ? {0} and A'^/^^J) ± {0} for JeM(A'(T)), then

kerTTiiT. l*ijtf)^kerTTiiTi /î(J)=C

(5.3.10) lfA'(Ti)/$ii(J) = {0} for some i: l<i<n, and J e J W ^ T ) ) , then

kerTT.iTJ$ik(J) = kerTTkTJ$ki(J) = {0}, fc = 1,2, • • • ,n.

Theorem 5.3.8 LetT = ® Tfc, t/ien /or eoc/i Ji£M(A'(Ti)), there is a fc=i

unique JeM{A'{T)) such that $ U ( J ) = Jx. Proof We only prove the theorem when n = 2 and T = Ti®T%-The general case can be proved similarly. By Lemma 5.3.1, there is an ideal J0 of A'(T) such that $u(J0) = Ji. Set J' = J0 + radA'{T).

Then J' is still an ideal of A'(T) and $n(J') = Jx. By Corollary 5.3.6, radA' (T2)C.§22{J'')- Therefore we may assume that Jo = J ' , ^ ' ( T i ) / $ i 1 ( J 0 ) = C and A'{T2) / $22{JQ) is semisimple. By Theorem 4.4.3, A'{Tk)/radA'{Tk) is commutative, k = 1,2.

Note that

A'(T)/J0 = { Sll S\2

S'21 'S'22 Sij£kerTT,TJ<f>ij(Jo), l<i,j<2}.

Denote ekk = [eA,{Th)]j,k = 1,2. Since A'^/^uiJo^C, en = 1. {Case 1} Assume that there are

ei2€kerTTiiT2/$i2(J0), e2i€kerTT2iTJ$2i{Jo)

such that ei2e2i = 1. Set Qi = e2iei2 and Q2 = e22 — Qi, then Qi and Q2 are idempotents in A'(T2)/$22(J0) and Q1Q2 = Q2Q1 = 0. Let

A' = { [ | ; Q^22 2 : Syekerr^/S^Jo), l<i,j<2}

and

.4" = { 0 0

0 Q2S22 : S22£A'(T2)/$22(Jo)}-

{Claim 1} A'(T)/J0 = A'®A". It is obvious that for each S = (Sij)2x2&A'{T)/J0,

5*ii S'12

S21 Q1S22 +

where

S n S12

S21 Q1S22

0 0 0 Q2S22

GA'

and

For

0 0 0 g2s22

eA".

S n S12

S'21 Q1S22 eA'

and

0 0 o g2522

eA",

we have that

tr = 0 S12Q2S22

0 0 rt = 0 0

Q2S22S12 0

To verify Claim 1, we need only to show that S12Q2 — Q2S21 — 0. For arbitrary S^e/cerr^ T 2 /$ i 2 ( Jo) and S2iekerTT2 Tl/$2i(Jo), we

can find a A € C such that S ^ f ^ e n = Aei2e2i. By (5.3.4), we have (S12Q2 - Aei2)ei2 = 0 and <r(e2i(5i2<32 - Aei2)) = {0}.

Since A'(T2)/$2:2 (i7o) is semisimple and commutative,

e2i(<Si2<22 - Aei2) = 0 .

This implies that (e2iSi2)<52 = AQi and therefore A = 0 and e2iSi2 = 0. Thus,

S12Q2 = eu(S12Q2) = e12e2i(5'i2<52) = ei2(e2i512(52) = 0.

Similarly, we can show that Q2S12 = 0. Thus rt = tr = 0 and this proves Claim 1. {Claim 2} A'^M2(C).

Let

A, = {Sn : SneA'inyQuiJo)}

and

A = {Ql522 : 522e>4'(r2)/*22(Jo)}.

Note that *4i=C. We define a map 0 : ^2— y Ai as follows:

0(6) = ei26e2i for all 6e.42-

Clearly, 0 is a homomorphism. Since <t>{Q\) = en = 1,0 is surjective. If 0(6) = 0(6'), then e2i0(6)ei2 = e2i<t>(V)el2. Thus QxbQx = Qib'Qy

and A2—A1=C. Similar to the proof of Theorem 5.3.7, we can get •4'=M2(C). Now we define a surjective map n : A'{T) —> A' as follows:

4 ( ^ ) 2 X 2 ) - [ ( [ 5 2 l ] U ( [5 2 2 ]U

Then n is a homomorphism. Since A'=M2(C), J = kern <E M{A'{T)) and

* n ( J ) = J i .

{Case 2} If there is no ei2£kerrTi T2/$i2(Jo) and e2i£kerTT2Ti/$2i(J0)

such that e12e2i = en = 1. Since A'(Ti)/<frii(J0)=C, for each Si2£kerTTi T2 /$i2(J0) and each S21£kerrT2]Tl/$2i(Jo) we have S12S21 = 0. By Lemma 5.3.4,

fcerrTir2/$12(Jo) = {0}

and

*erT^ i T l /*2i(Jo) = {0}.

Thus

A\T)/J0^A\Tl)/§u{JQ)®A'{T2)/<5>22{J0)^C@A'{T2)/<5>22{Jo).

Similar to the proof in Case 1, we can find a JeM(A'(T)) such that

Now we prove the uniqueness. Suppose that there are J and J'£M(A'(T)) such that $n(J) = $ n ( J ' ) = J i . Let

J = J + J' = {S + S' : SeJ, S'GJ'}.

Then J is an ideal of A'(T). Since $n(~J) = Jlt J = J' = ~J•

Theorem 5.3.9 LetT = 0 Tk, then for each SGA'(T), fc=i

a(S)= (J ^ ( S ) ) , j e M ( T )

u//iere $ j - is a canonical map: A'(T)—>A'(T)/J,a($j(S)) is the spectrum of$j(S)inA'(T)/J. Proof We only prove the theorem when n = 2 and T = T\®T2. The general case can be proved similarly.

If ^ ' ( T 1 ) / $ l x ( l 7 ) S C and A'(T2)/$22(J) = {0} for every JeM(A'(T)), then

*J(S) = Aen 0

0 0 , AeC (see Theorem 5.3.8)


If Al{T1)/$n(J)^A'(T2)/*22(J)2iC, then

where ei2e2i = en,e2iei2 = e22- This means that

A l i e n Ai2ei2

A21C21 A22C22

|J o-($j(S))ca(S). J£M(T)

If Xea(S), consider the maximal two-sided ideal J generated by (MH®H - S) in A'(T), clearly \£a($j(S)) and this completes the proof of the theorem.

2 Lemma 5.3.10 Let T = 0 Tfc, then the following are equivalent:

fc=i (i) There exist a positive integer n and Xi€kerrT T , yi£kerrT T ,

n where i = 1,2, • • • , n , such that J2 Xit/i = IA'(T)>

i = l (ii) There exists an idempotent e£M„(A'(T2)) such that

-f^'(r1)©°~a0®e in * Mn(A'(T2))\

Proof (i)=»(ii). Let

'y\

Vn

[xi ••• xn] =

yixi • • • y„xi

_Vn^l ' ' ' yn%n _

&Mn{A'{T2)).

By (i),

2/1

.Vn

[xi ••• xn]

2/1

Vn.

[xi %n — e.

Now set

0 [zi ••• xn] 0 0

( n + l ) x ( n + l )

0

~yi'

.Vn.

0"

0

( n + l ) x ( n + l )

Then

and

uv * Mn(A'(T2))

JU'(r,)©0 =

This implies that 7^1©0~a0©e in

vu.

•4'(Zi) * * M„M'(T2))

(ii)=>(i). If IAli 3e in

'^'(Ti) * * M„(^'(r2))

then by the basic properties of if-theory, we find

u,v€ A'(T!) * * M2(A'(T2))

such that /^'(T!)©0 = uv and 0©e = vu, and

« = (^ ' ( roMOQeJ. t ; = {0®e)v(IA.(Tl)).

n Since /A- (TI )©0 = uv, J2 xiVi = ^Ai-

i = i

Proposition 5.3.11 Lei T = A^ and {Pi,--- , P m } be an (SI) decomposition, then m — l and Ai = A^\p-^(i)&Bn(fl). Proof We first show that m<l. By Theorem 4.4.3, A'(A)/rodA'(A) is commutative. It follows from Gelfand Theorem that there exists a continuous natural homomorphism tp : A'(A)—*C(M(A'(A))) and (s)(J) = (<P('ij)(J))ixh

where 5 = (Sij)lxieA'(A),JeM(A'(A)). Set

Pk = (Plkj)ixi,k = 1,2, •••,m.

Then

1>(Pk)(J) = MP&iJVixi-

I Denote tr{ip{Pk){J)) := £ <p(P£)(J). Thus ir(-) defines a continu

a l ous function on M{A'{A)). Since A'(A)/radA'(A) is commutative and J4G(S7),M{A'(A)) is connected to Proposition 1.17 of [Jiang, C.L. and Wang, Z.Y. (1996a)]. Since ip(Pk)(J) is an idempotent, tr(ip(Pk)(J)) =

m m nk>l. Note that £ pk = / a n d PkPk> = Skk,Pk, thus £ tr(iP(Pk)(J)) = I

fc=i fc=i m m

or £ tr(*l,(Pk){J)) = £ nfc = /. Therefore m<Z. fc=i fc=i Now we prove that yliGi?n(^)- Otherwise, assume that AiGBk(Q,), and

k < n. Let 5 = ASAi. Since k < n, a simple calculation indicates that

0 kerr. 7 1 1 *

.keTTA2,Al °

By the arguments similar to that used in the proof of Theorem 5.3.7, we can find Ji€M(A'(S)) such that A'(S)/J1^C. Set

T1=A®T = AV+1).

By Theorem 5.3.7 A'(T)/j£*Ml+1(C) for all JeM(A'(Ti)). Note that Ti=A®Ai®---®Am and m<L Repeating the proof of Theorem 5.3.7 and using Lemma 5.3.10 we can find J2&M.{A'{Tx)) such that A'{Ti)/J2=Md(C), where d<l + l. This contradicts -4'(Ti)/ 'J=M l+ l(C) for each j7e.M(.4'(Ti)). Since every Ai£Bn(Q.),m = I.

5.4 The Commutant of a Classes of Operators

FIR algebra. Let A be a Banach algebra, n is a representation of A on a Banach space X, dimX>l, if 7r is a nontrivial continuous homomorphism from A onto C{7i). If a linear subspace y of X satisfies ir(a)ycy for all a£A, y is said to be an invariant subspace of ir{A). A representation ir is said to be irreducible if a subspace y of X satisfying n(a)ycy for all a^A is either y = {0} or y = X. An ideal J'cA is called a Primitive ideal of A if J = kern, when 7r is an irreducible representation of A.

Definition 5.4.1 A Banach algebra A is called an FIR algebra if for every irreducible representation ir : A—>£(X),TT(A) is finite dimensional, i.e., dimX < oo. A Banach algebra A is said to be n-homogeneous if there

GA'(S).

is a natural number n such that TX(A)=M„(C), where n is a irreducible representation of A.

Definition 5.4.2 Let A be a unital Banach algebra, A is said to be essentially commutative if A/radA is commutative.

Proposi t ion 5.4.3 The com/mutant of every strongly irreducible Cowen-Douglas operator is essentially commutative.

Definition 5.4.4 Let AC(7i). A is called typical strongly irreducible operator AG(SI) and A is essentially commutative. A is called a type-1 operator if A has a finite (SI) decomposition (Pi,P2,--- ,Pn) and each A\PiH *s a typical (SI) operator.

Proposi t ion 5.4.5 Every Cowen-Douglas operator is a type-1 operator.

By Gelfand Theorem, we know that an essentially commutative Banach algebra is 1-homogeneous. On the other hand, every 1-homogeneous algebra is essentially commutative. In fact, we have the following stronger result.

Proposi t ion 5.4.6 Let A be a unital Banach algebra and let x(-4) be the set of all nonzero multiplicative linear functionals. For x£A, r(x) denotes the spectral radius of x. Then the following are equivalent:

(i) A is essentially commutative; (ii) A is 1-homogeneous; (iii) a(x) = {<p(x) : ip&x(A)} for each x£A; (iv) o~(xyz) = a(xzy) for x, y, z£A.

Proof (i)=*>(ii). It follows from Gelfand Theory. (ii)=^(iii). First we consider A = A/radA. For xGradA, <p(x) = 0 for

all ip£x(A). Thus x(A) = x(-4)- Since a(x + radA) = a(x), using Gelfand Theory again, (ii)=^(iii).

(iii)=>(iv). For arbitrary x, y, z£A,

a(xyz) = {p(xyz) : ipGx(A)}

= {<p(x)<p{y)(p(z) : ipex(A)}

= {<p(x)<p(z)<p(y) • p£x(A)}

= W(xzy) '• <P£X(A)} = o(xzy).

(iv)=>(i). Given x,y£A, cr(xy) = a(yx).

Note that for each n > l and 0<fc<n — 1,

a({xy)n~kxkyk) = a((xy)n~k~1 xyxkyk) = a{yk{xy)n~k~lxyxk)

= a(yk{xy)n-k-1xk+1y) = a{(xy)n-k-1xk+lyk+1).

This implies that cr((xy)n) = a(xnyn). Thus

r(x,y) = (r((xy)n))i = (r(xnyn))± = | |*n | |*| |y| |*.

Let n—>oo, we have r(xy)<r(x)r(y). It follows from the Corollary 5.2.3 of [Aupetit, B. (1991)] that A is essentially commutative.

Free matrix algebra. Let Xi,-- • ,Xn be Banach spaces and X = X\@---®Xn be the product Banach space. For each i, let || • ||* be the norm oiXi. Then the norm ||-|| of X is denned by ||x|| = max {||xi||j}, wherex =

l < i < n {xi, • • • , xn}eX. Let SeC(X), then 5 = (Sjj)„x„), where Sij£C(Xi,Xj).

Definition 5.4.7 Let X = Xi@ • • • ®Xn be the product Banach space of {Ai}™=1. An algebra AcC(X) is called a free matrix algebra of order n if S = (Sij)nxn£A implies that

S(i,j) :=

"0 ••

0 ••

. 0 • •

• 0 •

• &ij •

• 0 •

• • 0 "

• 0

• o.

6.4;

A is called a unital free matrix algebra of order n if I, the identity on X, is in A.

Let AcB(X) be a free matrix algebra of order n. Denote

•sxij — l ^ i j •

0 • • • 0 • • • 0

0 • • • Si.

0 0 • • • 0

GA}

and Ai = An, then it has the following properties: (5.4.1) Ai is a Banach algebra and A is unital if and only if each Ai is

unital, l<i<n. (5.4.2) Aij is a commutative additive group for l<i,j<n.

(5.4.3) If SijGAij and Sjk&Ajk, then SijSjk&Aik- Particularly, if Sij&Aij and Sji^Aji, then SijSjiGAi.

Definition 5.4.8 The map 7Tjj : A —> Aij is called an entry map if

Ttij \\&ij Jnxn) == ^ij

for {Sij)nxn£A.

In most cases, we take great care of the diagonal elements Ai's in a free matrix algebra. So we denote the free matrix algebra A of order n with the diagonal elements A\, • • • , An by A~diag(Ai, ••• , An)-

Example 5.4.9 Let Hi be a separable Hilbert space, l<i<n, and H =

© Hi. Let Ti£jO.(Hi) andT = © Ti&C(H). Then A = A'(T) is a typical

free matrix algebra, where Ai = A'(Ti),Aij = kerrT_ T . .

Definition 5.4.10 Let A be a unital Banach algebra. Two idempotents e, feA are called orthogonal, denoted by e_L/, if ef = fe — 0. A set of elements {e i}^1c^4; n < +oo is called a finite decomposition of A if et±ej for l<i ^ j < n and

ei H he„ = 1.

Example 5.4.11 Let {e\, • • • ,em}cA be a finite decomposition of A. Define

M := {(Sij)nxn '• Sij = eiSej,SeA},

Xi = {eiSei : SeA} \<i<n.

Then Xi *s o, Banach space and Ai is a free matrix algebra of order n,Ai£C(X), acting on X = Xi® • ••©<¥„.

Define a map <f>: A—*Ai by

<f>(S) = (Sij)nxn, SSA,

then 4> is a continuous isomorphism.

Lemma 5.4.12 LetA~diag(A\,A2,---,An)andJ'cAbeanideal. Then J is a free matrix algebra and 3~diag{J\,Ji,-- • ,Jn)- Furthermore, either Ji = Ai or Ji is an ideal of Ai, l<i<n. Proof For S = {Sij)nXneJ,S(i,j) = IA{i,J)SlA{iJ)eJi, where S(i,j) is defined in Definition 5.4.7. Thus J is a free matrix algebra of order n.

Since A is a free matrix algebra, a simple computation shows that either Ji = Ai or Ji is an ideal of Ai, l<i<n.

Lemma 5.4.13 Let A~diag(A\, A2, ••• , An) be a unital free matrix algebra of order n acting on X = X\@ • • • ®Xn. If A is an irreducible subalgebra of C(X), then Ai is an irreducible subalgebra of C{Xi),l<i<n. Proof For l<i<n, since A is an irreducible subalgebra of C(X), for arbitrary x,y€zX there exists (Sij)nxnQ.A such that (Sij)nxn x = y. A simple computation shows that

OiiXi ^ IJit V Xi^yiGzA-i.

Thus Ai is an irreducible subalgebra of £(Xi), l<i<n.

Definition 5.4.14 Let

A~diag(Ai, A2, • • • , An)

and

B~diag(B1,B2,-..,Bn)

be two free matrix algebras of order n. The map <j> : A —> B is called a freely matrical morphism if

(i) 4>(A)CB; (ii) (friaSx + j3S2) = a</>(Si) + /fy(S2) for SUS2€A and a, /?eC; (hi) ^(5iS2) = </»(51)<A(52) for S1,S2eA; (iv) If <f>(S) = T, then 4>(S(i,j)) = T(i,j).

If in addition 4>{IA) = LB, then <j> is called a freely matrical homomor-phism.

If 4> is a freely matrical morphism from A to B, then 4> induces a morphism from Aij to B^. In general, we will not distinguish <f> with the morphism induced by it from Aij to B^.

Lemma 5.4.15 Let A be a unital free matrix algebra of order n and JcA be a closed ideal of A. Then A/J is a unital free matrix algebra of order n. Furthermore, there is a freely matrical morphism (j>j from A to A/J. Proof Let <f>j be the canonical map from A to A/J. Denote ei = I^.(i, i) and ft = 4>j{ei) for l<i<n. Then {/1, • • • , fn}cA/'J is a decomposition of Aj' J• By Example 5.4.11, A/J is a free matrix algebra of order n. It is easily seen that <j>j is a freely matrical morphism from A to Aj J.

Theorem 5.4.16 Let A~diag(Ai, A2, • • • , An), then A is an FIR algebra if and only if each Ai is an FIR algebra. Proof "4=" Suppose that -K is a continuous irreducible representation of A acting on a Banach space X. Let Pi = n(ei), then n(A) is a uni-tal free matrix algebra and -K(A)~diag(Tr(Ai),Tr(A2), • • • ,n(An)) acts on X = P\X® • • • ®PnX. By Lemma 5.4.13, every n(Ai) is a unital irreducible algebra, ir(Ai)c£(PiX) or PiX = 0, l<z<n. Since Ai is FIR for l<i<n,dimPiX < 00. Thus

n

dimX = y^dimPjX < 00. x=i

"=*>" Suppose that n is a continuous irreducible representation of Ai and J\ = kern. By Kaplansky [Bonsall, F.F. and Duncan, J. (1973)], there exists a unique primitive ideal JdA such that J\ — Jf\A\. Since A is FIR, A/J is a finite dimensional algebra and Ai/Ji is a subalgebra of A/J. Thus A\/J\ is finite dimensional and Ai is FIR. Similarly, we can prove that Ai is FIR for all i.

By Example 5.4.11, we can restate Theorem 5.4.16 as follows.

Theorem 5.4.16' Let Abe a unital Banach algebra and

{ei,e2 ,--- ,en}cA

be a decomposition of A. Denote Ai = eiAei, l<i<n. Then A is FIR if and only if Ai is FIR for all i.

An FIR algebra ,4 is said to be stable finite if Mn(A) is FIR for each natural number n.

Corollary 5.4.17 Let A be a unital FIR algebra, then for each natural number, Mn(A) is FIR. Proof This is a straightforward corollary of Theorem 5.4.16'.

Similar to the discussion in Section 5.3, we can get the following result n

about the type-1 operator T = 0 Tk-fc=i

Theorem 5.4.18 Let Ji&M(A'(Ti)), then there exists a unique

j£Ai(A'{T)) such that nn(J') = J\, where TTU is defined in Section 5.3.

Theorem 5.4.19 Let T = T\®T2 be a type-1 operator. If Ji is a subalgebra generated by kerrT T and kerrT T , then

(i) A'(T) is 1-homogeneous if and only if J\CradA'(Ti). Moreover, if A'(T) is semisimple, i.e., radA = {0}, then A'(T) is 1-homogeneous if and only if

A'(T)*A'(T1)®A'(T2).

(ii) A'(T) is 2-homogeneous if and only if there exist a positive integer n and Xi£kerrT T , yi£kerrT T ,l<i<n, such that

n

Y^xiVi =lA'(T1) i=\

and n

Y^Vixi = lA'(T2) + R, i=l

n where R&adA'iT^)- Furthermore, if A'(T) is semisimple, then Ylyixi ~

i=l -U'(T2)-Proof The first part of the theorem is obvious. We need only to show the second part.

Suppose that A'(T) is 2-homogeneous. {Claim} There exist a positive integer n and XiGkerrT T ,

n yi£kerrT Ti,l<i<n, such that £) x%Vi — IA'{TX)- Otherwise, J\ gener-ated by kerrTiT and kerrT T is an ideal of A'{T\). By Kaplansky Theorem, there is a JeM(A'(T)) such that nn(J) = J\Dj\. Thus for arbitrary x£kerrT T and y€kerrT , xyGiru(J'). Then it follows from Corollary 5.3.6, kerrT T2/iri2(J) = kerrT2T /iT2i(J) = {0}. This implies that A'{T)/J^C. A contradiction.

n Therefore, there exist Xi€kerrTiT2 and yi£kerrT2T such that ^ xtyt =

n -TA'(TI)- Note that £2 Vixi ~ ^A'(T2) *s a nilpotent and idempotent, thus

i=i

n

^yiXi - IA^T2)£radA'(T2) [Antonevich, A. and Krupmk, N. (2000)]. j=i

Conversely, if there exist a positive integer n and Xi£kerrT T , n n

ytekerr , l<i<n, such that £ xtyi = IA'^) and £ yiX{ - IA>(T2) + i=l i=l

R, R£radA'(T2). Similar to the Proof of Theorem 5.3.7, we can get A'(T)/J^M2(C) for all JeM(A'(T)). Thus A'(T) is 2-homogeneous.

5.5 The (SI) Representation Theorem of Cowen-Douglas Operators

In this section, we always assume that T££(H) is a Cowen-Douglas operator with index n.

Lemma 5.5.1 Let TeB„(Q) and T = A@B, T = A+D be two decompositions ofT, then D~B. Proof We may find three idempotents PA,PB,PD&A'(T) such that

-* \ranPA ~ -**> -*• \ranPs — &•> ± \ranPo = ^ t

then PA + PB = I, PA + PD = I- Thus PD is can be regarded as an invertible operator from ranPg to ranPo-

Suppose AeBm(Q), then B,DeBn^m{n). Let (ef(A),--- ,e£(A)) be the holomorphic frame of ker(A — A) and (/^(A),--- ,/,f_TO(A)) be the holomorphic frame of ker(B - A). Then (PDff(X), • • • , Pofn-mW) i s t n e

holomorphic frame of ker(D — A). Denote {PD\ranPB)~X = XD, then

XDDPDf?(\) = \XDPDf*{\) = A/f(A) = Bf?(\).

So B~D.

Lemma 5.5.2 Let

T = A(™l)®A2m2)®---®A{™k\ Aie(SI)nBni{£l), i = l,2,---,k

and Ai^Aj for i ^ j , then M {A'(T))^N^ if and only if k

V M ' ( ® A ))=N ( fc ), where {mi, • • • ,mk} and {ni, • • • ,nk} are two or-i=l

dered sets of positive integers. Proof We need only prove the necessity of the lemma. By Theorem

4.2.1, \/(-4'(^))=N ( fc ) implies that 0 A\mmi) has a unique finite {SI) 2 = 1

decomposition up to similarity, where m = YLnni- Set T\ = 0 A\" , i=l i=l

k k then T[n) = ® ^ n n i ) . Since mmi>nnu l<i<k and 0 ^ m m , ) =

i=l i=l

Ti ( n )e © ^ m m , ~ n n , ) . By Theorem 4.2.1 again, T ^ has a unique (SI)

decomposition up to similarity and

V(i'(®AK)))=N(fc».

»=i

Remark 4.5.3 By Theorem 4.2.1 and Lemma 5.5.2, if Ai£(SI), l<i<k k

and Ai'/'Aj (l<i ^ j<k), then ( © j4j" i ;)(n) has a unique (SI) decomposi-i = l

it tion up to similarity if and only if ( 0 Ai)^ has a unique (SI) decompo-

i = i sition up to similarity. Lemma 5.5.4 Let A be a finite irreducible algebra and JQA be a closed ideal and 0 —> J -^ A —> A/J —> 0 fee an exact sequence. If \J{A)=N and \I_A\ = 1, t/ien the induced map

nt : K0(A) -» tf0(-A/J)

is injective. Proof Suppose that n is a natural number and p, q£Mn(A) are two idempotents. Since \J(A)=N, [p] = [er] and [q] = [es], where efc = diag(I^,--- ,1^, 0, •••) with k Ij^'s in the diagonal for k = r,s. If 7T*([p]) = 7r»([g]), then [7r(er)] = [7r(es)]. Since A is FIR, A/J is FIR. Moreover, ^4/v7 is stably finite by Corollary 5.4.17, thus r = s and [p] = [q]. This proves that ir» : /^o(»4) —> KQ(A/J) is injective.

Lemma 5.5.5 Lei .4 6e o unital FIR algebra and J\ ^ J2 be two maximal ideals of A. Denote J = Jif)J2, then A/J=A/Ji®A/J2. Proof Suppose that fa be a quotient map from A to A/Ji, i = 1,2. Define 0 : A—>A/Ji®A/J2 as follows:

<f>(a) = fa(a)®fa(a), aeA.

Then <f> is a homomorphism. By Chinese Remainder Theorem, </> is surjec-tive. Since ker<j> = kerfankerfa = JiC\J2 = J, A/J=A/Ji®A/J2.

Lemma 5.5.6 Let Ai (i = 1,2, • • • , fc) 6e strongly irreducible operators and Ai'/'Aj for l<i y^ j<k. Let {ni,n2, • • • ,rik} be an ordered set of positive

integers and T = ® A\n ®B be a Cowen-Douglas operator. Denote j = i

H > 51 = 4ni)©5,52 = 0 ^ i=2

i.e., T = S1®S2. If\/(A'{S2))^N(k-1\ then

n

J\ = {^XiVi, Xi&kerTsus2, yi&kerTs2,Si, ±<i<n, n = 1,2, •••} i = l

is a proper ideal of A'(Si). Proof If J\ = A'(Si), then there exist

such that

So

xi,x2,--- ,xnekerTSus2, 2/i, 2/2, - - - ,yn^kerTs2,s1

zi2/i H 1- a;„j/n = /^'(sx)-

P =

y i

. ? / « .

[n-- - xn]eMn(A'(S2))

is an idempotent. By the similar argument used in the proof of Lemma 5.3.10, we have l ^ ( S l )®0~ a 0©P in A'iS^S^). Let S i € £ ( £ i ) , S^n)eC(fC2). Then by Lemma 4.2.4,

A[n)®B = Si

= (5i©5^" )|(/_4/(Si)©o)(AC1e/c2)~(5'i®5'2n )l(o©P)(/c1ffi/c2) = - V IPK:2

Since V(<4'(S'2))=iV(A:_1), it follows from Theorem 4.2.1 that S<n) has a unique (SI) decomposition up to similarity. Since A\~ ®B~S2 \PK2, AI~AJ, where 2<j<k. This contradicts to our assumption that Ai / Aj for lA/J, 7Ti : Ai—>A\/J\

are quotient maps and A/J={Ai/J\, 0, • • • , 0), then

Proof Define a* : ^\*{KQ{A\))—*-K„{KQ{A)) as follows

a*(7ri*([e])) = 7r„([e©0©- • •©()]) for all e&Mk(Ai).

We first show that a* is injective. If 7r*([e©0© • • • ©0]) = 0, then

7r»(e©0©---ffi0)~s0.

Thus there exists r such that 7r,(e©0ffi- • • ©0)©r~a0©r. Then

7ri*(e)©°© • • • ©0©r~a0©r.

Note that 0©r~ a r for each r. Let r' = 0© • • • ©0©r, then

7ri*(e)ffir'~a0©r'.

This shows that 7Ti*(e)~s0. Therefore, [7i"i*(e)] = 5Ti*([e]) = 0.

Remark For unital Banach algebra A, let p,q£Mcc{A) be two idempo-tents, we say p~3q if there exists an idempotent r in Moo (A.) such that p<$r~aq®r.

Second, we will prove that a* is surjective. It follows from A/J=A\/J\ that for every (3£Ko(A), there is an e such that

*l.(H)=7T.([/3]).

Similarly, for (Pij)nxn€Mn(Ko(A)), w e have \eij]nxn

such that

*"i*([eij]) = 7r* ([/%])•

By the basic K-theory, we have

[(7r*(e i ; ;-)®0---e0)nX„] = [7T,((ey)nxn)©0].

Thus a» is surjective and so a , is an isomorphism.

Lemma 5.5.8 Let T = Ax®A2, where A\ and Ai are strongly irreducible Cowen-Douglas operators and Ax^Ai- Suppose that n is a positive integer and

is another finite decomposition of T^n\ where mi,m2,m>0,Bi£(SI) and Bi'/'Aj for \<i<m and j = 1,2. Then mi + m = n,i = 1,2. Proof Since T = AX®A2, T^ = A^eA^K

{Claim 1} m,i + m<n, i = 1,2. Otherwise assume that mi + m > n. Let B = £ i©S 2 ® • • • © S m , R = T{n^®A2, then

A'(R) = A'{TW) fcerrr(„,iAa

Suppose that J is a subalgebra generated by kerTT{„)A2 and kerrMTin). By Theorem 5.2.2, \/A'(A2)^N. By Lemma 5.5.6, J is a proper ideal of A'(T^). Let J\ be the closure of J, then J\ is a closed ideal of A'{T™).

Jx kerTTi„)iA2 SetJ =

kerr A2,Ti") A'(A2) cA'(R), then J" is a closed ideal of

A'(R).

when rw = 4n)®4n),we have:

A;errT(„)>A2 =

kerTAuA2

kerTA1,A2

A'(A2)

A'(A2) 2nx l

:={ xn

2/1

J/n

: a^e/cerr^ A2, ?/iG^'(yl2),i = 1,2, ••• ,n}

and

fcerT^2iTC„)

= [kerTA2>Al, ••• , /ser-TA2,A,, ^ ' ( ^2 ) , ••• , <A'{A2)]lx2n

= {(x[,--- ,x'n,y'ir-- ,y'n) \x!i£kerTA A , y'i&A!{A2),i = 1,2, • • • , n } .

Thus

kerrTln) tA2-kerTA2tT^)

where

[kerTAuA2-kerTA2tAl}nxn * * Mn(A'(A2))

(5.5.1)

kerTAuA2-kerrA2,A1 = {xx' : xekerTAuA2,x' ekerrA^Al},

[kerTAuA2-kerrA2iAl}nxn = <

X\Xi • • • X\Xn

XnX-^ • • • XnXn

viy'i ••• yiy'n

Xi€kerrAuA2

x^Gkerr^^

yi,y'ieA,(A2)

i = 1,2,--- ,n j

> .

yny\ ••• yny„

Consider another decomposition of T^™':

T(")~4mi)©4m2)©51®- • -@Bm = 4mi)©4m2)©£~4mi)©£®4m2)-

Similarly, we have that

kerrT(n) tM -kerrA2 T(n)

= diag([kerTAuA2-kerTA2:Al]miXmi,kerTB1,A2-kerTA2iBl,

kerTBm,A2-kerTA2tBm, • • • , Mm2(A'{A2))) (5.5.2)

Note that J can not be a maximal ideal of A'(R). Otherwise, by the construction of J, J\ is a maximal ideal of A'(T^ and

A'(R)/J = A'{Ai(l)®A2n+1))/J^A!{T^/Jl^Mn{C).

But from (5.5.1) and (5.5.2),

A'(R)/J=A'(AlTl)®B®A{p+1))/J^Mmi+m(C).

This contradicts TOJ + m > n. Now consider

and

A'(T™) = dm5(^(4n)).^'(4n)))

Ji~diag(J^,J{2).

Denote A = A'(T^)/Ji. By Theorem 5.2.2, \J(A'(A2))^N. Moreover, by Lemma 5.5.6, J{[ is a closed ideal of A'{AP), and J£2 = A'(A2

n)) = Mn(A'(A2)). Therefore,

A={A'{A((l))/J^)®0. (5.5.3)

On the other hand, we consider

A'iT^-diagiA'iA^), A {By), • • • , A'(Bm), A'(A{™3))).

Now,

Jl = Jl~diag(Jn, J22, ••• , Jm+2,m+2)-

By Theorem 5.2.2 and Lemma 5.5.6 again, Jn is a closed ideal of .4'(A^mi)), 3u is a closed ideal of A\Bi—\), 2<i<?7i + 1, and ^Jm+2,m+2 ==

Therefore,

A = A\T^)IJX = {A'{A[mi)®B)/Jl)®Q, (5.5.4)

where J[ = J{ = diag(Jn,J22, ••• , Jm+i,m+i). Without loss of generality, we can assume m;, m2 > 0. Otherwise, we can consider that:

T(2n) = T (" )©T("^A[" + m i ) ©A^ + m 2 ) ©B 1 ©- • -@Bm

and

r<2n> = 4 2 n ) e 4 2 n ) . By (5.5.4), there exists a surjective homomorphism

<A : A'(A[mi)®B)-^A.

By Theorem 4.4.3, A'(Ai)/radA'(Ai) is commutative. Thus ^ ' ( A ^ 0 ) and A both are n-homogeneous. Therefore A is an FIR algebra.

Furthermore,

A = .4~cfoa3(.4i, A2, • • • , Am+i)

= ^ ~ ^ a f f M ' ( 4 m i ) ) / ^ n , A'{Bi)/J22,--- ,A'(Bm)/Jm+i,m+i).

Since each Jii is a proper ideal, Ai ^ 0, i = 1,2, • • • ,m + 1. Suppose

J ^ is a maximal ideal of A\. By Kaplansky theorem [Bonsall, F.F. and

Duncan, J. (1973)], there exists a unique maximal ideal J2<ZA such that

* n ( J 2 ) = Jii- Then

A/J2 = A/J2~diag(A1/J(1, A2/$22(J2), ••• , Ai+m/$i+m,i+m(J2))-

Since A is n-homogeneous, A/J2=Mn{C). Note that mi + m > n, by the arguments similar to that used in the proof of Theorem 5.3.7, we can see that there exist mi+m — n natural numbers {hi, k2, ••• , fcm+mi_„} in {1,2, • • • , m + 1} such that

•Aj/$jj(J2) = 0, je{ki,k2, • • • , / e m + m i _ n } .

Without loss of generality, assume that when j = m + 1,

Al+m,l+m/$l+m(J2) = 0.

Suppose that J{+m 1 + m is a maximal ideal of Ai+m, using Kaplansky theorem again, we can find a unique maximal ideal J% of A such that

®l+m,l+m(J3) = Jl+m,l+m-

Thus J2 ^ J3. Since A is n-homogeneous, A/Jz=Mn{C). Denote J4 = Ji^Jz- By Lemma 5.5.5, there exists an isomorphism

*i : A-Â/J^Mn{C)®Mn{C)

such that

$ i C U © 0 © • • • ©0) = (1©0© • • • ©0)©P, P ^ 0,

$1 (o© • • • ©o©/.A1+m) = o©(o© • • • ©o©i).

Set $ = $!•(/>. Then $ is a surjective homomorphism from A'(Ai ^ © 5 ) to Mn(C)®Mn(C) such that

^ ^ ' c ^ r 1 ' ) ® 0 0 • • • ©°) = C1©0© • • • ©o)© > PÔ,

$(0© • • • ®0®IA,{Bm)) = 0©(0© • • • ©0©1).

Since A'(R)/J = A'{T^)/JX®Q = A®0, there exists a closed ideal ED J such that A'{R)/E = A/Ji®0 = A'{A(™')®B)/ker§®0.

Suppose that TT : A'{R)->A'(R)/J, 7r2 : A'(A{(ni)®B)—Âl{A{™l)®B)lker$ and

TTI : A'{Ai)^A'(Ai)/J" are canonical maps. Then by Lemma 5.5.7,

irlt(Ko(A'(A^))))^,(K0{A'(R)))^nit(K0{A'(A^ni)®B))).

By Lemma 5.5.4, 7rl!#! is injective. So

n24K0(A\A^l)®B)))^4K0(A\R)))^n14Ko(A'(A[n))))^Ko(A'(A^)))^Z.

Furthermore, $ induces a homomorphism

* : A'(A(Tl]®B)/ker^—+Mn(C)®Mn{C).

Therefore,

* . = *,-TT2+ : K0(A'(A[mi)(£B))—>K0(Mn(C)@Mn(C))^Z@Z.

Since \t* is an isomorphism,

$»(tfoU'(4mi)©5))) = **(^*(^o(^'(4mi)©5))))^Z. (5.5.5)

Since

$ (^ ' (A ( m i , )®°® • • • ©o) = (i©o© • • • ©o)©p

and

$(0© • • • ®0®IA'{Bm)) = 0©(0© • • • ©0©1)

we have

$•{[!* (Afi))®0® • • • ©0]) = [l©0© • • • ©0]©[P] = 1©[P];

$*([0© • • • 0Oe/A'(Bm)]) = [0]©[0© • • • ©0©1] = 0©1.

By (5.5.5) again, there exists n£Z such that

* * ( [ ^ ( ^ i ) ) © 0 © • • • ©0]) = ™$*([0© • • • ®0®IA,{Bm)}),

i.e. 1©P = n(0©l) = 0(Bn£Z®Z. But this is impossible and so we verifies out claim that m4 + m<n, i = 1,2. {Claim 2} rrii + m = n.

By the Claim 1, we need only to prove that: rrii + m>n, and it suffices to prove that m\ + m>n.

By Theorem 4.4.3, A'(Bi)/radA'(Bi), l<i<m is commutative. Since

A = A'{T^)/J! = A'{A^)/Jn

is n-homogeneous and since

A = A^diag(A'(A^)/Jlu A\B1)/J22, ••• , A'(Bm)/Ji+m,i+m, 0).

By the similar arguments used in the proof of Theorem 5.3.7, we have

A/J'=Mi(C)

for each J'£M.(A) and l<mi + n. Since A is n-homogeneous, m\ + m>n. Similarly, we may obtain that m2 + m>n.

Therefore, m* + m = n for i = 1,2.

Lemma 5.5.9 Let Ai,A2 be two strongly irreducible Cowen-Douglas operators. Assume that A\^f>A2 and T = A^'QA^2, where ni,n2 are two natural numbers. If A'(T)fradA'{T) is commutative, then\J(A'{T))^N^ and K0{A'(T))^ZW. Proof Since A'{T)/radA'{T) is commutative, A'(T) is 1-homogeneous and

A\T)/radA\T)^(A\A[ni))/radA\A{ni)))®(A'(A2n2))/radA'{A{

2n2)))

Thus, K0(A'{T))^Z^.

Lemma 5.5.10 Let A\,A2 be two strongly irreducible Cowen-Douglas operators. Assume that A\^A2 and T = A± 1'®A2

n2 , where n-\_,n2 are two natural numbers, then

\J{A\T))^NM andK0(A'{T))^Z^.

Proof By Remark 5.5.3, we may assume that T = Ai®A2. By theorem 4.2.1, we only to prove that for each natural number n, T^ has a unique (SI) decomposition up to similarity.

Let T<"> = A^QA^ and assume that

T ^ W ^ W ^ 2 ' © . ^ © - • -®Bm

is another finite decomposition of T^n\ where mi , m2, m>0, Bj£(SI), Bj'/'Ai for i = 1, 2 and l < j < m .

By Lemma 5.5.9, mi + m = n, i.e., rrii = n — m for i = 1, 2. {Claim} m = 0, i.e. m, = n, i = 1,2.

Since T = A1@A2, A'{T) = A'(T)~diag(A'(Ai), A'{A2)). By Theorem 4.4.3, A'(Ai)/radA'(Ai) is commutative for i = 1,2. From the proof of

Theorem 5.3.7, for every JeM(A'{T)) A'(T)/J^Mt(C), where I = 1 or 1 = 2.

Without loss of generality, we can assume that there is a J&M.(A'(T)) satisfying A'(T)/J^M2(C). Then

A'(T^)/Mn(J)=Mn(A'(T))/J)^M2n(C).

Since TO* = n — m,

T(n) „ A[m i )©A(m 2 )eBi©- • -®Bm

= 4 n _ m ) ® 4 " _ m ) © 5 i © • • • ®Bm = T("-m)eBi©- • -®Bm,

Thus A'(TW) = A'(T™) = diag(A'(T^-m^),A'(B1),---,A'(Bm)). Therefore, for every J&M(A'{T^)), $n(j) = A'(T^-m^) or $ n ( J ) is a maximal ideal of _4'(T("-m)). This implies that A'(T^)/J^MS(C) for each JeM(A'(T^)), where s<2(n -m) + m. So 2n<2(n - m ) + m , i.e. 2m<m and m = 0.

Repeating the arguments in the proofs of Lemma 5.5.8, Lemma 5.5.9 and Lemma 5.5.10, we have the following theorem.

Theorem 5.5.11 Let A\,A2, • • • , Ah be strongly irreducible Cowen-Douglas operators. Assume that A^Aj for i ^ j , and T = Aj ®A2

2 ©• • •®A£lk , where (ni, • • • ,nfc) is a tuple of natural numbers, then

\J(A'(T))^N^k\ K0(A'(T))*ZW.

From Theorem 5.5.11, we can get the following theorem.

Theorem 5.5.12 Let T be a Cowen-Douglas operator, then for each nat

ural number n, T^n' has a unique (SI) decomposition up to similarity.

Theorem 5.5.13 Let A,B£t3n(Q,) and assume that

A~4ni)e4"2)©---©4nt),

where 0 ^ n ^ N , AiG(SI) for i = 1,2, • • • , k and A^Aj for i ^ j , then A~B if and only if the following two conditions are satisfied:

(i) (K0(A'(A®B)), \/(A'(A(BB)), J ) ^ ( Z W , N W , 1 ) ;

(ii) The isomorphism h from \J(A'{A®B)) to N (&) satisfies

h([I]) = 2n\e\ + 2n2e2 H h 2nfcefc,

where I is the identity of A'(A®B) and {ei}f=1 are the generators of N^) . Proof "<$=": Since B is a Cowen-Douglas operator, we know

5 = 7^Sl)®52/S2)©---©74H

where each Bi is strongly irreducible Cowen-Douglas operator for i = 1,2, • • • , m and B^Bj for i ^ j .

Without loss of generality, we can assume that

A _ A^) a, A^) ffi . . .ffiA(nk)

B = B{1Sl)®B{

2S2)®---®B^\

{Claim 1} VBi, i = 1,2, •• • ,m, there exist Aj,j = 1,2, •• • ,k such that Bi~Aj.

Otherwise, without loss of generality, we may assume that B\,- • • ,Bi, for l<l<m, are not similar to each Aj, {l<j<k), and each Bi(l < i<m) is similar to some Aj, l<j<k. Thus

(>i©5)~(yi[ti)©4t2)® • • • ©4 t fc )©£is i )© • • • ©5,(s,)).

By Theorem 5.5.11,

\J{A\A®B)) = yiA'iA^^A^® • • • ®A%k)®B{1Sl)® • • • ©£/ (s , )))=iV ( fc+i).

This contradicts (i). {Claim 2} m = k.

It follows from Claim 1 that m<k. If m < k, without loss of generality, we assume that B\~Ai, B2~Az, • • • , Bm~Am. Then

(A©£)~(4 n i + S l W 2 " 2 + S 2 ) © • • • ®Atm+Sm)®A{Z+V]® • • • ®A{knk)).

By Theorem 4.2.1, the isomorphism h satisfies

h([I}) = (ni + si)ei H h (nm + sm)em + nm+i

By (ii),/i([7]) = 2niei+2n2e2H Ylnke-k- A contradiction. Thus m = k. Now we may assume that B\~A\, B2~^42, • • • , B^^Ak, i.e.,

{A®B)~(A<f1+Sl)®A2n2+S2)® • • • ®A^k+'k)).

Repeating the proof of Claim 2, we have Si = rii, i = 1,2, • • • , k. "=»" Since A~B, B = fl{ni)®fl<n2)© • • • ®B^k\ where B^SI), i =

1,2, • • • , k and ^ ~ B i for i = 1,2, • • • , k. Thus (A®B)~(A<?ni) ® • • • 0

42"l))-The remainder of the proof of the theorem is a consequence of Theorem

4.2.1.

Theorem 5.5.13 is the Jordan canonical theorem for Cowen-Douglas operators. By the proofs of Theorem 5.5.11, Theorem 5.5.12 and Theorem 5.5.13, we have the following result.

Theorem 5.5.14 Let Ti ,T2e£(ft) and T i~ ® Afi], where At is a i=l

strongly irreducible Cowen-Douglas operator for every i, andTi is not similar to Tj if i j£ j . Then T\~T2 if and only if:

(i) (^oM'(Ti©T2)), VM'CTieTa)), J)S(Z(">, N<">, 1); (ii) The isomorphism h from \J (A'(Ti®T2)) to N'fc) satisfies

h([I}) = 2kiei + 2k2e2 + ••• + 2knen,

where I is the identity of A'(Ti®T2) and {ej}™=1 are the generators ofN^nK

In the following we will consider a more general form. Let T = ® T> ,

where T^S^^'iT^/radA'iTi) is commutative, \J{A'{Ti))^N,i = 1,2, •• • , n, and Ti is not similar to Tj if i ^ j . By the arguments used in the proofs Theorem 5.5.11, Theorem 5.5.12 and Theorem 5.5.13, we have the following theorem.

Theorem 5.5.15 V ( -4 ' (T ) ) - N ( n ) andKQ{A'{T))^ftn\ FuHhermore, T has a unique (SI) decomposition up to similarity and T~G€:£(H) if and only if the following are satisfied:

(i) (K0(A'(T®G)), \J(A'(T®G)), /)s*(Z<">, N<">, 1); (ii) The isomorphism h from \J(A'(T®G)) to N^ satisfies

h([I]) = 2fciei + 2k2e2 + • • • + 2knen.

Corollary 5.5.16 Let T = ($ A\ k)@® B)'', where Ak's, Bj"s are k=i j = i

Cowen-Douglas operators, Ak,Bj£(SI) and A^AK, Bj^Bji if k ^

k'J^f. Then

\f{A'(T))^N^mi+m^ and K0{A'(T))^Z{mi+m2).

This corollary implies such T has a unique (SI) decomposition up to

similarity.

5.6 Maximal Ideals of The Commutant of Cowen-Douglas Operators

In this section we will use the iô-group to characterize the commutant of Cowen-Douglas operators. The main theorem of this section is as follows.

Theorem 5.6.1 Let Ai,A? be strongly irreducible Cowen-Douglas operators. Assume that AiÂ2 and T = A^'QA^12, where ni,ri2 are natural numbers. Then for each JGA4(A'(T)), J must be one of the following two forms.

(ii)

J = J, i i kerr

kerr<„„, <ni) A\Aî])

( » l ) 4 ( " 2 ) - 1 1 ^ * 1

("2h

J

l ( " - i ) . A'{AF>) kerr,ni) (na)

kerr ("2> 4 ( " l ) J'. 22

where Ja is a maximal ideal of A'(A\n''),i = 1,2. Proof First, we assume that T = A\®A2-

Since A\,A2 are strongly irreducible Cowen-Douglas operators, by Theorem 4.4.3, A'(Ai)/radA'(Ai) and A'(A2)/radA1 {A2) are commutative.

Assume that J has neither form (i) nor form (ii). Then

J = Jl\ Jll

According to the discussion in Section 5.4,

Jii£M(A'{Ai)),{i = 1,2), Ji2 % kerTAuA2!J2i C kerTMtM.

Then A'(T)/' J=M2(C) and we have the following exact sequence:

O^J - U A'{T) ^ U A'(T)/J-^0,

which induces the following six-term cyclic exact sequence:

K0{J) -d\

K!(A'{T)/J) «-

> K0(A'(T)) - ^ K0(A'(T)/J) di

- KM'(T)) <— K,{J)

Since A'(T)/J^M2(C),K0(A'{T)/J)^Z, Ki(A'(T)/J)*0. By Theorem 5.5.11, K0(A'(T))^ZM. Note that

n*([diag(IA,{Al),0)]) = 1, TT*([diag(0,IA,{A2))}) = l .

Thus 7T» is a surjective map from Z©Z to Z, and we get a split exact sequence:

0-^K0(J) -±> Z0Z £± Z ^ O .

Since 7r»(Jfr0(^ /(T)))SZ, K0{J)=Z. Consider the following split exact sequence:

0—^J—>J+1^(J+1) / J—>0 ,

Using the six-term cyclic exact sequence again and by the fact that d : KQ((J+1)/J)—>Ki(J) is a zero map, we have the following split exact sequence:

0—+Ko(J)-^Ko(J+l)^Ko((J+l) /J)—>0.

Since ( j 4 - l ) / J ' S C © C f # o ( ( J + l ) A 7 ) = Z ( 2 ) . Combining it with K0(J)=Z, we have K0{J+1)=Z©Z©Z.

Note that

and

P2 =

•TA'(^I) ° 0 0

0 0

0 IA'(A2)


are two minimal idempotents of Moa(A'(T)) and P i / Q P 2 in Moo(A'(T)). It is obvious that Pi,P2 are also two minimal idempotents of M00(l7-j-l) and P i / a P 2 in M0 0(J-i-l) . Since K0{J+1)=Z®Z@Z, there exists a minimal idempotent P in Moo(J+l ) such that PâP\, Pâ

p2 in M o ^ J + l ) . Without loss of generality, we assume that P£A'(T). {Claim} I - P~aP in M^J+l).

Otherwise, / - PâP in MîJ+l). Then / - P~ Q Pi ox I - P~aP2

in MooiJ+1). Thus P~aP2 or P ~ a P i in MooU+1). This contradicts the choice of P.

Since M 0 0 ( J + l ) c M 0 0 ( ^ ' ( T ) ) , we have P ~ a P i and (I - P ) ~ a P i in Moo(.4'(T)) or P~QP2 and (J - P ) ~ a P 2 in MooM'(T)). By Lemma 4.2.4, there exists a natural number n such that

T(n'\ranPm = T\ranP~T\ra,nP1 (oi T\ranP2) = A\ (pi A2)

and

r ( n ' | r O n ( / - P ) e 0 = ^ | r o n ( / - P ) ~ y | r o n P 1 (or T|ronf>2) = A\ ( o r ^ ) -

This implies that T = AX®A2 or T = A2®AX. By Theorem 5.5.13 T has a unique (SI) decomposition up to similarity. Thus A\~A2. The contradiction indicates that J must have the form (i) or (ii).

Now we consider the general case, i.e., T = A" ®A2n 2 •

It follows from the proof of Theorem 5.3.7 that A'(T)/ JS*Mk(C) for every JeM(A'(T)), where A; = n\ or k = n2 or k = n\ + n 2 . Thus we need only to show that k^ m + n2. Otherwise, there exists J^M.(A'{T)) such that

A'(T)/J^Mni+n2(C).

Observing

A\T) = A\T)~(diag(A'(A1®A2)),diag(A'(A{ni-1)(BA2n2-1)))),

J = J~{diag{irn(J),ir22(J))).

Then fm(J)£M(A'(Ai@Aa)) and

*u(J) = J\\ J\2 Jl\ J22

where Jn£M(A'(A1)),J22eM(A'(A2)),Ji2 £ kerrAuM, J2i$kerTA2iAl. This contradicts the conclusion got in the first part. Thus, for each JeM(A'(T)), A'(T)/J is not isomorphic to M„1 +„2(C).

By the arguments similar to that in the proof of Theorem 5.6.1, we have the following theorem.

Theorem 5.6.2 Let Ai,A2,---,Ak be strongly irreducible Cowen-Douglas operators. If Ai is not similar to Aj if i ^ j and T = Ay11'Q)A2 '®- • -®Ak

nk', where n\,--- ,rik are positive integers, then for each JeM(A'(T)),

J =

J\\ J\2 J21 J21

.Jk\ Jkl

J\k J2k

Jkk

("i)->

where Jij satisfies the following properties. (i) Jij = kerrAint) A(nj), i ^ j ;

(ii) There exists a unique i, \<i<k such that Ju€M(A'(A\ni))) and

Jjj=A'(A^))forj,j^i.

Corollary 5.6.3 Let T = 4 n i ) © 4 " 2 V • -®Aknk), where AUA2, ••• ,Ak

are strongly irreducible Cowen-Douglas operators and Ai is not similar to Aj if i ^ j . Then for each JeM(A'{T)), A'(T)/J^Mi(C), where le{ni,n2,--- ,nk}.

5.7 Some Approximation Theorem

Let fi be an analytic Cauchy domain, T = dft. Denote L2(T) the Hilbert space of all square integrable complex functions with respect to the arc length measure on I\ Denote M\(T) the multiplication by A operator acting on L2(T). Denote H2(T) the subspace of L2(T) generated by the rational functions with poles outside fi. Then H2(T)GLatM\(T) and

MX(T) = M+(T) Z

0 M_(r) H2(T) L2(T)eH2(T).

It is easy to see that M£(r)e#i(fi*) and a{M+(T)) = Q.. If Q is the unit disk D, M+{T) = Tz. Furthermore, A'{M+{T)) = H°°(n) [Conway, J.B. (1990)].

Proposition 5.7.1 Let 0, be a connected analytic Cauchy domain, fii,Ti2 be two nontrivial invariant subspace of M+(T). Then Hir\H2 ^ {0}. Proof Given AeO. Since (M+(T) - \)H2(T) is closed subspace with codimension 1, (M+(T) - \)H\ is closed. Clearly, (M+(T) -\)HieLatM+(r)£Hi. Thus

( M + ( r ) - A ) W i ^ W i .

In fact, if (M+(r) - \)Hi = Hi, then for arbitrary f£Ki,f(z) = (z ->^)n9n(z),gn€H1. This implies that /<n)(A) = 0, (n = 1,2, • • •). So / = 0, a contradiction.

Given a nonzero function 4>€HiQ[(M+(T) - X)Hi]. For each heR(T),

0 = < (M+(T) - A ) A ( M + ( r ) ) ^ > = y (z - X)h{z)\cj>\2d^

where R(T) is the set of rational functions with poles outside fi. This implies that ^L{g^ReR{T) : g{\) = 0}{cReL2(T)). By [Fisher, S.D. (1983)], the orthogonal complement of ReR(T) in ReL2(T)) is the linear combinations of finite functions, denoted by Qi, • • • , Qm in C°°. Clearly, constant functions is orthogonal to {g£ReR(T) : g(X) = 0}. Thus \(j)\2 is a linear combination of Qi, • • • , Qm and 1 and sup |<£(z)| < oo. Therefore

zer </>h = h(M+(T))4> <= Hi for each heR(T). Since R(T) is dense in H2(T) and since cj> is bounded, c/>H2(T)cHi.

Similarly, we can find a nonzero bounded function 4'€'H2 such that

^F2(r)cw2.

This means that

ninn2D^H2(T) # {o}.

Lemma 5.7.2 Let Q be a connected analytic Cauchy domain, then

lranTM+m,M+m}nlkerTM+(r)iM+m} = {0},

i.e., if there exists an X&C(H2(T)) such that

XM+(T)=M+(r)X

and

M+(r)Y-YM+(T) = X

for some Y&H%{T), then X = 0.

Proof Since A'(M+(r)) = H°°(Q,), there exists g€i?°°(n) such that

X = Mg{Y),

i.e., M 9 ( r ) / = ff/ for all feH2{T). Thus

M + ( r ) y / = YM+(T)f = <?/, / e t f 2 ( r ) .

Set / = 1, AF(1) -Y(X) = goi Y(X) = Xh-g, where h = Y(1)GH2(T).

Set / = A, AF(A)- r (A 2 ) = Xg or F(A2) = X(Xh-g)-Xg = X2h-2Xg. Generally, set / = A" - 1 , we have

Ar(A"-x) - Y(Xn) = Xn~lg

Y(Xn) = Xnh-nXn-1g (n = 1.2,---)-

Without loss of generality, we can assume that QcD. Thus it follows from lAI^IAI2*"-1) (A <= fl,n = 1,2,- • •) that

w2/r |A|3<w-1>dm / r \X\2ndm

•oo, (n — • oo).

HnA" This implies " i|A"|i ""2(r)—>oo, (n —• oo). Therefore, if g ^ 0, then Y is

unbounded, a contradiction. This means that g = 0 and X = 0.

For each natural number n, define

M„(r) =

M+(r)

0 / M+(r)J

e£((tf2(r))(">).

Theorem 5.7.3 Lei M n ( r ) be defined as above. Then for each AeA'(Mn(T)),

A = Mh Mh

Mr. Mh Mh.

where fi£H°°(T) (i = 1,2, • • • , n). Moreover, A'(Mn(T)) is commutative; (ii) Mn(T)e(SI);_ (Hi) a(Mn{T)) = Cl, <Te(Mn{T)) = T, dimker(\ - Mn(T)) = 0 and

ind(X - Mn{T)) = -n for all A G 0 ; (iv) M*(T)eBn(n*);

(v) A'(A) is commutative.

Proof (i) Assume that

A = An ••• Aln

_-^*nl " ' " Ann

eA'{Mn{T)),

i .e.

An ••• Ain

•™-nl ' ' * -Ann

M+(T) I M+(T)

M+(T) I M+(T)

0 / M+(T)_

At the (1, n), (1, n — 1) entries, we have

AlnM+(T) = M+(T)Ain

i M+(r)

An ••• A i n

A , . . . A

-"^nl **-nn

and

Ai, n_iM+(r) + Ai„ = M + ( r )Ai , n _i .

By Lemma 5.7.2, Ain = 0 and Ai<n-ieA' (M+(T)). Similarly, we can conclude that Aij = 0, (l<i < j<n). Consider the (i,i) entries of both sides, we have

Aiti€A'{M+{T)), (l<i<n).

Compare the (i + l,i) entries ( l < i < n — 1), we get

A i + M M + ( r ) + A i + i , i + i = A i : i + M+(r )^ i + i , i .

By Lemma 5.7.2 again, A i + i , i + i = Aiti and Ai+itieA'(M+(T)).

At the (i + 2, i) entries (l<i<n - 2),

Ai+2,iM+{Y) + Ai+2,i+i = A+i,i + M+(r)Ai+2,i.

Thus Ai+2,i+i = -4i+i,j and Ai+2,i&A'(M+(T)). Using this argument and Lemma 5.7.2 respectively, we get the general form of A.

(ii) Let P be an idempotent commuting with Mn(T). By (i),

P =

Mfl

Mh Mh

Since P2 = P, f\ = fa. Since fi is connected, / i = l or / i = 0 . In either case, fa = h = ••• = / „ = 0, i.e., P = / or P = 0 and M n ( r ) e ( S / ) .

(iii), (iv) and (v) are obvious.


A2

R =

0 Qll Ql2 • • • Qlm

Q21 Qii--- Q 2m

ci 0

C2

with respect to the orthogonal direct sum

n m

« = ( ® W i ) © ( 0 « i ) (n,m<oo), j=l i=\

where (i) {a(Aj)}^=1 and {a(ci)}^l1 are two families of pairwise disjoint com

pact sets such that

n m

a(R) = [\Ja(Aj)M\Ja(ci)} j=l i= l

and cr(R) is connected; (ii) Aj,Ci£(SI) and kerr A. = {0} (l<j<n, l<i<m);

(Hi) Either o-(Aj)n<r(ci) = 0 and Qji = 0 or a(Aj)r\a(ci) ^ 0 and QjigranTA.Ct.

ThenRe(SI). Proof Let P be an idempotent commuting with R and assume that P admits the block matrix representation P = (Ptj)"^Ti with respect to the decomposition of the space "H.

Since kerrc. A, = {0} ( l < j < n , l<i<m), computations indicate that

Pjt — 0 (n + l < j < n + m and l< i<n ) .

Since {o~(Aj)}™=1 and {o^Cj)}?^ are two families of pairwise disjoint compact sets, kerrA. A. = {0} and kerrc, = {0} (i ^ j). Thus computation shows that

Pji = 0 (n>j > i>l or n + m>j > i>n + 1),

i.e., P has an upper triangular matrix representation with respect to the given decomposition of the space. Moreover, the same arguments indicate that

Pji = 0 (l<j < i<n or n + l<j < i<n + m).

Since Pjj is an idempotent and Aj, Cj£(S7), Pjj = 0 or 1, (j = 1,2, • • • , n+ m). Without loss of generality, we can suppose that P\\ = 0. Since cr(R) is connected, there must exist an integer i (l<i<m) such that a(Ai)Da(ci) ^ 0. It follows from PR = RP that

* l,n+iCi == -^l-* l,n-H i S;H-*n+i,n-f-i

or

Since Qu£ranTAi ,Pn+itn+i = 0. If a(Aj)Cia(ci) ^ 0, by the same argument, Pjj = 0.

Since cr(R) is connected, after a finite number of steps Pjj = 0 ( l<j<ra+ TO). It follows from P2 = P that P = 0, i.e., Rz(SI).

Lemma 5.7.5 Lei fi be a bounded and connected Cauchy domain, then there exists an (SI) operator B satisfying

(i) a(B) = ft; (it) pF(B)r\o-(B) = ft and ind{\ - B) = 0 (A € ft); fm,) min-ind(\ — B) = 1 (A € ft); (w) A'(B)/radA'{B) is commutative.

Proof Denote

B Bi I 0 B2

H

where 5 i = A/+(r*), B2 = M+(T), T* = dCl* and T = d£l. Then B satisfies (i), (ii) and (iii). Suppose that P is an idempotent commuting with B. By Lemma 3.2.3,

P = P1P12 0 P2

n n'

Since B1,B2£(SI),P1 = 0 or / , P2 = 0 or / . If Pi =I,P2 = 0, it follows from P S = £ P that / + P12P2 = P1P12, or l£ranTBiB2. But this is impossible. Thus Be(SI). Note that A'(Bi) commutes with A'(B2) and for each AeA'(B),

A = Ai An 0 A2

where Ai&A'(B{),A2&A'(B2). Thus A'(B)/radA'(B) is commutative.

Lemma 5.7.6 Let T be in C(H) with connected spectrum cr(T). Ifaire{T) is the closure of an analytic Cauchy domain (I. Then there exists an (SI) operator A such that the spectral picture A(A) of A is the same as the spectral picture A(T) of T and

min-ind(A - X)k <min-ind(T - X)k, (fc>l, A € ps-p(A)).

Proof Let (Cli,ki),(Q2,k2),- • • ,(Cln,kn) be the finitely many components of a(T)\U, where ki = ind(T - A), A <E Qi (i = 1,2, • • • , n). Thus $7* are pairwise disjoint and each fii is a connected Cauchy domain.

If \ki\ < 00, by Theorem 5.7.3 and Lemma 5.7.5, there is an (SI) operator Ai — A(Q.i,ki)eC(H) such that a(Ai) — H,, ind(Ai - A) = ki (A € fii) and either min-ind(Ai — A) = 0 (if k{ ^ 0) or min-ind(Ai — A) = i (if ki = 0) and A'(Ai)/radA'(Ai) is commutative.

If \ki\ = 00, by Proposition 3.13 of [Jiang, C.L. and Wang, Z.Y. (1998)], we can find an operator Ai = A(Qi, ki)££(H) such that

a(Ai) = fij, ind(Ai — A) = kit min-ind(Ai — A) = 0 (A € fij),

A'(Ai)/radA'(Ai) is commutative and V ker(Ai - X) —H.

Let $ i , $ 2 , - " !$m be all the components of Cl. By Theorem 3.7 of [Jiang, C.L. and Wang, Z.Y. (1998)], we can find (57) operators ci, C2, • • • , cm such that a(ci) = aire(c,) = $£ and A'{ci) is commutative.

Set

M Qll Ql2 ••• Qlm

M Q21 Q22 • • • Q2m

A An Ujnl V n 2 ' ' ' Sinm

~~ C2 0

0

where {Qij : l < i < n , l < j < m } are defined as in Lemma 5.7.4. By Lemma 5.7.4, AG(SI) and satisfies all the requirements of the lemma.

Lemma 5.7.7 Let A be given in Lemma 5.7.6, then A'(A)/radA'(A) is commutative.

n n m m Proof Since A'{@Ai)/radA'(0 Ai) and A'(0 Cj)/radA'(0 Cj) are

i= l i = l j=l j=X commutative and since kerr „ m = {0}, A1(A) IradA!(A) is commu-

i = l J = l

tative.

Theorem 5.7.8 Let Te£(W) with connected spectrum a(T), then there exists a sequence of (SI) operators {Tn}^L1 satisfying

(i) A'(Tn)/radA'(Tn) is commutative for all n; (ii) lim \\Tn - T\\ = 0.

n—*oo

Proof By Theorem 1.27 of [Jiang, C.L. and Wang, Z.Y. (1998)], we can find a sequence {^n}^.]^ of operators such that for each n

(i) °~ire(An) is the closure of an analytic Cauchy domain; (ii) cr(An) is connected; and (iii) lim \\An - T\\ = 0.

n—>oo

By Lemma 5.7.6 and Lemma 5.7.7, for each An, there exists a Bne(SI) such that A(B„) = A(An) and

min-ind(Bn — \)k<min-ind(An - X)k, (k>l, A e PS-F{A)).

Moreover, A'(Bn)/radA'(Bn) is commutative. It follows from Similarity Orbit Theorem that AneS(Bn)~• Thus we

€£(H{n+m)),

can find a sequence {Tn}^Ll of (SI) operators satisfying the requirements of the theorem.

Applying Theorem 5.7.3 and repeating the arguments above, we have the following theorem.

Theo rem 5.7.9 Given A£Bn(Cl), there exists a sequence of operators

{Ak}cBn(£l)n(SI)

such that (i) A'(Ak) is commutative for each k; (ii) lim \\Ak - A\\ = 0.

k—>oo

Definition 5.7.10 Let fl\,0,2 be two bounded connected open subsets of C and let n and rn be two natural numbers. An operator TG£(H) is said in the operator class Bn,m(fli,Q,2) if

(i) QiCpF(T)ruT(T), (i = l,2); (ii) dimker(T - A) = n for A s fli and dimker(T — fi)* = m for

/i € 1 2; (hi) V {fcer(T-A), ker(T-^)*} = H. [M.J. Cowen and P.R. Douglas

[I]]-

Proposition 5.7.11 Given TeSi, i(f i i , f i2) , A'(T)/radA'(T) is commutative. Proof Denote Hr{T) = \J ker(T - \),Hi(T) = \J ker(T - / * ) * . By

Apostol's triangular representation, Hr(T)±.Hi(T) and

Tr T\2

0 T[

Hr{T) Hi(T)'

where Tr = T |^ r( j ) and T; = (r*|-^((^))*. A simple computation shows that T r€Bi(fii) and T^eB^fl^).

Thus A'(Tr) and A'(T{) are commutative. This implies A'(T)/radA'{T) is commutative.

Proposition 5.7.12 Given T£Bm,m(Q,i,Q,2), for each e > 0, there exists a compact operator K with \\K\\ < e such that A'(T + K)/radA'(T + K) is commutative.

Proof By the argument of Proposition 5.7.11,

T Tr T\2

0 Ti nr{T)

where Tr£Bm(Qi) and T^eB^il^)- By Theorem 3.2.1, we can find compact operators K\ and K2 such that maa;{||ii'i||, H-ft H) < § and

Tr + JFf1eBm(fi1)n(57), (T, + K2)*GBn(n*)n(SI).

Set

K Kx 0 0 K2

By Theorem 4.4.3, T + K satisfies the requirement of the proposition.

In this section we proved that for "almost every" strongly irreducible operator T, A'(T)/radA'(T) is commutative. For the matter of that, we conjecture that every (SI) operator T has the property, i.e., A'(T)/radA'(T) is commutative.

5.8 Remark

The results in Sections 5.1 and 5.2 are contributed by [Jiang, C.L. (1994)]. The work in Section 5.3 are due to [Fang, J.S.(2003)] and [Jiang, C.L. (1994)]. The work in Section 5.4 belongs to [Fang, J.S.(2003)]. The results in Section 5.5 were proved by [Fang, J.S.(2003)], [Jiang, C.L., Guo, X.Z. and Ji, K.], [He, H. and Ji, K.]. The work in Section 5.6 are due to [Jiang, C.L., Guo, X.Z. and Ji, K.]. The work in Section 5.7 are given by [Jin, Y.F. and Wang, Z.Y.(l)], [Jiang, C.L. (1994)] except that Proposition 5.7.1 is due to [Fong, C.K. and Jiang, C.L. (1993)]. Here we must point out that the work of classification of Cowen-Douglas operators was inspired by the work of [Elliott, G. and Gong, G. (1996)], [Elliott, G., Gong, G. and Li, L.], [Dadarlat, M. and Gong, G. (1997)].

5.9 Open Problem

1. Let Ts0„iTn(f2i,172)) given a necessary and sufficient condition for Te(SI). 2. Let T&Bn,m(fii,fi2)n(S'J). Is A'(T)/radA'(T) commutative?


3. If TeBn,m(Qi,n2)ri(SI), is K0{A'{T)) isomorphic to Z? 4. Let T~{wfc} be an injective unilateral weighted shift. If for each M^LatT,T|x~T, then is T similar to aTz for some positive a? 5. Let Tf be an analytic Toeplitz operator. Is the following statement true? Tfe(SI) if and only if A'(Tf)/radA'{Tf) is commutative. 6. Let fl be an analytic Cauchy domain. Does there exist an (57) operator A satisfying the following conditions?

(i) A e B i . i ^ n ) ; (ii) A'(A) is commutative.

7. Let Q be an analytic Cauchy domain. What is Ki(H°°(fl))7 8. Given TGC(H) with connected cr(T), does there exist a sequence {Tn} of (57) operators satisfying

(i) lim \\Tn - T\\ = 0; n—*oo

(ii) Tn — T is compact; (iii) A'(Tn)/radA'(Tn) is commutative.

9. Let Tf be an analytic Toeplitz operator. Is the following statement true? Tf£(SI) is equivalent to Tfe{RI). 10. If A is a unital subalgebra of 77°°, is Ko(A) isomorphic to the integer group Z?

Chapter 6

Some Other Results About Operator Structure

6.1 i^o-Group of Some Banach Algebra

Theorem 6.1.1 Let fi be a bounded connected open subset of C with (f2)° = Q,. Let H°°(£l) be the unital Banach algebra consisting of all bounded analytic functions on Q. Then K0(H°°(fi))^Z and V(#°°(f t))=N, where (il)° denotes the interior of closure Cl of Q.

To prove Theorem 6.1.1, we need some lemmas. For a bounded connected open set fi, if (fi)° = fi, then there exists a probability measure JJ, such that support /x = T := dfl satisfying

f(z) = f fdfi for all / analytic on H [Herrero, D.A. (1974)].

Denote N(T) :=the "multiplication by A" on L2(T,fi) and H2(T) := "the subspace generated by all analytic functions on fi". Then H2(T)£LatN(T) and

N(T) = N+(T) Z

o w_(r) H2(T) L 2 ( r » a H " 2 ( r ) •

Lemma 6.1.2 (i) N(T) is normal, and N+(T),N-(T) are essentially normal;

(ii) ^ ; ( r ) 6 B i ( f i ) ; (in) A!(N+{T))^H°°{Q). [Conway, J.B. (1978)]

Lemma 6.1.3 Let G = Z and (G,G+) be an ordered group. Then there exists an isomorphism <j>: G—»Z such that (/>(G+)cN. Proof By the definition of ordered group, we can assume that there is an n, 0 < n£G+. If 0 > raeG+, then {-m)GeG+ and mneG+.

203

Note that {—m)n + mn = 0, by the definition of ordered group again, we have {—m)n = 0 and mn = 0. Thus G + c N . Suppose that (f> is the identity isomorphism from G onto Z, then </>(G+)cN.


Proof of Theorem 6.1.1 For the given Q, by Theorem 5.1.6 K0(A'(N+(r)))S£Z. It follows from Lemma 6.1.2 that K0(H°°(Q,))^Z. By Lemma 6.1.3,

\/(A'(N+(r))) = \J(H°°(n))cN.

We need only to prove that \J(A'(N+(T)))^N. Set P = diag{I,0,0,---)€Moo{A'{N+(T))) and

r = [P}£\f(A'(N+(T))). Then r is a positive integer, since A'(N+(T)) is stably finite.

Suppose that q£Mn(A'(N+(T))) is a nonzero idempotent, then

0^[q] = ss\/(A'(N+(T))).

Suppose that B = (N+{T))^\pnn, then A'(B) is ^-homogeneous and k>l. Note that rs = r[q] = s\p], there exists n'>n such that

Q = diag(q,q!,--- ,qr,0,--- ,0)~adiag(P,Pi, • • • ,PS ,0 , ••• ,0) = P

in Mn(A'(N+(T))). By Lemma 4.2.4,

p>{r) _ A(n')\ , , ^ 4 ( n ' ) | , „ — 4( s )

Thus ^ ' (B( r ) )^^ ' (A( s ) ) , i.e., M r (>t ' (B))SM,( / (y l ) ) . Since M r(^t'(B)) is rfc-homogeneous and M.,(.4'(.<4)) is s-homogeneous,

s — rk. Therefore,

\/(A'(A)) = {kr: k = 0,1,2,- • • }.

Since (KQ(A'(A)),\/(A'(A))) is an ordered group, r = 1. Thus

V(.4'(;v+(r)))=N.

In Theorem 5.1.6, we have directly proved that \/(A'(N+{T)))^N. In the proof of Theorem we give a new proof by using a different method.

Let 0 be an analytic connected Cauchy domain and W22(Q) denote the Sobolev space W22(ST) = {f&L2(Cl,dm) : the distributional partial derivatives of first and second order of / in L2(Q,dm)}, where dm us the planar Lebesgue measure.

Some Other Results About Operator Structure 205

Set

W(Q) = {Mf : feW22(Q)},

where Mf is "multiplication by / " on W22{Q). Denote R(£l) the subspace of W22(fi) generated by the set of all rational functions with poles outside U. Note that R(Q)£LatM\. Denote M\(Q) = MX\R(Q.). By Proposition 4.5.1,

^'(MA(Q))=i?(fi).

Theorem 6.1.4 K0(R{n))^Z and \/(R{Q))^N. Proof Since M^EBi(Q*), by asimilar argument of the proof Lemma 6.1.2 we can prove Theorem 6.1.4.

For an analytic connected Cauchy domain fi, denote

A(fi) = {/ : / is analytic in ft and /€C(fi) .

Then A(£l) is a unital Banach algebra with the norm | | / | | = max|/(.z)|. zen

A(D), when Cl = D, is called disk algebra. Given an f£A(Q,), there exists a sequence rn of rational functions with

poles outside Q, such that

lkn- / |U(f i )—»0 (n—• oo).

Theorem 6.1.5 K0(A{D))^Z and \/(A{D))^N.

Proof We need only to show that for each idempotent P£Mn(A(D)),

P~adiag(l,l,--- , lfc,0, •••)

in M00(A(D)), where k>n. Since PtMn{A(D)),

'fn(z) fuiz)--- fm(z)-P=

Jnl{z) fnl{z) ••• fnn(z).


Let y = ^ + z and define

7 n (£ + * ) / « ( £ + *)• •

./nl(£ + *)/»2(£+*)--

7 n ( y ) / i 2 ( y ) ••• / i n ( 2 / ) "

P' fin(m + z)

fnn(rn + z)

Jnl(y) fn2(y) ••• fnn(y)

Since ||P'—P\\ can be arbitrary small when m is big enough, P' is homotopic to P , denoted by P'~hP in Mn(A(D)).

Note that P'eMn(A(D)). It follows from \/(R(D))^N and the tf-theory that

P'~hdiag(l,l,--- ,lk,0,--- ,0)

in Min{A{D)).

Let fti(t) : [0,1]—>Mn(A(D)) be a continuous map satisfying

h1(0) = P, /ii(l) = P '

and h,2(t) : [0,1]—>Min{A{D)) be a continuous map satisfying

h2(0) = P', h2(l) = diag(l, 1, • • • , lfc, 0, • • • , 0).

Denote hi{t) = (h(£)©0(3">), then

/n(t) :[0, l]—M4„(i4(2?))

is a continuous map and

/H(0) = P©0( 3"\ /n( l ) = P'©0 (3").

Thus /i3(t) = (fe2o/ii)(*) : [0,1]—>M4n(A{D)) satisfies

h3{0) = P, h3{l) = diag{l, 1, • • • , lfc, 0, • • • ,0).

Therefore, \/(A{D))^N and K0(A(D))^Z.

6.2 Similarity and Quasisimilarity

We have the following question: Is A~B if A^n^B^n\ where n is a natural number.

Theorem 6.2.1 Let A,BeMk(C). Then A~B if and only ifAln)~BW. Proof By matrix theory, we can easily obtain the theorem.

Theorem 6.2.2 Let A,B££(H)n(SI). If

V(A ' (A) )S \ / (A ' (B) )SN,

A'(A)/radA'(A) and A'(B)/radA'(B) are commutative, then A^B if and only if A^~B(n^ for each natural number n.

Proof By Theorem 5.5.15, A~B if and only if K0(A'(A@B))^Z and by Theorem 5.5.15 again,

\f(A'(A®B))^ \/{A'(A®B)(n))^NV\

where I = { J J Af^ . Therefore, A~B if and only if A^^B^. ^ 0 it 7l~X3

Theorem 6.2.3 Suppose that A = 0 A<-ni),B = 0 J9Jmj), where {Ai} t=i i = i

and {Bj} are two families of (SI) and pairwise not similar operators. If foreachi andj, (l<i<k,l<j<l),A'(Ai)/radA'(Ai) andA'(Bj)/radA'(Bj) are commutative, and \J(A'(Ai))=\/(A'(Bj))='N, then A~B if and only if j4.(")~£(n) for each natural number n. Proof Applying Theorem 5.5.15 and repeating the arguments in the proof of Theorem 6.2.2, we can prove the theorem.

Corollary 6.2.4 Given A,BeBn(fl), A~B if and only if A^^B^ for each natural number n. Proof By Theorem 4.4.3, Theorem 5.5.15 and Theorem 6.2.3, we get the corollary.

K. Davidson and D.A. Herrero raised the following question in [Davidson, K.R. and Herrero, D.A. (1990)].

Suppose that Ae£(H)n(SI) and A~q.s.B, is BG(SI)?

In general, the answer to the question above is "no".

Proposition 6.2.5 Suppose that AeBn(tt)n(SI) with a(A) = T), and

W(In + A) + (In + A)X^In

for each XGC(H), then

T = In + A In

0 -In - A l^SI)

and ifker(IH + A) = {0}, then T^q.s.(In + A)®(-In - A). Proof In + AeBn(D(l, 1)), -(IH + A)&Bn(D(-l, 1)), where

D(l, 1) = {zeC : \z - 1| < 1}, D(-l, 1) = {z£C : \z + 1| < 1}.

Thus

I>(1, l)Cap(In + A), D(-l, l)c<rp(-(In + A)).

Since £>(1, l)r\D(-l, 1) = 0 and

\J ker(Iu + A - z) = \J ker{-In - A ze£>(i,i) zeD(-i,i)

•z) = n,

by Lemma 3.2.3,

Let P be a nontrivial idempotent in A'(T) and

P n P12

P21 P22

Since (I+A)P2i+P2i(I+A) = 0 and P2\ = 0, P n and P22 are idempotents in A'(A). Since Ae(SI), Pn = 0 and P2 2 = / or Pn = I and P2 2 = 0. Assume that P n = 0 (otherwise consider I — P). If follows from PT = TP that

P 1 2 ( / + ^ ) + (7 + A)P12 = - P 2 2 .

By the assumption of the proposition P2 2 = 0. This shows that Te(SI). Set

and

X

Y =

2(1 +A) I 0 I

I I 0-2(1 +A)

n n

H

w Then XT = ((I + A)®(-I - A))X,TY = Y((I + A)®(-I - A)). This implies that T~ g . s . ( / + A)®(-(I + A)).

Example 6.2.6 Let S be the injective unilateral shift on H = I2. Then

S*£Bi(D). Let {e n }~ = 1 be the ONB ofH and Sen = en+1 (n = 1,2, • • •).

If for some Xe£{H), X(I + S*) + {I + S*)X = I, then

XS* + S*X + 2X = I.

n A simple computation indicates that Xen = \ ^2, (—l) fc_ len+i-fc ( n =

fc=i

1,2,- ••) and\\Xen\\ = ^ —>oo (n—»oo). Thus X(I+S*) + (I+S*)X ^ I for all XeC{H). By Proposition 6.2.5,

I + S* I 0 -(I + S*)

n H

and

Y = I I 0 - 2 ( 7 + A) ne{SI)-

Theorem 6.2.7 Suppose that AeBn(n)r\(SI) and B~q.s.A, then B<=(SI). Proof Without loss of generality, we assume that the minimal index of A is n. Since B~qs.A, there exist operators X and Y, with trivial kernels and dense ranges, such that AX = XB and YA = BY.

A simple computation indicates that Q, £ crp(B) and dimker(B-z) = n for all zeQ.. Note that AXTY = XBTY = XTYA for each TeA'(B). This implies that XTY&A'(A). Since n is the minimal index of A, by Theorem 4.4.3, cr(XTY\ker(A_z)) is connected for each z€Q. Since YA = BY,Y(z) is a linear transformation from ker(A — z) to ker(B — z), where Y(z) =

y\ker(A-z)-Similarly, X(z) = X\ker(B-z) *s a linear transformation from ker(B-z)

to ker(A — z). Since

AXY = XBY = XYA,

XYGA'(A)

and

T{XY){z) = X(z)Y(z)

for all z£tl. Similarly, YXeA'{B) and T{YX)(z) = Y(z)X(z) for all zeQ. By Theorem 4.4.3, a(X(z)Y(z)) = {X(z)},X(z) ^ 0. Since X,Y are

injective, Y{z)X(z) is invertible. Thus a(Y(z)X(z)) = {X(z)}.

For arbitrary TeA'(B), since XTY&A'(A),

T(XTY)(z) = X(z)(T\ker(B_z))Y(z)

and a(X(z)(T\ker(B-z))Y(z)) is connected. Assume that

a(X(z)(T\ker(B_z))Y(z)) = {»(z)},

then

a(Y(z)X(z)(T\ker(B_z))) = {/x(2)}U{0}.

{Claim 1} If n(z) = 0 for some zed, then cr(T\ker(B_z)) = 0. The claim can be proved as follows. Since Y(z) and X(z) are invertible, 0£a(T\ker(B-z))- If 0 ^

Xea(T\ker(B-z)), then XIker(B-z) -T\ker(B-z) i s n o t invertible. Repeating

the argument above, we have

Thus

But

a(X(z)(\Iker{B_z) -T\ker{B_z))Y(z)) = {0}.

tr(X(z)(XIker{B_z) -T\ker(B_z))Y{z)) = {0}.

<r(X(z)(A/fcer(B_,) - T\ker{B_z))Y(z))

= tr(AX (z)y(z) - x(z)(r|fcer(B_,))y(z))

= \tr(X(z)Y(z))-tr(X(z)(T\ker(B„z})Y(z))

= n\X(z) ^ 0

A contradiction. {Claim 2} If there exists z£Q, such that /J,(Z) ^ 0, then <^{T\kerrB_z\) is

connected. Suppose that /3(z)£a(T\ker{B-z))> then

P(z)Iker(B-z) ~ T\ker(B-z)

is not invertible. Repeating the proof of Claim 1, we have

a(X(z)((3(z)Iker{B_z) -T\ker(B_z))Y{z)) = {0}.

By Claim 1, (r{T\ker{B_z)) = {(3{z)}. It follows from V ker(B ~ z) = z£D

U that V ker(T - /3(z))n = H. This implies that a(T) is connected. z€D

Therefore <x(T) is connected for each TeA'(B). {Claim 3} B£{SI).

Otherwise, there exist M,NeLatB such that MnAf = {0} and M +

M = H. Denote Bx = T\M,B2 = T\#, then B = By+Bi. Set T = IM+2I//,

then TeA'(B), but a{T) is disconnected. This contradiction implies that Be (SI).

Proposition 6.2.8 Suppose that AeBn(Q)n(SI) and B~q.s.A, then

A'{B)/radA'{B)

is commutative. Proof Without loss of generality, we still assume that the minimal index of A is n. Then Qcap(B), dimker(B — z) = n (z€fi) and \J ker(B — z) =

H. For X,Y€A'(B),

{XY — YX)\ker(B_z) = X\ker(B-z)Y\ker(B-z) — Y\ker(B-z)X\ker(B-z)

for all zefl. It follows from Claim 2 in the proof of Theorem 6.2.7 that

a((XY-YX)\ker{B-z)) = {0}

for all zeO. Thus ((AT - YX)\ker{B_z))n = 0 for all zed and

A'(B)/radA'(B) is commutative.

Next we will discuss the similarity of irreducible operators. Since irre-ducibility is unitarily invariant, some important behavior of C*-algebras and irreducible operators can be described in terms of irreducible C*-algebras and irreducible operators. F. Gilfeather proved that if Ne£(H) is a normal operator with empty point spectrum, then N is similar to a irreducible operator. [Jiang, Z.J. and Sun, S.L. (1992)] proved that each self-adjoint operator with infinitely many point spectrum is similar to a irreducible operator. D.A. Herrero posed the following conjecture in [Herrero, D.A. (1979)]:

An operator QeCCH) is similar to an irreducible operator if and only if (i) X — Q is not finite rank for each complex number A;


(ii) Q does not satisfy any quadratic equation ax2 + bx + c = 0, where |a| + |6| + | c | ^ 0 ;

(iii) A — Q is not a direct sum of a finite rank operator and an operator satisfying a quadratic equation for A G C.

In the following we will answer Herrero's conjecture in the case of nilpo-tent operators, normal operators and Cowen-Douglas operators.

Denote Afk = {TeC{H) : Tk = 0 and T ^ 1 ^ 0}. For TeAfk(H), T admits a kxk operator matrix representation of the following form:

T =

0 T1 2 T13 • 0 T 2 3 -

0 '

0

• Tifc-i Tifc

• Tak-i Tzk

0 Tk-ik 0

kerT © kerT0

kerT2 © kerT1

kerT3 © kerT2

kerT''-1 0 kerTk

kerTk © kerTk

where fcerT0 = {0},kerTk = H, kerTj-i j = {0},j = 1,2, ••• ,fc [Davidson, K.R. and Herrero, D.A. (1990)]. Using Lemma 7.8 of [Herrero, D.A. (1990)], we can prove that TGAfkCH) (k>3) is similar to an operator of the form

0 Ti Ti2 • • • Ti fc_2 Ti k-i 0 T2 • • • T2 fc_2 T2 fc_i

0 T3 k-i T-j, k-\

0 Tk-x 0

W2

W3

Tik-i Hk

(6.2.1)

where fcerT, = {0} (z = 1, 2, Or

, k — 1) and ranT\,ranTi are dense.

T~0 H l (

0 Ti T12

0 T2

0

Ti fc_2 Ti fc_i

?2 fc_2 T2 fc_l

T3 k-2 T3 fc_i

0

(6.2.2)

where 0-^ is the zero operator acting on H\,dimrii = n, ( l<n<oo), kerTi = {0} (i — 1,2, • • • , k — 1) and ranT\, ranT2 are dense.

Or

T~Ti + F, (6.2.3)

where Ti£.N2(H) and .F has finite rank.

Theorem 6.2.9 Given TGAfkCH), T is similar to an irreducible operator if and only if the following conditions are satisfied:

(i) T is not finite rank; (ii) T2 # 0; (Hi) T ^ T\®F, where TiGAfk{H) and F is finite rank.

Before we prove Theorem 6.2.9, we need several lemmas.

Lemma 6.2.10 [Kato, T. (1984)] Suppose that A,BeC(H). If A is positive with kerA = {0}, then there exists a positive number S such that ker(A + XB) = {0} for |A| < S.

Lemma 6.2.11 Given A, BGC(H). If A is positive with kerA = {0}, then there exists a positive number A such that ran(XA + B) is dense in "H. Proof Since ran[XA + B]~ = [ker(XA + B*)]-L, we need only to show that ker(XA + B*) — {0} for some positive number A. Note that XA + B* = X(A + iJB*). Thus Lemma 6.2.11 follows from Lemma 6.2.10.

Lemma 6.2.12 Suppose that A,B&C(H), A is a positive operator with kerA = {0} and cr(A) is uncountable. Then there exists an operator E&C{H) such that the C*-algebra generated by A,AE + B and I is irreducible. Proof Since cr(A) is uncountable, there exist disjoint uncountable Borel uncountable sets CTi,0"2 such that cr(A) = aiL)a2 and 0 ^ 2 - Suppose that E is the spectral measure of A satisfying E{a{)E{a2) = 0. Then

dim(E(ai)H) = oo, and dim(E(a2)H) = oo.

Denote

A = Ai 0 0 A2

E{a1)H E{a2)H'

where A\ = A\E(ai)H> ^2 = ^|£(a2)W- Then A\,A2 are positive operators and kerA\ = kerA2 = {0}. Since 0 ^ 2 ) A2 is invertible.

Assume that

B = Bn B12

B21 B22 E(a2)H'

By Lemma 6.2.11 ran(XAi + B12) is dense in E(ai)Ti for some A > 0. Set

E = 0 A

-A2-1B2i-A2-

1(B22-(d + V))

where d > \\Bu || + 1 , V is the Volterra operator. Thus V is irreducible and a(d+V) = {d}.

AE + B = 0 XAi

-B21 d + V-B22_

?n A A i + 5 1 2 " 0 d + V

+ Bn B\2 B21 B22_

We are now to prove that the C*-algebra generated by / , A and AE + B is irreducible.

Suppose that P is a projection commuting with A and AE + B, and

P = P11 Pia P21 P22

Since PA = AP, P21A1 = A2P21 and A1P12 — P12A2. It follows from E{ai)E{<r2) = 0 that P2\ = P i 2 = 0 and

P11 0 0 P22

Since P(AE + B) = (AE + B)P, P2(d + V) = (d + V)P2. Since d + V is irreducible, P2 = 0 or / . Without loss of generality, we assume that P 2 = 0 (Otherwise consider I - P). Since P(AE + B) = (AE + B)P,

PiO 0 0

Bn XAi + B\2 0 d + V

Bn XAi + B12

0 d + V PiO 0 0

Thus Pi(A^4i + P12) = 0. Since the range of XAi + B12 is dense, Px = 0 and P = 0. Therefore the C*-algebra generated by A, AE + B and / is irreducible.

Lemma 6.2.13 Suppose that A,BG£(H), A is a positive operator with kerA = {0} and a(A) is uncountable. Then there exists an operator


E££(H) such that the C*-algebra generated by EA + B,A and I is irreducible. Proof The lemma follows from Lemma 6.2.12.

Lemma 6.2.14 Let Ai,A2,B£C(7i). If Ai is a positive diagonal self-adjoint operator, A\ is a positive operator and kerA\ = kerA^ = {0}. Then there exist X, E££(H), X is invertible, such that:

(i) A2X is a positive diagonal self-adjoint operator with trivial kernel

ker(A2X) = {0};

(ii) The C*-algebra generated by AiE+BX, A2X and A\ is irreducible. Proof Since A2 is a diagonal operator, A2en = Xnen for some ONB {enj^Li- Since kerA2 = {0}, An^0 for all n > l . Assume that A\ admits the following matrix representation with respect to the ONB {e„}J°=1)

Ai =

an ai2

0.21 a i i

Onl 0-n2 - '

0-2r>

and each row vector (aji,Oj2> • * • ,Qjn, • • • )^0. Otherwise, if (a.,1,0,2, • • • , a.jn, • • •) = 0. Since A\ = A\,

(o-ij, &2j, • • • , a,nj , • • • ) = 0 .

This implies that 0£ap(Ai) and contradicts kerA\ = {0}. Without loss of generality, assume that aij1^0.

Set

X =

u h 0

0 tn

ei

e2

where tj£[l, 2] satisfying Xit^Xjtj {i^j). Clearly, X is bounded.


Assume that

BX =

~bu &12 •

&21 &22 -

bnl bn2 •

• b i n • • • '

• b 2 n •••

' ®nn

• :

Set

en e i 2 • • e2i e22 • •

e n i e„2 • •

e i„ •• e2« • •

^nn

where ejin = y°+ny, an € (0,1] such that ahejin + 6 l n ^0 . If (j,k)^(j\,n), then e^ = 0. It is easy to see that E is bounded. Since

AoX =

« i

a2 0

0 a„

where an = Xntn and {a„} are pairwise distinct. Thus ker(AzX) = {0} and A2X is a diagonal self-adjoint operator.

Assume that

^ i E + S X =

^ 1 1 ^ 1 2 •

h\ h2 •

Inl ln2 -

• Wn • • hn •

' 'un *

where lXn = ai^e^n + bln^0 n = 1,2, • • •. We now prove that the C*-algebra generated by Ai,A2X and A\E+BX

is irreducible.


Suppose that P is a projection commuting with A2X, A% and AiE+BX. Since P commutates with A%X,

Pi

Vi 0

0 pn

Since P2 = P,Pi = 0 or / , i = 1,2, • • •. Without loss of generality , we assume that pi = 0 (Otherwise, consider I — P). From P commutates with AiE + BX, i.e.,

P i

P2

0 j?n

' l l ^12 '

'21 ^22 •

hi 'n2 '

• ' i n - -

• hn • •

' Inn ' '

hi hi •

hi I22 •

hi hi

hn hn •••

*nn ' '

Pi

P2

0 Pn

We have pi/12 = '12^2- Since p\ — 0 and '127^0, we get pi = 0. Similarly, since pihn = hnPn and Zi„^0, pn = 0 (n = 1,2, • • •). Thus P = 0. This implies that the C*-algebra generated by A\,A2X and ylii? + BX is irreducible.

Lemma 6.2.15 Assume that T = [0 Ai

0 0 0 0

5 ] A2

0 .

w H, where Ai and Ai are H

positive operators and kerA\ = kerA2 = {0}. Then T is similar to an irreducible operator. Proof (1) If o~{A\) is uncountable, by Lemma 6.2.12 there exists an operator Ee£(H) such that the C*-algebra generated by Ai, A\E + B and / is irreducible.


Define

X

then

x-

10 0 0 1 -E 0 0 I

7 0 0 01 E 00 I_

Denote

Ti = XTX-1 = 0 Ai AiE + B 0 0 A2

0 0 0

We are now to prove that T\ is irreducible. Suppose that P is a projection commuting with Ti and

P n P12 Pis

P21 P22 P23

•P3I A 2 P33

It is obvious that kerT^LatP, kerTi = W©080, fcerTj2 = ft©W©0. Thus

Pn = P31 = P32 = 0.

Since P is a projection, Pi 2 = P13 = P23 = 0. Therefore we may assume

P = Pi 0 0 0 P2 0 0 0 P 3

Since PTi = TiP,Pu4i = AiP2 . So A : Pi = P2Ax. Thus

Pi A? = Pi^ i^ l i = A1P2A1 = A\PX.

By functional calculus, P ^ i = A\Pi. Thus AiPi = A ^ , i.e., A\{Pi -P2) = 0. Since A\ is positive and kerA\ = {0}, Pi — P2 = 0. Thus pi = p 2 . Similarly, we have Pi = P2 = P3 and

P = Pi 0 0 0 Pi 0

_ 0 0 P i .

Since the C*-algebra generated by A\, A\ E + B and 7 is irreducible, Pi = 0 or 7, and therefore P = 0 or 7. Thus T is similar to the irreducible operator T\.

(2) If <r{A2) is countable, by Lemma 6.2.14, there exist an invert-ible operator X£C(H) and an operator EE£(H) such that the C*-algebra generated by Ai,A2X and A\E + BX is irreducible and A2X is diagonal.

Set

Gi =

7 0 0 0 7 0 0 0 X " 1

then

G;1

Thus,

7 0 0 0 7 0 0QX

GxTGl - l 'QAi BX 0 0 A2X 0 0 0

Set

G2 = 7 0 0 0 7 - £ 0 0 7

then

Therefore,

G? 7 0 0 0 7 £ 0 0 7

G2G{TGX G2

OAi AXE + BX 0 0 A2X 0 0 0

Clearly, G 2GiTGf ^ J 1 is irreducible. Thus T is similar to an irreducible operator.


Lemma 6.2.16 Let

T =

0 Ax A12

0 A2

0

0

A13 • A23 • A3 •

• Aln

• A2n

• A3n

' • An

0

If Ax,A2 are positive operators and kerAk = {0} (A; = 1, 2, • •-,n). ThenT is similar to an irreducible operator. Proof Set

Tx OAx Ax2

0 0 A2

0 0 0

By Lemma 6.2.14 and Lemma 6.2.15,there exists an invertible operator Xx such that

XxTxX^ = 0 Ax M2 0 0 ~A2

0 0 0 e(Sl),

where Ax and A2 are positive injective operators. Therefore, there exists an invertible operator X such that

XTX - l

0 Ax Ap Ay3

0 A2 A23

•• Mn

•• A2n

0 A3 '•• A3n

An

0

where Ax and A2 are positive and kerAk = {0}, k = 1,2, • • • , n. Now we prove that XTX~l£{RI). Suppose that P is a projection com

muting with XTX-1. By the argument similar to that in the proof of


Lemma 6.2.15, we can prove that

Pi Pi 0

P 3

0 p„ +1

Since XTiXf 1 £ ( # / ) , Px = P2 = P 3 = 0 or / . Without loss of generality, assume that Pi = P2 = P3 =_0. Since PXTX~l = XTX~XP, P3A3 = I3P4 and A3P4 = 0. Thus kerA3 = {0}. This implies that P4 = 0.

Similarly, Pi = P2 = P3 = P4 = • • • = Pn+1 = 0. Therefore T is similar to an irreducible operator.

Lemma 6.2.17 Suppose that TeJ\fk{H) is of the form (6.2.1), then T is similar to an irreducible operator. Proof Without loss of generality, assume that

T =

OTi 0 T2 *

0 T3

'••Tk

0

where kerTj = {0} (j = 1,2, •• • , k) and ranT\ and ranT2 are dense. By the Polar Decomposition Theorem, T\ = A\Ui and U\T2 — A2U2, where Ai is positive, kerAi = {0} and Ui is a unitary operator (i = 1,2).

Set

G =

C/j 0

U2

I

0


then

GTG-1 =

OAi 0 0 A2 * 0 0 0 T3

o o o o ' - .

I I I '•• '•• Tfc_i 0 0 0 0 ••• 0

By Lemma 6.2.16, T is similar to an irreducible operator.

Lemma 6.2.18 Suppose thatT is similar to 0-H^TI, where dirriHi = oo.

\QA1

0 0

.° °

B~\

A2

0 .

n n, n

where A\ and A2 are positive operators, and kerA\ = kerA2

T is similar to an irreducible operator. Proof Without loss of generality, we can assume that

{0}, then

T = 0Hl®Ti =

0 0 0 0 OOAx B

0 0 0 A2

0 0 0 0

Set

E

I - 7 0 0 0 7 0 0

- 7 0 / 0 0 0 0 7

Then

E~

7700 0700 7770 0007


and

ETE-1 =

-Ai -Ax -Ax -B Ai Ax Ax B 0 0 0 A2

0 0 0 0

Suppose P be a projection commuting with ETE 1 and

Pn P\i Pn Pu Pi\ P22 P23 P24

P31 P32 P33 P34

P41 P42 P43 -P44

It follows from PETE'1 = ETE~lP that

-PxxAx + P12Ax -PxxAx + Px2Ax -PxxAx + P12Ai -PnB + P12B + P13A2--P31A1 + P22AX -P21Ax + P22AX -P2xAx + P22A1 -P21B + P22B + P23A2

-P3lAx + P32AX -PsxAx + P32Ax -PzxAx + P32A1 -P31B + P 3 2 P + P33A2 -PAXAX + P42AX -PixAx + P42AX -PAXAX + P 4 2 ^i -P41B + P 4 2 P + P43A2.

S T

W V (6.2.4)

Where

S = -AxPxx - AxP2x - AxP3x - 5 P 4 i -AxP12 - AxP22 - AxP32 - BP42

AxPxx + AxP2x + AxP3x + BP41 AxP12 + AxP22 + AxP32 + BP42

T = -AxPxs - AxP23 - AxP33 - -BP43 A1P13 + A i P 2 3 + ^ i P 3 3 + BP43

-AXPXA - AxP2i - AxPM - PP44 AxPx4 + AxP24 + AxPzi + BP44

W = A2P41 A2P42

0 0

V = A2P43 A2P44 0 0

Compare the (4, 1) entry of (6.2.4), we have

-PixAx + P42AX = 0.


It follows from [ran^i] = Ti that

P41 = P42 . (6.2.5)

Compare the (4, 4) entry of (6.2.4), we have -P41B+P42B + P43A2 = 0. By (6.2.5), P43A2 = 0. Since [ranA2}" = H,

P43 = 0. (6.2.6)

Compare the (3, 3) entry of (6.2.4), we have -P31A1+P32A1 = A2P43 = 0. Thus

P31 = P32. (6.2.7)

This indicates that

w=\00' [ 0 0 J '

A2P41 = A2P42 = 0 a n d

P4 1 = P4 2 = 0. (6.2.8)

Note that the sum of the (1, 3) entry and (2, 3) entry of the right side of (6.2.4) is 0. Thus

-(P11 + P12 ~ P21 + P22)B + P13A2 + P23A2 = 0

and

P13A2 + P23A2 = 0,

i.e.,

{P13 + P23)A2 = 0

or

P13 + P23 = 0.

By (6.2.7) and A* = A,P13 = Pai.fta = P32 = P31. Thus P13 = P23 = 0.


Therefore the (6.2.4) can be written as follows:

-PuAi + P12A1 -PuAi + P12Ai - P n A i + P12Ai -PuB + P12B -P21A1 + P22A1 -P21A1 + P22Ai - P 2 i A ! + P22A1 -P21P + P22B

0 0 0 P33A2

0 0 0 0

-AxPxi - A1P21 -AxPi2 - AXP22 -A1P33 -PP44 A1P11 + A1P21 A1P12 + A1P22 AxP33 BP44

0 0 0 A2P44 0 0 0 0

:= (/?«)

Since a2 i = #21, -P21A1 + P22A1 = AiPn + A1P21 or

AiP2i + P2iAi = P22AX - A i P u . (6.2.9)

It follows from a 12 = /?i2 that

-P11A1 + Pi2Ai = -A1P12 - AXP22.

Taking the adjoint of both sides, we get —AiPu + A\P2\ = —P2\A\ — P22A\ or

A1P21 + P21A1 = -P22A1 + A i P u . (6.2.10)

Prom (6.2.9) and (6.2.10) we get P22Ai = A i P u . Repeating the proof of Lemma 6.2.15, we can prove that P\\= P22.TI1US

P22Ai = AXP22. (6.2.11)

Since A2+/S22 = 0, a?12+a22 = 0 or ~PnAi +P12A1-P21A1 +P22At = 0, i.e., P12A1 - P21A1 = 0. Thus

P21 = P12- (6.2.12)

Since a22 = (322, -P2lAx + P22AX = A1P12 + A1P22. By (6,2,11),

P 2 1 A 1 + A 1 P 1 2 = 0 .

By (6.2.12),

A1P12 + P12A1 = 0. (6.2.13)

Note that P* = P. Thus P{2 = P2i and P i 2 is self-adjoint. Thus P?2 is a positive operator. From (6.2.13) we know Pi2Ai = —P12A1P12 = AiP^2. Thus P12AX = AiP12. From (6.2.13), PuAi = 0. Thus P12 = 0 and P2i = 0. The (6.2.4) can be written as

-P i i i4 i -PuAi -PuAi -PuB P22A1 P22A1 P22A1 P22B

0 0 0 P33A2

0 0 0 0

-AtPu -AXP22 -A1P33 -BP44 AiPn AXP22 AXPSA BP44

0 0 0 A2P44 0 0 0 0

By a proof similar to the proof of Lemma 6.2.15, Pi 1 and XTX~1G(RI) for some invertible X.


P22 = Pi 33 44

T = (W 0 i 4 i A12'

0 0 A2

0 0 0

n n, H

where A\ and A2 are positive operators, and kerA\ = kerA2 = {0}. Then T is similar to an irreducible operator. Proof If dimHi = °°, then by Lemma 6.2.18 the conclusion is true.

If dimHi < 00, say dimHi = n. Denote

0 Ai A12 0 0 i 2

0 0 0

Then by Lemma 6.2.15 there is an invertible operator X\ such that

A = X1AX11 = 0 Ai An 0 0 3 2

0 0 0 e(Ri),

where Ai and A2 are positive operators, and kerAi = kerA2 = {0}.

Thus

XTX~l =

for some invertible X. Let

"0 0 0 0 " 0 0 Ai An

0 0 0 A2

.0 0 0 0

m H H H

= Ti 0 0 0A w<3>

e i , e 2 ) -be an ONB of Hu / i , / 3 , • • • Jn&kerA QkerA,EeL{n{3),Hi) such that

EAh = eu EAf2 = e2, • • • , Elfn = en and £([ V 3 /* ] x ) = 0. fc=i

Denote

Y = IE 0 /

and

T2 = Y T j y - 1

Hi W<3)

0E_A 0 3 WW

Now we are to prove that T2£(RI). Suppose that P is a projection commuting with T2. Denote

M = (0®kerA)f)PkerT2, Af = {0®kerA)n{I - P)kerT2,

M' = PkerT2eM, N' = (I - P)kerT2QN.

It is easy to see that dimM.1 = h<n, dimAf' = l2<n and

n n

V x£M', x = ^caei + zx, zx£M, V y€Af', y = ^faei + z2, z2GAf. i = l i = l

Thus, for ei£Hi, et = x + y, x£M', y&A/"', where

n n

x = ^ a , e i + z\, zxeM, y = ^ f t e j + z2, z2eN. i=l i=\

Since A4JJV, we obtain x = 0 or y = 0. This implies that

ei£PkerT2

or

eiG(7 - P)kerT2 (i = 1,2, •• • , n).


Without loss of generality, we assume that

e i , e 2 , • • • ,ek€PkerT2, €(I-P)kerT2.

By a routine proof, P admits a block upper triangular matrix representation with respect to the decomposition H\®H®'H®'H.

Since P is a projection,

P =

Pi 0 0 0 0 P2 0 0 0 0 P3 0

. 0 0 0 PA

Note that Ae(PJ) , thus P2 = P3 = P4 = 0 or I. Without loss of generality, assume that P2 = P3 = P4 = 0. Since PT2 = T2P, P\EA = 0. But .EM is surjective, thus P\ = 0 and P = 0. So T is similar to an irreducible operator.

The proof of next lemma is similar to the proof of Lemma 6.2.17.

Lemma 6.2.20 Suppose that TeNk(H) and T = 0Wl©A is of the form (6.2.2). Then T is similar to an irreducible operator.

So far we have proved the sufficiency of Theorem 6.2.9. As for the necessity, we need only to show that if T2 = 0 or ranT is finite rank, then TftBI).

(1) If T2 = 0. We may assume that

Q 0 A 0 0

H H

and dirriH = oo. By the proof of Lemma 6.2.15,

Q= QA 0 0

where A is a positive operator. Suppose that Pi is a nontrivial projection commuting with A. Set P = Pi©P2, then P commutes with Q.

(2) If T is finite rank, then T* is finite rank. Denote M = y{ranT, ranT*}, then M. is a reducing subspace of T. Thus the proof of the necessity of theorem 6.2.9 is now complete,

Next, we are going to discuss normal operators.

Theorem 6.2.21 Suppose that N€.C(H) is a normal operator. Then N is similar to an irreducible operator if and only if the following conditions are satisfied.

(i) X — N is not finite rank for all X £ C; (ii) N does not satisfy any quadratic equation ax2 + bx 4- c = 0, where

\a\ + \b\ + \c\?0.

In order to prove Theorem 6.2.21, we need several lemmas.

Lemma 6.2.22 Suppose that AE£(H) and each operator in the similarity orbit S(A) of A is irreducible, then each operator in S(A"(A)) is also irreducible, where S{A"(A)) denotes the similarity orbit of A"(A). Proof Assume that Be A" (A) and X is invertible. Since XAX~X <£ (RI), XAX~l commutes with some nontrivial projection P, i.e., P&A'{XAX~l). Thus X^PXeA'iA) and therefore BX~lPX = X~lPXB or XBX^P = PXBX-1. Namely XBX'1 0 {RI).

Lemma 6.2.23 Suppose that Ai€C(H) (i = 1,2,3) and A = Ai®A2®A3

satisfying kerrA A. = {0} (i ^ j). Then A is similar to an irreducible operator. Proof By Lemma 6.2.22, it is sufficient to show that there is some operator in A"(A) that is similar to an irreducible operator. Since kerrA_ A, = {0} (i •£ j), calculations indicate that

A'{A) = {Bi©B2©-B3 : Bi&£(H),i = 1,2,3}.

Therefore, it is sufficient to prove that B = ai©a2©«3 is similar to an irreducible operator, where {«i}f=1 are pairwise distinct complex numbers.

Define

"<*i 1 D' T= 0 a2 1

0 0 a3_

where D is an irreducible operator. Suppose that P&A'(T) is a projection and

Pll Pl2 P l 3

P21 P22 P23

P31 P32 P33

Since {ai}:f=1 are pairwise distinct, Ptj = 0 (i ^ j) and P n = P22 = P33.

It follows from PUD = DPn and De(RI) that P n = P2 2 = P33 = 0 or / . Thus TG(RI). It is easy to see that T^a1®a2®as.

Lemma 6.2.24 Suppose that A i € £ ( C m ) (m < 00), A2, ^ S ^ W ) satas-/2/m# kerrA. A. = {0} (i ^ j ) , t/ien A = Ai®A2®A% is similar to an irreducible operator. Proof By Lemma 6.2.22, it is sufficient to prove that B = a\@a.2@a3

is similar to an irreducible operator, where {&i}f-i are pairwise distinct complex numbers. Without loss of generality, we assume that % = L2(0,1). Define

ai F 0 0 a2 M 0 0 a 3

where M is "multiplication by the idempotent variable", i.e., (Mf)(t) = tf(t), F is a surjective operator from L2(0,1) to C m given by

Ff = ([ tf(t)dt, f t2f(t)dt,-.- , / tmf(t)dt), /eL2(0,l). Jo Jo Jo

It is not difficult to see that T ~ P . It is sufficient to prove that T£(RI). Suppose that PGA'(T) is a projection. Since on 7 aj (i ^ j), P =

Pi©p2©P3, where each P is a projection and P2M = MP3,PiF = FP2. Thus

P2M2 = MP3M = M{MP3)* = M(P2M)* = M2P2.

Therefore

P2M2k = M2kP2 for fc>l.

Using functional calculus we have P2M = MP2 and P2 = P3. The spectral theory of self-adjoint operators asserts that there exists a Borel subset Ec[0,1] such that P2f = XB! for all f£H, where \E denotes the characteristic function of E. If the Lebesgue measure of E, 11(E) = 1, then P2 = P3 = 1. Since P\F = FP2 and since F is surjective, P\ = 1, i.e., P = 1. If n(E) < 1, let Ai,A2,--- ,Am be nonzero, pairwise distinct Lebesgue points of F = [0,1]\ (AeP is a Lebegue point of F if limn{Fn[\ - e, A + s})/2e = 1).

Since the matrix

Ai A2 Aj A2

• • A m

1 • A_,

A"

is invertible, there is a 8 > 0 such that the matrix (aij)mXm is invertible provided that |ay — AM < 5 (i, j = 1,2, • • • , m. Choose a sufficiently small £ > 0 such that

'2e / fdt-XU <6, i,j= 1,2, • ,m.

[\i-e,\i+£]nF

S e t fj = i^XlXi-e.K+e^F- Since EnF = 0,

P1Ffj = FP2fj = FxEfj = 0 (j = 1,2, • • • , m).

On the other hand, if {efc}jj=1 is an ONB of C m , we have

0 = P1Ffj = P ^ tfjdt, fi t2fjdt, ...,fi tmfjdt)

fc=l [Ai-e,Ai+e]nF

E(i / tkdt)Pxek. k=l [\i-e,Xi+e]nF

Define

Oti 2e I tldt,

{\i-E,Xi+e]r\F

then (ay)mxm is invertible. Thus Piek = 0 (A; = 1,2, •••,m), i.e., Px = 0. Since FP 2 = P\F,FP2 = 0. Note that FP2e = (^tXEdt,J^t2XEdt,--- ,^tm

XEdt), where e&H2,e(t) = 1. Therefore fi{E) = 0, i.e., P2 = -P3 = 0 and P = 0. Thus T€(PJ").


Proof of Theorem 6.2.21 By the arguments similar to that used in the proof of Theorem 6.2.9 we can prove the necessary condition of the theorem. Now we prove the sufficient condition. If (i) and (ii) of the theorem are satisfied, then there are three possibilities:

(a) a(N) consists of infinitely many points; (b) ae(N) consists of at least three points; (c) cr(N) is a finite set, cre(N) consists of two points and a(N)\cre(N) ^

0.

In cases (a) or (b), there exist pairwise disjoint Borel sets <Ji,(T2 and 03 such that

a(N) = 0-1U0-2U0-3

and

dimE(ai) = 00 for i = 1,2,3,

where E(-) is the spectral measure of N. Set Nj = N\E(ai)H (* = li 2,3). Then kerrN^N, = {0} (i ^ j). By Lemma 6.2.23, N is similar to an irreducible operator.

In Case (c), assume that cre(N) = {Ai, A2}. Thus

dimE({\i})H = 00 (i = 1,2)

and

0 < dimE{a(E)\ae(N)) < 00.

By Lemma 6.2.24 N is similar to an irreducible operator.

Proposition 6.2.25 Given T££(H) such that A'{T) is abelian, T is similar to an irreducible operator. Proof (i) Suppose that there exists a projection P&A'(T) such that

dimPTi = 00

and

A = T\PHe(RI).

Fix a \<jLa{A). Then

\®Ae£({PH)s-®PH)£A'(T).

Since A'{T) is abelian, A'{T) = A"(T). Thus

X®AGA"{T).


Choose D£JC(PH, (PH)~L) such that D is surjective, then

\®A~B XD 0 A

{PH)L

PH '

By Lemma 6.2.22 it suffices to prove that BE(RI). Suppose that

P = Pll Pl2

P21 P22 €A'(B)

"Ai

_ 0 A2

0 '

As.

is a projection. Since A ^ CT(J4),P2I = P12 = 0- Since A is irreducible, P22 = 0 or I. It follows from DP22 = PnD and D is surjective that P2 2 =Pn=0oi I and Be(RI).

(ii) If there is not any projection PGA'(T) with infinite rank such that T\PT-[£(RI), then we can find projections Qi, Q2 and Qz&A'(T) such that

dimranQi = 00 (i = 1,2,3).

Set Ti = T\ranQi (i = 1,2,3), then T = T1®T2@T^. Thus

ranQi ranQ2e„4'(T) = ,4"(T) ranQz

where {A4}?=1 are pairwise distinct numbers. By Lemma 6.2.23, F is similar to an irreducible operator and by Lemma 6.2.22, T is similar to an irreducible operator.

Proposition 6.2.26 Every Cowen-Douglas operator is similar to an irreducible operator.

In order to prove Proposition 6.2.26, we need some lemmas.

Lemma 6.2.27 Given B&Bn(fl) and A 6 fl. Denote

Hk = ker(B - X)kQker(B - A)*"1 (fc = 1,2, • • •),

then

H = @Hk

fc=i


and

B =

A B12 B13 *

A S23

0

where dirnHk = n and each Bkk+i is invertible. Furthermore, there is a unitary operator U such that

UBU* =

XB12 * 0 A £ 2 3

0

H2 (6.2.14)

where Bkk+i is a positive invertible operator for each i. Proof Without loss of generality, we assume that A = Oefl. It is easily seen that dimHn = n and

B

0 B12 B13 *

0 B 2 3

0 ' • •

Let ko = min{k : kerBkk+i 7 {0},&>1}. Then there exists a vector fco

Xk0+i 7 0 such that Bk0k0+ixk0+i — 0- Note that M = 0 Hk = kerBko

and dimM = &on. But y = (0,••• ,0,Xfco+i,0,• • -)&kerBk° and y $. M. This contradiction indicates that each Bkk+i is invertible. Thus B12 — UiB[2, where U\ is a unitary operator and B[2 is a positive invertible operator. Similarly,

U1B23 = ^2-B23 , U2B34 = UzBM, • • • , UkBk+i fc+2 = Uk+iB'k+1 fc+2, •

and each Uk is a unitary and each B'k k+1 is positive and invertible.


Define U = ® Uk, where U0 = I, then U is a unitary operator and fc=0

UBU* =

0 S i2 *

0 0 B23

0

0

where Bkk+i = ^kB'kk+lUk (k>l) is positive and invertible.

Lemma 6.2.28 Given B£Bn(Q) with the representation (6.2.14). If £ l f c oe(i?J) for some k0>2, then BG(RI).

Proof Suppose that PGA'(B) is a projection, then P admits the following oo

expression with respect to the decomposition H = © Hk, k=l

P =

Pi Pl2 * P 2 P23

0

Since P* = P, P = Pi©P2© • • • . Since PkBk fc+i — -Bfcfc+i-Pfc+i and since Bkk+i is positive and invertible, Pk = Pi for k>2. Thus

Blk0Pko = PlP>lfc0-

Since Blkoe(RI),Pk = Pi = 0 or I (k = 2,3, • • •), i.e., P = 0 or / . Therefore Be(RI).

Proof of Proposition 6.2.26 without loss of generality, assume that Oefi. Because of Lemma 6.2.27, we may assume that

B =

where each Bkk+i is positive and invertible. Set

B\ = {B12, B13, • • • )£C{H1 , Tii)

0 5 i 2 * " 0 B23

o '•• 0

Hi

n2 W3'


and

B2

0B23 * 0 £34

0 '•• £C{Hi,Hi).

Let Jn be the nxn Jordan nilpotent. Then Jn£C(Cn)t~\(SI). Set

and

X1

E — B\2 (-^13 — Jn)

1 £ 0 * 1 0

1 '•• &C{Hi,Ht)

Set

X--

Then X is invertible and

XBX-1 =

1 0 OXi

€C(H).

"1 0 ' o i l

'0 Bi 0B2

'0 B1X11

OXtl h xr>\

"1 0 OXf 1

0 B12 Jn * 0 B23

0 '••

By Lemma 6.2.28, XBX^^RI).

6.3 Application of Operator Structure Theorem

Theorem 6.3.1 Let SI be a connected open subset ofC and (Sl)° be finitely connected. Assume that {Pk(z)}™=i,z^ *s a class of Mn(C)-valued holo-morphic functions such that:

(i) Pk(z)eMn(H°°(Sl)), l<k<m; (ii) Pi{z)Pj(z) = SijPj(z),zen,l<i,j<m;

m (Hi) Y.Pk{z)~In-

fc=l

Then for fixed ZQ^SI, there exists an Mn(C)-valued holomorphic invert-ible function X(z)€Mn(H°°(Sl)) such that

X{z)Pk{z)X-\z) = Pk(z0),

where l<k<m and X(ZQ) = In-Proof Let H = L2

a(ST) denote the Bergman space on SI* and Bz be the "multiplication by z" on L2

a(SV). Then B*z£Bi(SI) and Al(Bz)^A'(B*z)^H°°(Sl). Set A = B*z and T = A™, then A'(T) is iso-metrically isomorphic to Mn(H°°(Sl)). Thus there exist Pk€A'(T) such that Pk\ker(T-z) = Pk(z) for k : l<k<m,

m

PiPj = 5ijPi,l<i,j<m and £ Pfc = JW(»).

Denote Hk = PkH{n) (l<k<m), then

Denote Tk = T\uk- Since T^ has a finite (57) decomposition up to similarity (Theorem 5.5.11), there exist invertible Xi such that

XiTiXr1 = A(ni\

Set X = X:-i- • • • +Xm, then XeA'(T) and

XPkX~l = 0Wl© • • • ®QHk-M-Hk®Qnk+l® • • • ©0Wm.

Denote Y(z) = X| f c e r ( T_2 ) , then Y(z)&M„(H°°(Sl)) and

Y(z)Pk(z)Y(z)-1 = 0 f c e r ( T l _ z ) © • • • ®Iker(Tk-z)®Oker(Tk+1-z)®- • " © 0 f e e r ( T m _ z ) .

Set X(z) = Y~1(z0)Y(z), then X(z)eMn(H°°(Sl)) and

X(z)Pk(z)X-1(z) = Pk(z0)

and X(zo) = /„ , l<k<m.

In the following, we will discuss the winding number of some analytic functions.

Let D be the unit disk and / be an analytic function on D. For a&D, the winding number of / (e l t ) with respect to / ( a ) is given by

1 Z"1 f'(eu) W(f,f(a)) = — / f ^ > dt.

Lemma 6.3.2 Given feH°°. If there is an aeD such that the inner function inn(f — f(a)) of f — / ( a ) is a finite Blachke product, then there exists a natural number n such that Tf~Tg , where g&H°° and Tg€(SI). Proof Denote h = inn(f — / ( a ) ) , which is a finite Blachke product. By Corollary 2.1 of [Cowen, C.C. (1978)], there exists a finite Blachke product <p such that

A'(Tf) = A'(Tv).

Note that Tv is an isomorphic operator with codimranTv = n < oo, thus

A'(Tv)^A'(Tzn)^Mn(H°°).

By Theorem 5.5.11, V(A' '(T/))SN. If

codimranTu — 1,

then

A'(TV)^A'(TZ)^H°°.

Thus

TMSI).

If codimranTv > 1, by von-Neumann-Wold theorem, there exists a unitary operator U such that UTVU* = Tz, and A'{UTVU*) = A'iT^^M^H00). Let Tz

(n) act on (H2)^ and Pt be the projection from (H2)^ to the i-th copy of subspace H2, then Tz \P.(H2)(.n) = Tz. Let Ti = UTfU*\P.^H2yn), then

n

UTsU* = @Ti and T^{SI). i=l

Clearly, T{TZ = TZT{ (t = l , 2 , - - - , n ) . Thus there exist / i , /2, • • • , fn€H°° such that T* = T/4. This implies that

T l

UTtU* = @Tti, TMSI). i=l

Note that V(-4 ' (7»)=N. By Theorem 4.2.1, Tft~Tfl (i = 2,3,- •• ,n) .

Thus Tf~T* . Denote f\ = g, the proof of the theorem is complete.

Theorem 6.3.3 Let f be an analytic function on D, if Tf $ (SI) and

W(f,f(ao)) = p, a prime number, for some cto&D, then for each a&D, W(f, f(a)) = kp, where k is a natural number. Proof Since W(f,f(a0)) = p is a prime number, inn(f — f(a0)) is a finite Blachke product. Since Tf £ (SI), it follows from Lemma 6.3.2 that T /~Tg(n) for some geH°° and TgG(SI). Thus

codimran{Tf — f(cto)) = ncodimran{Tg — g(ao)) = p.

But p is a prime number, thus codirnran(Tg — g(ao)) = 1 and n = p. Therefore W(f, f(a)) = pW(g,g{a)) = kp for all a<=D.

6.4 Remark

Lemma 6.1.3 is given by [Fang, J.S.(2003)]. Theorem 6.1.1 is due to [Fang, J.S.(2003)], [Jiang, C.L. (1991)]. Section 6.2 is the work of [Jiang, C.L. and He, H. (2004)], [Jiang, C.L., Guo, X.Z. and Ji, K.]. Example 6.2.6 is given by [Fong, C.K. and Jiang, C.L. (1993)]. All the contents of Section 6.3 are given by [Fang, J.S.(2003)], [Jiang, C.L. (1994)].

6.5 Open Problems

1. Let A£C(H). If A2 is irreducible, does A have non-trivial invariant subspaces? 2. Give the necessary and sufficient conditions for A~B if A^~B^2\ where A,B££(H). 3. Conjecture: Given Ae£(H), if o-(A) is uncountable, then A is quasisim-ilar to an irreducible operator.

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Index

Banach Algebra, 1 Banach reducing decomposition, 109 Blaschke product, 10 bundle

holomorphic bundle, 5 Hermitian holomorphic

bundle, 5

commutant, 3 cyclic

strictly cyclic, 11

eigenspace, 13 generalized eigenspace, 13

entry map, 174 equivalent

algebraic equivalent, 3 similar equivalent, 3 stably equivalent, 3 unitary equivalent, 5

locally unitary equivalent, 5 essentially commutative, 170

finite decomposition, 172 Fredholm domain, 7

semi-Fredholm domain, 8 free matrix algebra, 171

free matrix algebra of n, 171 function

inner function, 10 outer function, 10

group Grothendieck group, 3 ordered group, 3

idempotent, 14 minimal idempotent, 14

index, 8 minimal index, 10 minimal index of Cowen-Douglas

operator, 85 invariant

completely similarity invariant, 15 completely unitary invariant, 43 similarity invariant, 15 unitary invariant, 43

Jocobson radical, 2 Jordan block, 13

multiplicity, 127

n-homogenuous, 169 normal eigenvalue, 9

operator Cowen-Douglas operator, 5 essentially normal operator, 9 Fredholm operator, 8 holomorphic idempotent, 85 irreducible operator, 7 nilpotent operator, 212 normal operator, 43

247


strongly irreducible operator, 7 subnormal operator, 43 Toeplitz operator, 10

analytic Toeplitz operator, 10 type-1 operator, 170 typical strongly irreducible

operator, 170 unilateral weighted shift ,69

operator weighted shift, 127 bilateral operator weighted shift,

127 unilateral operator weighted shift,

126

representation, 169 resolvent set, 1

left and right resolvent set, 1

similarity orbit, 8 closure of similarity orbit, 8

six-term exact sequence, 4 Sobolev disk algebra, 98 spectral family, 40 spectral picture, 8 spectrum, 1

left spectrum, 1 point spectrum, 6 right spectrum, 1 Wolf spectrum, 8

strongly irreducible decomposition, 40 unique strongly irreducible

decomposition, 41

structure of Hilbert Space Operators This book exposes the internal structure

of non-self-adjoint operators acting on

complex separable infinite dimensional

Hilbert space, by analyzing and studying

the commutant of operators. A unique

presentat ion of the theorem of

Cowen-Douglas operators is given. The

authors take the strongly irreducible

operator as a basic model, and find

complete s imi la r i ty invariants of

Cowen-Douglas operators by using K-

theory, complex geometry and operator

algebra tools.

5993 he

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