Lifting Module Maps Between Different Noncommutative Domain Algebras

Transcription

1 University of Denver Digital DU Electronic Theses and Dissertations Graduate Studies Lifting Module Maps Between Different Noncommutative Domain Algebras Jonathan Von Stroh University of Denver Follow this and additional works at: Recommended Citation Von Stroh, Jonathan, "Lifting Module Maps Between Different Noncommutative Domain Algebras" (2010). Electronic Theses and Dissertations This Dissertation is brought to you for free and open access by the Graduate Studies at Digital DU. It has been accepted for inclusion in Electronic Theses and Dissertations by an authorized administrator of Digital DU. For more information, please contact jennifer.cox@du.edu.

2 Lifting Module Maps Between Different Noncommutative Domain Algebras A Dissertation Presented to the Faculty of Natural Sciences and Mathematics University of Denver in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy by Jonathan C. Von Stroh August 2010 Advisor: Alvaro Arias

3 c Copyright by Jonathan C. Von Stroh, All Rights Reserved

4 Author: Jonathan C. Von Stroh Title: Lifting Module Maps Between Different Noncommutative Domain Algebras Advisor: Alvaro Arias Degree Date: August 2010 Abstract The classical Carathéodory interpolation problem is the following: let n N, a 0, a 1,..., a N C, and D the unit disk. When does there exist an analytic function F : D C and complex numbers a N+1, a N+2,... such that F (z) = a 0 + a 1 z + a 2 z a N z N + a N+1 z N+1 + and F < 1? In 1967, Sarason used operator theory techniques to give an elegant solution to the Carathéodory interpolation problem. In 1968, Sz.-Nagy and Foias extended Sarason s approach into a commutant lifting theorem. Both the theorem and the technique of the proof have become standard tools in control theory. In particular, the commutant lifting theorem approach lends itself to a wide range of generalizations. This thesis concerns one such generalization. Arias presented generalizations of the original commutant lifting theorem relating to the full Fock space. Popescu then refined the approach by introducing domain algebras. While Arias and Popescu focused on module maps from one noncommutative domain algebra to itself, we present a unitarily equivalent representation which allows us to easily generalize their results to module maps between different noncommutative domain algebras. We use a renorming technique to prove that as long as the formal identity map is bounded, we can lift module maps between different noncommutative domains generated by finite positive regular free holomorphic functions. Finally, we apply the lifting theorem to create projective resolutions. ii

5 Acknowledgements First and foremost, I would like to express my deepest gratitude my advisor, Alvaro Arias, for his guidance, support, patience, and encouragement during my time at the University of Denver. My appreciation for his constant belief in my abilities, from when he hired me as a graduate teaching assistant to my finishing of this dissertation, is beyond words. In addition, thanks to the members of my committee, Jim Hagler, Frédéric Latrémolière, and Nic Ormes, for providing numerous suggestions on how to improve my dissertation. They all helped to shape this document into what it is today. In addition, I would like to thank every member of the Department of Mathematics. Special thanks goes to Rick Ball for his mentoring and for the example of excellent teaching he constantly provides. Thanks to Liane Beights, Sharon Bütz, and Don Oppliger for their support and friendship. Thanks are due to my graduate student colleagues, especially my officemates, Dan Daly, Kyle Pula, and Brett Werner. Their friendship, conversations, and participation in study sessions have been pivotal to my success in my studies. Lastly, my greatest thanks go to my family, who have helped to shape me into the person I am today. I am eternally grateful to my parents for providing me with the opportunity and encouragement to attend first-class institutions such as St. Olaf College and the University of Denver. Thanks go to my siblings, Christina and Justin, for always giving me support and encouragement. Most importantly, I would like to express my deepest thanks and love to my wife, Jenny Von Stroh, who has supported me in innumerable ways over the past 10 years. Her strength, encouragement, and constant love have been essential in helping me to achieve more than I ever thought possible. iii

6 Table of Contents Acknowledgements List of Figures iii vi 1 Introduction 1 2 Motivation Preliminaries Two Representations of a Hilbert Space The Banach Algebra H Unilateral Shift Operators Interpolation Problems The Full Fock Space The Full Fock Space of a Hilbert space H Left and Right Creation Operators The Noncommutative Disk Algebra A n Hilbert Modules Hilbert Modules Submodules and -Submodules Semi-invariant Subspaces Subquotients Module Maps Weighted Fock Spaces Domain Algebras Positive Regular Free Holomorphic Functions Weighted Shift Operators Weighted Fock Spaces iv

7 5.3 Module Maps Between F 2 (f) and F 2 (g) Examples Lifting Module Maps Lifting Theorems L-Bounded and R-Bounded Functions A Lifting Theorem When ε(f, g) is Bounded A Lifting Theorem When F 2 (g) is the Full Fock Space Submodule Correspondence Between Two Isomorphic Weighted Fock Spaces A Lifting Theorem When ε(f, g) is Bounded Projective Resolutions Future Problems Bibliography 79 v

8 List of Figures 1.1 The commutant lifting theorem of [19] Setup for a lifting theorem A lifting theorem between different noncommutative domain algebras The unilateral shift operator on l The multiplication operator on H The diagram for Theorem Diagram for Corollary The full Fock space for n = Correspondence between domain algebras and weighted Fock spaces Diagram for part (iii) of Proposition Commutative diagram for Theorem Commutative diagram for Theorem Commutative diagram for Lemma Commutative diagram for Theorem Introducing the isomorphic weighted Fock space F 2 (h) H Simplified diagram with weighted Fock space F 2 (h) H Commutative diagram for Theorem Exact sequence of partial isometric module maps Construction of the exact sequence of partial isometric module maps Projective Resolutions Constructing Projective Resolutions Lifting the module map T Completed projective resolution vi

9 6.14 Commutative diagram for Problem Commutative diagram for Conjecture Commutative diagram for Problem Commutative diagram for Conjecture vii

10 Chapter 1 Introduction In [21], Sarason used an operator theoretic approach to solve two different interpolation problems. The first of these problems, the Carathéodory interpolation problem, was introduced by Carathéodory in [6] and extensively studied by Schur in [22] and [23]. The second of these problems, the Nevanlinna-Pick interpolation problem, was posed and solved by Pick in [16]. Unaware of Pick s work, Nevanlinna posed and solved the problem in [13] and expanded his solution in [14]. By taking an operator theoretic approach to these problems, Sarason was able to solve both problems in almost identical fashion. This approach was refined by Sz.-Nagy and C. Foiaş in [24] and [25] into a commutant lifting theorem. The advantage of this approach is that it is very generalizable. This thesis is concerned with one such generalization. The main results of this thesis are contained in Chapters 5 and 6. In Chapter 4, the notion of a module map is introduced. This notion is tied to that of a weighted Fock space in Chapter 5, where we attempt to categorize when module maps exist. This is where the first major result, Theorem is 1

11 obtained. In Theorem 5.3.1, equivalent conditions to the existence of non-zero module maps are obtained. This result allows us to determine when non-zero module maps exist by asking easily computable questions of the underlying weighted Fock spaces. In chapter 6, a generalization of the commutant lifting theorem of [25] is suggested. However, the usual machinery of proving such lifting theorems, Parrott s Lemma ([15]), fails. The second major result of this thesis is obtained in the form of the lifting theorem of Theorem 6.1.7, which is proved using a renorming technique. This renorming technique allows us to partially sidestep Parrott s Lemma, and thus avoid the difficulty encountered using the usual methods of commutant lifting theorems. In [19], Popescu studied noncommutative domains D f (H) B(H) n generated by positive regular free holomorphic functions f. He proved that each such domain has a universal model (W 1, W 2,..., W n ) of weighted orthogonal shifts acting on the full Fock space with n generators. In this thesis, we will look at a unitarily equivalent version of the weighted Fock spaces used by Popescu, which we will denote F 2 (f). We will use a Hilbert module language similar to that of Douglas and Paulsen [8] and Muhly and Solel [12]. As stated in [19], a commutant lifting theorem for F 2 (f) follows directly from [2], which was adapted from a paper of Clancy and McCullough [7]. Assume M and N are both -submodules of F 2 (f) and X : M N is a non-zero module map. This commutant lifting theorem states that there exists a module map X : F 2 (f) F 2 (f) with X = X such that the diagram in Figure 1.1 commutes. 2

12 F 2 (f) P M M 0 X F 2 (f) X P N N 0 Figure 1.1: The commutant lifting theorem of [19]. Notice here that both N and M are -submodules of the same Hilbert module F 2 (f). In this thesis we ask a more general question. What if M and N are instead -submodules of F 2 (f) and F 2 (g) respectively? This gives the diagram shown in Figure 1.2. F 2 (f) F 2 (g) P M M 0 X P N N 0 Figure 1.2: Setup for a lifting theorem. In this case, there is certainly no reason why we should be able to lift X to a module map between F 2 (f) and F 2 (g) in general. Indeed, there are examples of spaces F 2 (f) and F 2 (g) such that there does not exist a non-zero module map X : F 2 (f) F 2 (g) of any kind. But if non-zero module maps do exist between F 2 (f) and F 2 (g), under what cases can a lifting theorem be proved? Assuming that the formal identity map ε : F 2 (f) F 2 (g) is a contraction, the commutant lifting theorems of [2] and [7] can be adjusted with some care to prove that there exists an X with X = X such that the diagram in Figure 1.3 commutes. 3

13 F 2 (f) P M M 0 X F 2 (g) X P N N 0 Figure 1.3: A lifting theorem between different noncommutative domain algebras. We follow the standard argument of lifting element by element, then iterating using Parrott s lemma [15]. However, this technique breaks down if the formal identity map ε : F 2 (f) F 2 (g) is not a contraction. In the case where F 2 (g) is the full Fock space and ε : F 2 (f) F 2 (g) is bounded, it immediately follows that ε is a contraction, and thus any module map X : M N can be lifted to an X : F 2 (f) F 2 (g) with X = X. This suggests that X : M N can be lifted (with some penalty in the norm) in general, as long as there exists a non-zero module map between F 2 (f) and F 2 (g). To prove that this is indeed the case, we apply a renorming technique to F 2 (f) to create the situation where we can apply our lifting theorem. It turns out that if the formal identity map ε : F 2 (f) F 2 (g) has norm ε = C, then there exists a module map X : F 2 (f) F 2 (g) with X C X such that the diagram in Figure 1.3 commutes. Chapter 2 will present some background material needed to approach the subject. This will include a brief look at both the Carathéodory and Nevanlinna-Pick interpolation problems. In order to present a solution to these problems, both Sarason s approach and the Sz.-Nagy-Foiaş refinement will be presented. Chapter 3 will introduce the full Fock space. In Chapter 4 we will give a brief overview overview of the Hilbert module language. Chapter 4

14 5 will discuss the weighted Fock spaces given in [2] and [19]. A unitarily equivalent description of these weighted Fock spaces will the be presented, which will allow some aspects of weighted Fock spaces to be addressed with greater transparency. Finally, a lifting theorem for a specific class of weighted Fock spaces will be given in Chapter 6. For convenience, we will assume that our functions f and g are finite. 5

15 Chapter 2 Motivation We will now introduce some preliminaries which will be necessary for this thesis. We begin by discussing two representations of the infinite separable Hilbert space as well as the Banach algebra H. We will then look at the unilateral shift operators and multiplication operators. These operators play a fundamental role in our study of noncommutative domain algebras. Finally, we will introduce some historical interpolation results. 2.1 Preliminaries Two Representations of a Hilbert Space We will begin by discussing two representations, l 2 and the Hardy space H 2, of the infinite separable Hilbert space. 6

16 Definition The Hilbert space l 2 is the collection of all square-summable sequences (a 0, a 1, a 2,...), where a i C: l 2 := { } (a n ) n=0 : a n 2 <. n=0 The norm of the sequence (a n ) n=0 is given by (a n ) n=0 2 = ( n=0 a n 2 ) 1 2. The space l 2 is an infinite dimensional separable Hilbert space, and thus has a countable orthonormal basis given by the set {e n } n=0, where e i = (0, 0,..., 0, 1, 0, 0,...) is the sequence with zero in every coordinate except the ith coordinate, which is 1. Definition The Hardy space H 2 is the collection of all analytic functions on the unit disc whose power series representation has square-summable complex coefficients: H 2 := { } f = a n z n : a n 2 <. n=0 n=0 The norm of the analytic function f = f 2 = ( n=0 a n z n is given by n=0 a n 2 ) 1 2. Let f : D C be an analytic function in H 2 and let θ T be given. 7

17 Define f r (θ) by f r (θ) = f ( re iθ). Now lim f r(θ) exists almost everywhere. r 1 Thus, define F (θ) = lim f r(θ). Since f r (θ) = r 1 n=0 a nr n θ n, we get that for r < 1, f r 2 2 = a n 2 r 2n < a n 2 = f 2 H 2 n=0 n=0 Thus, the functions f r are bounded for each r < 1. By Hoffman ([10], pg 32), it follows that F L 2 (T). From here we get the following alternative definition of the Hardy space H 2 : Definition The Hardy space H 2 can be thought of as the following subset of L 2 : H 2 := { f = L 2 (T) : f(n) } = 0 for n < 0. With the following norm: f 2 = ( 1 2π f(e iθ ) ) dθ. 2π 0 Define a map F : l 2 H 2 by (a n ) n=0 n=0 a nz n. It is easy to see that F defines an onto isometry between l 2 and H 2. Thus, H 2 is an infinite dimensional separable Hilbert space. Furthermore, the orthonormal basis for H 2 is 1, z, z 2, z 3,.... While the Hilbert spaces l 2 and H 2 are isomorphic, it is useful to have both definitions, as each definition will give a different set of insights into the Fock spaces discussed in Chapters 3 and 5. 8

18 2.1.2 The Banach Algebra H Another space which will be of importance is H, the Banach algebra of holomorphic functions on the open unit disc D = {z C : z < 1} with a finite essential supremum. The space H is given in the following definition: Definition The Banach algebra H is given by: H := = { f = } a n z n : ess sup z D f(z) < n=0 { f L (T) : f(n) } = 0 for n < 0. The norm of the bounded holomorphic function f = f = ess sup z D f(z). a n z n is given by n=0 While H is not a Hilbert space, it plays a fundamental role in the study of interpolation problems Unilateral Shift Operators The unilateral shift operators are the foundation upon which the study of domain algebras is based. We will now define the shift operators S : l 2 l 2 and M z : H 2 H 2 : Definition The unilateral shift operator S : l 2 l 2 is defined by Se i = e i+1 for i 0. Definition The multiplication operator M z : H 2 H 2 is defined by M z f = zf. 9

19 e 0 S e 1 S e 2 S e 3 S Figure 2.1: The unilateral shift operator on l 2. 1 M z z M z M z M z z 2 z 3 Figure 2.2: The multiplication operator on H 2. The unilateral shift operator and multiplication operator above can both be generalized. Let p be an arbitrary polynomial in H 2, and define M p : H 2 H 2 as multiplication by p. The unilateral shift operator S : l 2 l 2 can be extended similarly. The multiplication operator M p gives a strong connection between H 2 and H, as shown in the following lemma: Lemma Let p H be given. Then: M p = p Moreover, if f H 2, then M f is bounded if and only if f H. Proof. (Sz.-Nagy and Foiaş [24]) First, let p H 2. Then: p 2 = ( n=0 a n 2 ) 1 2 = ( n=1 a n 1 2 ) 1 2 = zp 2 = M z p 2 10

20 Let p H be given. Then for every f H 2, we have M p f 2 = 1 2π p(e 2π iθ )f(e iθ ) 2 dθ 0 ( ) 2 1 2π sup p f(e z D 2π iθ ) 2 dθ 0 = p f 2 Thus, it follows that M p p. Next, let ɛ > 0 be given. Now since p is a bounded holomorphic function, there exists U T with µ(u) > 0 such that p(z) p ɛ for all z U. There also exists an open set G U with µ(ug) < ɛ. Without loss of generality, we may assume that G = (a, b) T. Now define g C(T) by g(z) = 1 (b a) ɛ if z (a + ɛ, b ɛ) and g(z) = 0 if z (a, b)c. Define g(z) in the obvious linear fashion when z (a, a + ɛ) and (b ɛ, b). It is easy to see that g 2 = 1. Now we can find a polynomial q C(T) such that g(z) q(z) <. If q is analytic, let f(z) = q(z). Otherwise, q(z) = a n z n where ɛ (b a) n= m m > 0. Define f(z) = z m q(z). Now f(z) is analytic and thus f H 2. Then: 11

21 M p f 2 = pf = pz m 1 2π q = pq = p(e 2π iθ )q(e iθ ) 2 dθ 0 1 p(e 2π iθ )q(e iθ ) 2 dθ (a,b) 1 ( p 2π ɛ) 2 q(e iθ ) 2 dθ (a,b) 1 = ( p ɛ) q(e 2π iθ ) 2 dθ (a,b) 1 ( ( p ɛ) g(e 2π iθ ) 2 g(e iθ ) q(e iθ ) 2) dθ (a,b) 1 = ( p ɛ) g(e 2π iθ ) 2 dθ g(e iθ ) q(e iθ ) 2 dθ (a,b) (a,b) 1 2π ( p ɛ) g(e 2π iθ ) 2 ɛ dθ 0 (a,b) (b a) dθ = ( p ɛ) g 2 ɛ = ( p ɛ) 1 ɛ Since ɛ is arbitrary, it follows that M p p. Therefore, M p = p as desired. This leads immediately to the following corollary: Corollary H = { f H 2 : M f < } In this sense, the Banach algebra H can be viewed as an operator algebra. In particular, we will view H as the multiplier algebra of H 2. 12

22 2.2 Interpolation Problems We are now in a position to state some classical problems of interpolation theory. Problem (Carathéodory Interpolation Problem, 1907) Let a 0, a 1, a 2,..., a N be complex numbers such that the function f(z) is analytic at 0. When does there exist an analytic function F : D C with the following properties: F (z) = a 0 + a 1 z + a 2 z a N z N + F 1? Problem (Nevanlinna-Pick Interpolation Problem, 1916) Let N points λ 1, λ 2,..., λ N in the unit disk D and N complex numbers c 1, c 2,..., c N be given. When does there exist an analytic function f : D C with the following properties: f(λ i ) = c i, 1 i N f 1? Necessary and sufficient conditions for both the Nevanlinna-Pick and the Carathéodory interpolation problems were discovered relatively quickly, but the proofs were cumbersome. Furthermore, the effort to generalize these problems was difficult due to the nature of the proofs. However, Sarason ([20], [21]) developed an operator-theoretic approach which allowed these interpolation problems to be solved with greater ease. This approach was formalized by 13

23 Sz.-Nagy and Foiaş ([24], [25]) into the following commutant lifting theorem: Theorem (Sz.-Nagy, Foiaş, 1968) Let H and K be Hilbert spaces with H K, T : H H and W : K K be bounded, and W H = T. If X : H H is given with XT = T X, then there exists Y : K K such that: Y H = X Y W = W Y Y = X ie, the diagram in Figure 2.3 commutes. K P H H 0 Y K X P H H 0 Figure 2.3: The diagram for Theorem We immediately arrive at the following corollary: Corollary Let M be a subspace of H 2 that is -invariant under S. Let X : M M be a contraction that commutes with T = P M S M. Then there exists F : H 2 H 2 such that F M = X, F S = SF, and F = X. Furthermore, there exists φ H such that F = M φ. 14

24 H 2 P M M 0 F =M φ H 2 P M M 0 X Figure 2.4: Diagram for Corollary Using the abovce commutant lifting theorem, the Carathéodory interpolation problem can be addressed with considerable ease, as shown in the following theorem: Theorem (Carathéodory Interpolation Problem, 1907) Given complex numbers a 0, a 1,..., a N, there exists an analytic function f(z) = a 0 + a 1 z + + a N z N + in the unit ball of H (C) if and only if the Toeplitz matrix a a 1 a a 2 a 1 a a N a N 1 a N 2 a 0 is a contraction. Proof. (Sarason, 1968) Choose M = {z k : 0 k N 1}, the closed linear span of 1, z, z 2,..., z N 1. Apply Corollary The Nevanlinna-Pick interpolation problem can be solved with similar ease by using a different choice of M, but we will need one definition before we can state and prove the original Nevanlinna-Pick interpolation theorem. 15

25 Definition Let H 2 and λ D be given. The Szegő kernel at λ, denoted k λ, is the element of H 2 with the property that for all f H 2, f, k λ = f(λ) Note that the Szegő kernel exists since evaluation at any point λ D is a continuous linear functional on H 2. We will now state and prove the original Nevanlinna-Pick interpolation problem: Theorem (Nevanlinna-Pick Interpolation Problem, 1916) Given points λ 1, λ 2,..., λ N in the unit disk D and complex numbers c 1, c 2,..., c N, there is a function φ H satisfying φ(λ i ) = c i for 1 i N if and only if the Pick matrix ( 1 ci c j 1 λ i λ j ) N i,j=1 (2.2.1) is positive semi-definite. Proof. Choose M = {k λi : 1 i N}. Apply Corollary The beauty of the approach developed by Sarason and streamlined by Sz.-Nagy and Foiaş is that the proofs of both the Nevanlinna-Pick and the Carathéodory interpolation problems are virtually identical. More importantly, it allowed more general interpolation results to follow. In the remainder of this thesis, we will explore one such generalization. 16

26 Chapter 3 The Full Fock Space In this chapter, we will introduce the full Fock space F 2 (H) of a Hilbert space H. A formal definition of F 2 (H) will be given in Section 3.1. In Section 3.2, we will define the left and right creation operators. These operators are fundamental in the study of the domain algebras introduced by Popescu in [19] and discussed in Chapter 5. Finally, Section 3.3 introduces the noncommutative disk algebra A n, which turns out to be the universal model of all row contractions. 3.1 The Full Fock Space of a Hilbert space H Let H be an n-dimensional Hilbert space with orthonormal basis e 1, e 2,..., e n. The full Fock space of H is defined as F 2 (H) := k 0 H k = C H (H H) (H H H) 17

27 with the following inner product, defined on elementary tensors: ζ 0 ζ 1... ζ j, ξ 0 ξ 1... ξ k = 0 if j k n ζ i, ξ i otherwise i=0 If the underlying Hilbert space is clear, we shall denote F 2 (H) by F 2 and I H by I. One representation of the full Fock space is given using the free semigroup on n generators. Let F + n be the unital free semigroup on n generators, g 1, g 2,..., g n, together with the identity g 0. Define the length of an element α F + n as α := 0 if α = g 0 and α := k if α = g i1 g i2 g ik, where 1 i 1, i 2,, i k n. For each α F + n, define e gi1 e gi2... e gik if α = g i1 g i2 g ik e α := 1 if α = g 0 It is clear that the set {e α : α F + n } is an orthonormal basis for the full Fock space F 2. Thus, we can identify the full Fock space F 2 with the Hilbert space l 2 (F + n ). Finally, if (T 1, T 2,..., T n ) is an n-tuple of operators T i B(H) and α = g i1 g i2... g ik F + n, define T α to be the product T i1 T i2 T ik. 18

28 3.2 Left and Right Creation Operators The concept of a row contraction is fundamental in both the study of the full Fock space, defined in the previous section, and the domain algebras introduced by Popescu in [19]. Definition Let H be an n dimensional Hilbert space. A row contraction on H is an n-tuple of operators (T 1, T 2,..., T n ), with T i B(H) for every 1 i n, such that n i=1 T i T i I H It is usually beneficial to think of the n-tuple (T 1, T 2,..., T n ) as the operator T = [T 1 T 2 T n ] acting on H n. In this sense, (T 1, T 2,..., T n ) is a row contraction if and only if T is a contraction, since n i=1 T it i I H holds exactly when T T I H n. With this definition in mind, we can begin our discussion on the left and right creation operators. Definition For each i = 1, 2,..., n, the left creation operators L i : l 2 (F + n ) l 2 (F + n ) are defined by L i e α = e gi α for every α F + n. Similarly, for each i = 1, 2,..., n, the right creation operators R i : l 2 (F + n ) l 2 (F + n ) are defined by 19

29 R i e α = e αgi for every α F + n. The left and right creation operators have some important properties, as shown in the following proposition: Proposition The left (respectively, right) creation operators are isometries with orthogonal ranges. Furthermore, they are row contractions: n L i L i I, i=1 n R i Ri I. i=1 Proof. Since the left (right) creation operators map the orthonormal basis of l 2 (F + n ) into the orthonormal basis of l 2 (F + n ), it is clear to see that they are isometries. Now the left (right) creation operators have orthogonal ranges since for all α F + n and i j, L i δ α, L j δ α = δ gi α, δ gj α = 0 R i δ α, R j δ α = δ αgi, δ αgj = 0 Now to show that (L 1, L 2,, L n ) forms a row contraction, let δ α l 2 (F + n ) be given. If α 1, then α = g j β for some j n and β F + n with β = α 1. 20

30 Then: However, if α = 0, then: Thus, ( n ) L i L i δ α = L j δ β = δ α. i=1 ( n ) L i L i δ α = 0. i=1 ( I ) n L i L i δ α 0. i=1 Thus, the left creation operators form a row contraction. An identical argument will verify that the right creation operators also form a row contraction. The full Fock space for n = 2 together with the left creation operators L 1 and L 2 can be visualized as in figure

31 L 1 e 111 L 1 e 11 L 2 e 211 L 1 e 1 L 2 e 21 L 1 e 121 L 2 e 221 e 0 L 1 e 112 L 2 L 1 e 12 L 2 e 212 e 2 L 1 e 122 L 2 e 22 L 2 e 222 Figure 3.1: The full Fock space for n = The Noncommutative Disk Algebra A n The study of noncommutative domains is based on the study various creation operators. When looking at the left creation operators, as defined above, we get the following algebra: Definition The noncommutative disk algebra A n is the norm closure of the algebra generated by the left creation operators and the identity: A n = {L 1, L 2,..., L n, I}. 22

32 The noncommutative disk algebra A n is very important in the study of row contractions due to the following property discovered by Popescu: Theorem (G. Popescu, [17]) Let (T 1, T 2,..., T n ) be an n-tuple of operators on a Hilbert space H. Then (T 1, T 2,..., T n ) is a row contraction if and only if there exists a unique unital completely contractive morphism Φ : A n B(H) such that L i T i for 1 i n. The previous theorem states that the n-tuple of left creation operators (L 1, L 2,..., L n ) is the model of all row contractions. In [19], Popescu developed the notion of domain algebras, the natural generalizations of the noncommutative disk algebra A n with weighted shift operators. Domain algebras will be discussed in Chapter 5, where the weighted shift operators will be shown to be the models of row contractions in their corresponding domain algebras. 23

33 Chapter 4 Hilbert Modules 4.1 Hilbert Modules Let H be a Hilbert space, and V i : H H be bounded linear operators for 1 i n. Then (H; V 1, V 2,..., V n ) is called a Hilbert module over the free algebra generated by n-noncommutative variables. In this thesis, we will primarily consider Hilbert modules of the form (F 2 (f) H; L 1 I H, L 2 I H,..., L n I H ), where F 2 (f) is a weighted Fock space, which will be defined in chapter 5, H is an arbitrary Hilbert space, and the L i s are the left creation operators L i : F 2 (f) F 2 (f). For the purposes of this chapter, you may consider the weighted Fock space F 2 (f) to be the full Fock space of chapter Submodules and -Submodules If N F 2 (f) H is invariant under L i I H for 1 i n, we say that (N ; L 1 I H, L 2 I H,..., L n I H ) is a submodule of F 2 (f) H. Similarly, if M F 2 (f) H is invariant under (L i I H ) for 1 i n and V i = P M (L i 24

34 I H ) M for 1 i n, we say that (M; V 1, V 2,..., V n ) is a -submodule of F 2 (f) H. Every submodule induces a corresponding -submodule, and every -submodule induces a corresponding submodule, as shown in the following lemma: Lemma Let (F 2 (f) H; L 1 I H, L 2 I H,..., L n I H ) be a Hilbert module. Then (N ; L 1 I H, L 2 I H,..., L n I H ) is a submodule of F 2 (f) H if and only if (M; V 1, V 2,..., V n ), where M = H N, is a -submodule of F 2 (f) H Proof. Let (N ; L 1 I H, L 2 I H,..., L n I H ) be a submodule of F 2 (f) H. Then for x N and y M, x, L i y = L i x, y. But since (N ; L 1 I H, L 2 I H,..., L n I H ) is a submodule of F 2 (f) H, L i x N, and so L i x, y = 0. Thus, x, L i y = 0 and so it follows that (M; V 1, V 2,..., V n ) is a -submodule of F 2 (f) H, as desired. The other direction is identical Semi-invariant Subspaces Let a submodule N F 2 (f) H and an algebra A B(F 2 (f) H) be given. Then N is said to be semi-invariant under A if for every a, b A it follows thatp N ap N bp N = P N abp N. In [20] Sarason proved the following lemma, which we will state and prove in less generality: Lemma (Sarason, [20]) Let N F 2 (f) H be given. Then N is semiinvariant under L i I H for i n if and only if there exist two submodules N 1 and N 2 with N 1 N = N 2. Proof. (= ) Assume that N is semi-invariant under L i I H for i n. Then define N 1 and N 2 as follows: 25

35 N 2 := p(l)x : p(l) = c α L α I H, x N α F + n (4.1.1) N 1 := N 2 N, (4.1.2) Clearly, N 1 N = N 2, so we only need to show that both N 1 and N 2 are submodules. Now, it is easy to see from that N 2 is invariant under L i I H for i n, and thus, N 2 is a submodule of F 2 (f) H. Thus, we need only show that N 1 is a submodule of F 2 (f) H. This is equivalent to proving the following for every p(l) = α F + c n αl α I H : P N1 p(l)p N1 = p(l)p N1 (P N2 P N )p(l)(p N2 P N ) = p(l)(p N2 P N ) P N2 p(l)(p N2 P N ) P N p(l)(p N2 P N ) = p(l)(p N2 P N ) p(l)(p N2 P N ) P N p(l)(p N2 P N ) = p(l)(p N2 P N ) (4.1.3) P N p(l)(p N2 P N ) = 0 P N p(l)p N2 = P N p(l)p N (4.1.4) Note that follows from the fact that N 2 is a submodule of F 2 (f) H. Next we need to prove that holds for every y N 1. Combining equations and gives us that y = q(l)x, where q(l) = α F + c n αl α I H and x N. We get: 26

36 P N p(l)p N2 q(l)x = P N p(l)p N q(l)x P N p(l)p N2 q(l)p N x = P N p(l)p N q(l)p N x P N p(l)q(l)p N x = P N p(l)p N q(l)p N x (4.1.5) Now is obtained since N 2 is a submodule. Finally, since N is semiinvariant, equation is verified, and thus N 1 is a submodule, as desired. ( =) Assume that there exist two submodules N 1 and N 2 such that N 1 N = N 2. Then: P N pqp N P N pp N qp N = P N p(l)q(l)p N P N p(l)p N q(l)p N = P N p(l)p N2 q(l)p N P N p(l)p N q(l)p N = P N p(l)(p N2 P N )q(l)p N = P N p(l)p N1 q(l)p N = 0 The last equality follows from the fact that N 1 is a submodule of F 2 (f) H together with N 1 N = N 2. Thus, P N p(l)q(l)p N = P N p(l)p N q(l)p N, and N is semi-invariant, as desired. 27

37 4.1.3 Subquotients Now if N F 2 (f) H is semi-invariant under L i I H for i n and W i = P N (L i I H ) N for i n, then (N ; W 1, W 2,..., W n ) is called a subquotient of F 2 (f) H. This leads immediately to the following lemma: Lemma If (N ; L 1 I H, L 2 I H,..., L n I H ) is a submodule of F 2 (f) H, then (N ; L 1 I H, L 2 I H,..., L n I H ) is a subquotient of F 2 (f) H. Similarly, if (M; V 1, V 2,..., V n ) is a -submodule of F 2 (f) H, then it follows that (M; V 1, V 2,..., V n ) is a subquotient of F 2 (f) H. Proof. Assume (N ; L 1 I H, L 2 I H,..., L n I H ) is a submodule of F 2 (f) H. Define N 2 := N and N 1 := 0. Clearly N 1 and N 2 are both submodules of F 2 (f) H. Since N 1 N = N 2, we can apply lemma 4.1.2, giving us that N is a subquotient of F 2 (f) H as desired. Now assume that (M; V 1, V 2,..., V n ) is a -submodule of F 2 (f) H. Define N 2 := H and N 1 := H N. Now N 2 is clearly a submodule of F 2 (f) H, and by lemma 4.1.1, N 1 is also a submodule of F 2 (f) H. Thus, since N 1 N = N 2, we can apply lemma 4.1.2, giving us that N is a subquotient of F 2 (f) H as desired. The converse does not hold, as shown in the following example: Example Let n = 2, and let the full Fock space l 2 (F + n ) be given. Let N 1 = {δ α : α 2} and N 2 = {δ α : α 1}. Then define N := N 2 N 1. It is easy to see that N = {δ α : α = 2}. Now (N 1 ; L 1, L 2,..., L n ) and (N 2 ; L 1, L 2,..., L n ) are both submodules of l 2 (F + n ), and so lemma gives us that (N ; L 1, L 2,..., L n ) is a subquotient of l 2 (F + n ). However, it is clear that 28

38 for any i, j n, L i δ gj = δ gi g j / N, so (N ; L 1, L 2,..., L n ) is not a submodule of l 2 (F + n ). Similarly, for any i n, L i δ gi = δ g0 / N, so (N ; V 1, V 2,..., V n ) is not a -submodule of l 2 (F + n ). 4.2 Module Maps Let X : H K be a bounded linear operator between the Hilbert modules (H; V 1, V 2,..., V n ) and (K; W 1, W 2,..., W n ). If X(V i h) = W i (Xh) for every h H and 1 i n, we say that X is a module map. The Hilbert modules (H; V 1, V 2,..., V n ) and (K; W 1, W 2,..., W n ) are isomorphic if there exists an invertible map X : H K such that both X and X 1 are isometric module maps. For our purposes, it is important to determine when the orthogonal projection and the inclusion map are module maps. To this end, we give the following propositions: Proposition The orthogonal projection P N : F 2 (f) H N is a module map if and only if N F 2 (f) H is a -submodule of F 2 (f) H. Proof. (= ) Let N F 2 (f) H. Assume that the orthogonal projection P N : F 2 (f) H N is a module map. Let x N be given. Then: L i x = L i P N x = (P N L i ) x = (V i P N ) x = P N V i x = P N L i x Thus, N is a -submodule, as desired. ( =) Let N F 2 (f) H be a -submodule of F 2 (f) H. Let x 29

39 F 2 (f) H be given. Then: (P N L i ) x = L i P N x = P N L i P N x = V i x = P N V i x = (V i P N ) x Thus, P N is a module map, as desired. Proposition The inclusion map ι : N F 2 (f) H is a module map if and only if N is a submodule of F 2 (f) H. Proof. (= ) Let N F 2 (f) H. Assume that the inclusion map ι : N F 2 (f) H is a module map. Let x N be given. Then: L i x = L i ιx = ιl i x But since the domain of ι is N, it follows that L i x N. Thus, N is a submodule, as desired. ( =) Let N F 2 (f) H be a submodule of F 2 (f) H. Let x N be given. Then: L i ιx = L i ιp N x = L i x = ιp N L i x = ιl i x Thus, ι is a module map, as desired. 30

40 Chapter 5 Weighted Fock Spaces In this chapter we will first present the non-commutative domain algebras introduced by Popescu ([19]). Then we will present a unitarily equivalent description of the non-commutative domain algebras which will allow us to more easily lift module maps between these algebras. We will spend some time determining when module maps exist between two different weighted Fock spaces. Finally, we will present some examples which demonstrate different properties of the weighted Fock spaces. 5.1 Domain Algebras Positive Regular Free Holomorphic Functions Definition (Popescu, [19]) Look at the formal power series over n free variables (X 1, X 2,..., X n ) given by f = a f αx α for scalar a f α. Then f is a α F + n positive regular free holomorphic function on B(H) n if the following properties on a f α are satisfied: 31

41 sup k N a f g 0 = 0 a f g i > 0 1 i n a f α 0 α F + n, α > 1 1 2k a f α 2 = M < α =k In Chapter 3 we looked at n-tuples of operators of the form (T 1, T 2,..., T n ). Specifically, we asked the following question: when is (T 1, T 2,..., T n ) a row contraction? The question was answered in Theorem 3.3.2, where it was discovered that the unilateral shift operators (L 1, L 2,..., L n ) acted as the model of all such row contractions. In [19], Popescu asked a more general question. Given a positive regular free holomorphic function f, when did the n-tuple (T 1, T 2,..., T n ) B(H) n belong to the noncommutative domain algebra D f (H) = (X 1, X 2,..., X n ) B(H) n : a f αx α Xα I H? α F + n The answer to this question rests with weighted shift operators, which we will describe in the next section. 32

42 5.1.2 Weighted Shift Operators We will now briefly summarize the construction of the weighted shift operators W f i : l 2 (F + n ) l 2 (F + n ) which together will act as the model of n-tuples in D f (H). Let f = be a positive regular free holomorphic function. Choose r R α F + n such that rm < 1. Then: 2 a α r k S α α =k 2 = a α r k S α, a β r k S β α =k β =k = r 2k a α S α, a β S β = r 2k α =k α =k β =k β =k a α S α, a β S β = r 2k a α S α, a α S α α =k = r 2k a α 2 α =k (rm) 2k Therefore, it follows that a α r α S α < 1, so I a α r α S α is invertible in A n with its inverse given by b α r α S α for some b α C. Since α F + n α F + n α F + n 1 we know that = x k for any x B(H) n with x < 1, we get that 1 x k 0 b α r α S α = α F + n I α F + n a α r α S α 33 1 = k a α r α S α (5.1.1) α F + n k=0

43 In addition, we also know the following: I a α r α S α b α r α S α = 1 (5.1.2) α F + n α F + n α F + n b α r α S α I α F + n From equation we get several properties of the b α s: a α r α S α = 1 (5.1.3) b g0 = 1 (5.1.4) b α > 0 for all α F + n (5.1.5) b gi = a gi (5.1.6) b α = α i=1 γ 1 γ 2 γ i =α γ 1 1,..., γ i 1 a γ1 a γ2 a γi > 0 (5.1.7) b αβ b α b β for any α, β F + n (5.1.8) From equations and we get another important property of the b αs which holds when γ 1: b γ = βα=γ β 1 a β b α = αβ=γ β 1 a β b α (5.1.9) We can now define the weighted left creation operators W f i : l 2 (F + n ) l 2 (F + n ) for i = 1, 2,..., n associated with the noncommutative domain D f (H). 34

44 In particular, for α F + n, define W f i e α = b α b gi α e gi α. A few easy calculations give us the following: a α W α W α I α F + n W β e α = W β e α = W α = b α e βα b βα b γ b α e γ 1 bα if α = βγ; 0 else We can define the weighted right creation operators Λ f i : l 2 (F + n ) l 2 (F + n ) for i = 1, 2,..., n associated with the noncommutative domain D f (H) in a similar fashion. In particular, for α F + n, define Λ f i e α = b α e αgi. b αgi The weighted right creation operators have similar properties to the weighted left creation operators. Some of these properties are shown in Figure 5.1. We are now in a position to state the following theorem: 35

45 Theorem (G. Popescu, [19]) Let f = a f αx α be a positive regular free holomorphic function. Then: α F + n a f αw f α W f α I α F + n Furthermore, if (T 1, T 2,..., T n ) is an n-tuple of operators acting on a Hilbert space H, then a f αt α Tα I if and only if there exists a unique unital α F + n completely contractive morphism Φ from the algebra generated by the weighted { } left creation operators W f 1, W f 2,..., Wn f into B(H) such that W f i T i for 1 i n. A similar theorem holds for the weighted right creation operators. If we consider the Fock space l 2 (F + n ) together with the weighted shift operators W f 1, W f 2,..., W f n defined in the previous theorem, we get the Hilbert module (l 2 (F + n ) ; W f 1, W f 2,..., W f n ). 5.2 Weighted Fock Spaces It is more useful for us to work with a unitarily equivalent version of the weighted Fock space, which we will call F 2 (f). For a positive regular free holomorphic function f, set F 2 (f) to be the Hilbert space with a complete orthogonal basis { } δα f : α F + n, where the norm is given by δ f α = 1. b f α For each i n, define the left creation operators L i : F 2 (f) F 2 (f) by L i δ α = δ gi α. Similarly, for each i n, define the right creation operators R i : F 2 (f) F 2 (f) by R i δ α = δ αgi. We obtain this unitarily equivalent version via the unitary operator U : l 2 (F + n ) F 2 (f) defined as Ue α = 36

46 bα δ α. Furthermore, U δ α = 1 bα e α. Verifying that this unitary satisfies the conditions is easy. We will verify the most important relation: U L β Ue α = b f αu L β δ α = b fαu δ fβα = b f α e βα = W f b f β e α βα The correspondence with Popescu s version is shown in figure 5.1, with the left column containing Popescu s representation and the right column containing our representation. l 2 (F + n ) F 2 (f) e α δ α a α a α e α = 1 δ α = W β e α = bα b βα e βα L β δ α = δ βα W β = 1 b β L β = 1 b α 1 b β Λ β e α = b α b α β e α β R β δ α = δ α β Λ β = 1 b β R β = 1 b β Figure 5.1: spaces. Correspondence between domain algebras and weighted Fock If we consider the weighted Fock space F 2 (f) together with the weighted shifts L f 1, L f 2,..., L f n, we get the Hilbert module (F 2 (f); L f 1, L f 2,..., L f n). If the underlying positive regular free holomorphic function is understood, we will refer to L f α and δα f simply as L α and δ α, respectively. It turns out that not only is U a unitary operator, but it is also a module map, as shown in the following lemma: 37

47 Lemma Let (l 2 (F + n ) ; W f 1, W f 2,..., W f n ) and (F 2 (f); L 1, L 2,..., L n ) be given, and let Ue α = b α δ α as above. Then U : l 2 (F + n ) F 2 (f) is an isometric module map. Proof. To prove that U is a module map, let e α F + n be given. Then: UW β e α = U bα bβα e βα = b βα bα bβα δ βα = b α δ βα = b α L β δ α = L β Ue α Clearly U is isometric, since b α δ α = bα δ α = b α bα = 1 = e α Notice that the b α s end up being identical between spaces, as they are solely determined by the a α s. Let α = g i1 g i2... g ik be given. Denote by α the reverse of α, α = g ik... g i2 g i1. It is clear to see that if f(x 1, X 2,..., X n ) = a α X α is a positive regular free holomorphic function on B(H) n, then so α F + n is f(x 1, X 2,..., X n ) = this work: α F + n a α X α. We will now state a result we will need for Proposition (G. Popescu, [19]) Let (L 1, L 2,..., L n ), (R 1, R 2,..., R n ), and f = a α X α be given. Then: (i) β 1 (ii) β 1 α 1 a β L β L β I and a βr β R β I. (iii) V R f i V = L f i ( ) for 1 i n, where V : F 2 f F 2 (f) is the unitary operator defined by V δ α = δ α for α F + n. 38

48 Proof. We will first verify (i). To do this, we will take advantage of our unitary equivalence and use the following result from Theorem 5.1.2: Thus, it follows that a β W β Wβ I β 1 a β L β L β = a β UW β U UWβ U = a β UW β Wβ U β 1 β 1 β 1 = U a β W β Wβ U UIU = I β 1 as desired. The proof of (ii) is identical. The third is as follows. Let α F + n : V R f i V δ α = V R f i δ α = V δ αgi = δ αgi = δ gi α = L f i δ α This proves part (iii). This part of the proof is deceptively easy, and does ( ) not explain why the change from F 2 f to F 2 (f) is necessary. Here is a more cumbersome proof utilizing the unitary equivalence to the domain algebras of [19] and the result in that space that states that V Λ f i V = W f i with V eã = e α : (l 2 (F + n ); W f 1, W f 2,..., W f n ) V (l 2 (F + n ); W f 1, W f 2,..., W f n ) Ũ ( ) F 2 f V U F 2 (f) Figure 5.2: Diagram for part (iii) of Proposition

49 Since the diagram in figure 5.2 commutes, we can do the following computations: V Λ f i V e α = W f i e α V U R f i U V e α = Ũ L f i Ũe α Ũ V U R f i U V Ũ b f α δ α = L f i b f α δ α Ũ V U R f i U V Ũ δ α = L f i δ α side: The right side is easy to compute, since L f i δ α = δ gi α. Computing the left Ũ V U R f i U V Ũ δ α = = = = 1 b f α b f α b f α b f α b f α b f α b f α = δ gi α Ũ V U R f i U V e α = 1 b f α Ũ V U R f i Ue α Ũ V U R f i δ b f α α = Ũ V U δ αgi 1 b fαgi Ũ V e αgi = b f αg i δ αgi = b f αg i b f α b f α b f α b f α b f α 1 b fαgi Ũe αgi b f αg i δ gi b f α αg i The last equality comes from the fact that b f α = bf α for all α F + n. This ( ) is where the change from F 2 f to F 2 (f) is necessary, since it is not always the case that b f α = bf α. 40

50 5.3 Module Maps Between F 2 (f) and F 2 (g) Let (F 2 (f); L 1, L 2,..., L n ) and (F 2 (g); L 1, L 2,..., L n ) be two Hilbert modules. Define the formal identity map ε(f, g) : F 2 (f) F 2 (g) to be the linear map ε(δα) f = δα. g When f and g are clear, we will just refer to ε(f, g) as ε. By this definition, ε, if bounded, acts like the formal identity map. However, it is important to note that ε does not have to be an isometry, need not be bounded, and isn t necessarily well-defined. Now an element x F 2 (f) looks like x = x α δ α, so we can think of an element in F 2 (f) as the sequence α F + n x = {x α : α F + n }. We will therefore use F 2 (f) F 2 (g) to denote sequence inclusion. The main result of this section is Theorem Let (F 2 (f); L 1, L 2,..., L n ) and (F 2 (g); L 1, L 2,..., L n ) be two Hilbert modules. Then the following are equivalent: (i) F 2 (f) F 2 (g). (ii) ε : F 2 (f) F 2 (g) is well defined. (iii) ε : F 2 (f) F 2 (g) is bounded. (iv) ε : F 2 (f) F 2 (g) is a module map. (v) sup α F + n b f α b g α < Furthermore, if f has finitely many terms, then: (vi) There exists a non-zero module map X : F 2 (f) F 2 (g). We start with a lemma. 41

51 Lemma Let f = a α X α, where f has only finitely many terms. α 1 Then there exists C > 0 such that for all α F + n and i n, b α C b g i α b gi b α (5.3.1) b α C b αg i b α b gi (5.3.2) Proof. We will only prove 5.3.1, as has a similar proof. The right inequality of was proven by Popescu in [19]. Thus, it remains to prove the left inequality of Now b gi α = βγ=g i α β 1 a β b γ Since f is finite, there exists a k N such that for all α with α > k, a α = 0. Thus: b gi α = βγ=g i α 1 β k Let A = max 1 β k {a β}. Then: a β b γ max {a β } b γ βγ=g i α βγ=g 1 β k i α 1 β k b gi α A σγ=α b γ = A 0 σ k 1 = A σγ=α 0 σ k 1 Cb α b α b σ = Ab α σγ=α 0 σ k 1 σγ=α 0 σ k 1 b σ b γ A b σγ b σ σγ=α b σ 0 σ k 1 1 b σ 42

52 where C = A σ k 1 1 b σ. Thus, bα C b g i α, as desired. It is important to note that this result does not extend to arbitrary f s, as shown in the following example: Example Define f = a α X α by: α F + n a α = 1 if α = 1 2 k if α = 12 k 1 0 else Let α = 2 k 1 and i = 1. Let s calculate bα b gi α : b α b gi α = b 2 k 1 b 12 k 1 = βγ=2 k 1 β 1 βγ=12 k 1 β 1 a β b γ a β b γ = (a 2 ) k a 1 a 1 (a 2 ) k a 1 + a 12 k 1 = k This converges to 0 as k, and therefore, no C > 0 exists such that b α C and therefore no C exists such that bα b b gi α C g i α. We are now ready to complete the proof of Theorem Proof. The following implications are clear: (iv) = (iii) = (ii) = (i). We will now complete the loop by showing that (i) = (iv): Assume that x m x in F 2 (f) and that ε(x m ) y in F 2 (g). Now we know that x m = c m,β δ β. A quick calculation reveals that x = c β δ β, β F + n where c β = lim m c m,β for every β F + n : β F + n 43

53 lim m x m, δ α f = lim c m,β δ β, δ α m β F + n f = lim m c m,α δ α, δ α f On the other hand: x, δ α f = c β δ β, δ α = c α δ α, δ α f β F + n f Notice that the above calculations will look exactly the same in F 2 (g). Since ε(δ f α) = δ g α, it is clear to see that ε(x) = y. Thus, by the closed graph theorem, ε is bounded. Now it is trivial to show that ε is a module map: εl f β δf α = εδ f βα = δg βα = Lg β δg α = L g β εδf α As desired. Thus, (i) (ii) (iii) (iv). Now we will show that (v) (iii): Notice the following: ε = sup α F + n = sup α F + n g b ε g f αδ α = sup b f αδ α α F + n b f α b g α b f α δ α g = sup α F + n Thus, ε is bounded if and only if sup α F + n (ii) (iii) (iv) (v). b f α b g α <, as desired. Thus, (i) 44

54 Now assume that f has finitely many terms. Since (iv) = (vi) always holds, we need only prove (vi) = (v): Assume X : F 2 (f) F 2 (g) is a non-zero module map. Then X is completely determined by what it does to δ 0 : Xδ 0 = c β δ β. Then in general, we have: β F + n Xδ α = XL α δ 0 = L α Xδ 0 = L α β F + n c β δ β = β F + n c β L α δ β = β F + n c β δ αβ Let β 0 be such that β 0 0, but for all other β F + n such that β < β 0 it follows that β = 0. Then: Xδ α = c β0 δ αβ0 + β β 0 β β 0 c β δ αβ Now δ αβ0 is orthogonal to δ αβ for every β β 0. Thus: Xδ α g = c β0 δ αβ0 + β β 0 β β 0 c β δ αβ c β0 δ αβ0 g = c β0 δ αβ0 g = g c β0 b g αβ 0 By applying Lemma iteratively, we can peel off the β 0 on the bottom of the fraction. This gives us: Thus, 1 Therefore, it follows that b g α Xδ α g c β 0 K b g α C Xδ α g C X δ α f = C X 1. b f α 45

55 b f α b g α C X Since α was arbitrary, it follows that as desired. sup α F + n b f α b g α < This leads immediately to the following corollary, which gives a simple method to calculate the norm of the ε map: Corollary Let (F 2 (f); L 1, L 2,..., L n ) and (F 2 (g); L 1, L 2,..., L n ) be two Hilbert modules. Assume that f has finitely many terms. Then if ε : F 2 (f) F 2 (g) is a module map, its norm is given by: ε = sup α F + n b f α b g α 5.4 Examples In order to get some intuition about the b α s, we will look at some examples. These examples will show some nice combinatorial connections between the a s and the b s for certain classes of functions f: Example Let f = a α X α, where a α {0, 1} for all α F + n. Then b α = { α 1 α 2 α i = α : a αj α 1 = 1, 1 j i }. In other words, bα counts the number of ways you can decompose α into subwords α = α 1 α 2 α i such that a α1 = a α2 = = a αi = 1. 46