Lifting Module Maps Between Different Noncommutative Domain Algebras

Transcription

1 Lifting Module Maps Between Different Noncommutative Domain Algebras Alvaro Arias and Jonathan Von Stroh University of Denver December 6, 2010 Abstract In this paper we use a renorming technique to lift module maps between -invariant submodules of domain algebras. These algebras were recently introduced by Popescu as non-selfadjoint operator algebras generated by some weighted shifts on the Full Fock space. 1 Introduction In [12], 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 l 2 (F + n ). These algebras are natural generalizations of the algebras generated by the left creation operators on the Full Fock space. Popescu proved that the domain algebras satisfy commutant lifting theorems and have Poisson kernels, and as a result they satisfy the completely bounded version of Caratheodory and Nevanlinna-Pick interpolation problems [12]. This is an area of active research (see [1], [4], [8], and [6]). In the Hilbert module language of Douglas and Paulsen [9] and Muhly and Solel [10], Popescu studied the Hilbert modules (l 2 (F + n ); W 1, W 2,..., W n ) (see sections 2 and 3 for details). In this paper, we will look at a unitarily equivalent version of these Hilbert modules which we will denote by (F 2 (f); L 1, L 2,..., L n ) (see section 3 for details). As stated in [12], 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 following diagram commutes: 1

2 F 2 (f) P M M 0 X F 2 (f) X P N N 0. Notice here that both N and M are -submodules of the same Hilbert module F 2 (f). In this paper 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 following diagram: F 2 (f) F 2 (g) P M M 0 X P N N 0. 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 following diagram commutes: F 2 (f) P M M 0 X F 2 (g) X P N N 0. We follow the standard argument of lifting element by element, then iterating using Parrott s lemma [11]. 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 2

3 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 following diagram commutes: F 2 (f) P M M 0 X F 2 (g) X P N N 0. In section 2 we will give a brief overview of the Hilbert module language. Section 3 will provide a unitarily equivalent description of the Fock spaces given in [2] and [12] 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 section 4. For convenience, we will assume that our functions f and g are finite. 2 Hilbert Modules Let A be the algebra of polynomials in n-commutative variables {g 1, g 2,..., g n }. A Hilbert module H over A is completely determined by the n bounded linear operators L i x = g i x, for i n. Therefore we denote a Hilbert module over A by (H; L 1, L 2,..., L n ). All Hilbert modules in this paper have this form. 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, defined in the next section, the L i s are the left creation operators, also defined in the next section, and H is an arbitrary Hilbert space. If N F 2 (f) H is invariant under L i I H for 1 i n, and V i = P M (L i I H ) M for 1 i n, we say that (N ; V 1, V 2,, V n ) 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 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. A subspace N F 2 (f) H is semi-invariant under L i I H for 1 i n if for every i, j n, P N (L i I N )P N (L j I N )P N = P N (L i I N )(L j I N )P N. In [13] Sarason proved that N is semi-invariant if and only if there exist two submodules N 1 and N 2 with N 1 N = N 2. Now if N F 2 (f) H is semiinvariant 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. It is easy to see that every submodule and every -submodule is a subquotient, but not every subquotient is a submodule or a -submodule. A module map between the Hilbert modules (H; L 1,..., L n ) and (K; V 1,..., V n ) is a bounded linear map X : H K satisfying X(L i h) = V i (Xh) for every h H 3

4 and 1 i n. It is well known and easy to verify that 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. Similarly, 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. 3 Non-commutative domains as module maps - Weighted Fock Spaces In this section we give a unitarily equivalent description of the non-commutative domain algebras of Popescu. 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 { egi1 e α := e g... e i2 g ik if α = g i1 g i2 g ik, 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 l 2 (F + n ). 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 giα 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 R i e α = e αgi for every α F + n. The left and right creation operators are isometries n with orthogonal ranges. Furthermore, they are row contractions: L i L i I, n R i Ri I. i=1 For a Hilbert space H, let B(H) be the algebra of all bounded linear operators over H. If (T 1, T 2,..., T n ) B(H) n and α = g i1 g i2... g ik F + n, define T α to be the product T i1 T i2... T ik. Definition 1 (Popescu, [12]) 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 positive regular free holomorphic function on B(H) n if the following properties on a f α are satisfied: lim sup k α =k a f α a f g 0 = 0, a f g i > 0 1 i n, a f α 0 α F + n, α > 1, 1 2k 2 <. i=1 4

5 In [12], Popescu proved that the a α s given above induce a family of positive real numbers (b α ) α F + n with the following properties: b g0 = 1, b α > 0 for all α F + n, b γ = a β b α if γ 1, βα=γ b γ = αβ=γ a β b α if γ 1, b α b β b αβ for any α, β F + n. Theorem 2 (G. Popescu, [12]) Let f = a f αx α be a positive regular free holomorphic function. For every i n, there exists a bounded linear operator W f i : l 2 (F + n ) l 2 (F + n ) defined by W f i (e α) = bα b e gi g α iα. Furthermore, a f αwα f Wα f I, and if (T 1, T 2,..., T n ) is an n-tuple of operators acting on a Hilbert space H with a f αt α Tα I then there exists a unique unital completely contractive { } morphism Φ from the algebra generated by W f 1, W f 2,..., W n f into B(H) such that W f i T i for 1 i n. A short sketch of the proof of this theorem appears in [3]. If we consider the Fock space l 2 (F + n ) together with the weighted shifts W f 1, W f 2,..., W n f defined in the previous theorem, we get the Hilbert module (l 2 (F + n ) ; W f 1, W f 2,..., W n f ). 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 δα f = 1. For each i n, define the left creation operators L f i : F 2 (f) F 2 (f) by L f i δf α = δg f i α. Similarly, for each i n, define the right creation operators R f i : F 2 (f) F 2 (f) by R f i δf α = δαg f i. We consider the weighted Fock space F 2 (f) together with the weighted shifts L f 1, Lf 2,..., Lf n to construct the Hilbert module (F 2 (f); L f 1, Lf 2,..., Lf n), which is unitarily equivalent to (l 2 (F + n ) ; W f 1, W f 2,..., W n f ). The equivalence of the two Hilbert modules is obtained via the operator U : l 2 (F + n ) F 2 (f) defined as Ue α = δα. f We can easily verify that U δα f = 1 e α and that U is a unitary module map. 5

6 Lemma 3 The operator U : l 2 (F + n ) F 2 (f) is a unitary module map. The correspondence with Popescu s version is as follows, with Popescu s version on the left and our version on the right: l 2 (F + n ) F 2 (f) e α, e α = 1 δα, f δ α = W f β e α = bα b βα e βα L f β δ α = δ βα = = W f β 1 b β L f β 1 b β Λ f β e α = bα b βα e αβ R f β δ α = δ αβ = = Λ f β 1 b β R f β 1 b β 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 is f(x 1, X 2,..., X n ) = 1 b α a α X α. We will now state a result we will need for this work: Proposition 4 (G. Popescu, [12]) Let f = a α X α, (L f 1, Lf 2,..., Lf n), and (R f 1, Rf 2,..., Rf n) be given. Then: (i) a β L f ( β L f β) I, (ii) a βr f ( β R f β) I. 3.1 Module maps between F 2 (f) and F 2 (g) If the underlying positive regular free holomorphic function is understood, we will refer to L f α, Rα, f δα, f a f α, and simply as L α, R α, δ α, a α, and b α, respectively. 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 isometric, 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 se- quence x = {x α : α F + n }. This allows us to define F 2 (f) F 2 (g) as sequence inclusion. 6

7 The main result of this section is Theorem 5 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 b g <. + α n 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. Lemma 6 Let f = a α X α, where f has only finitely many terms. 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, b α C b αg i b α b gi. Proof. We will only prove the first inequality. The other one has a similar proof. The right inequality was proven by Popescu in [12]. Thus, it remains to prove the left inequality. Now b gi α = a β b γ. βγ=g i α Since f is finite, there exists a k N such that for all α with α > k, a α = 0. Thus: Let A = b giα = βγ=g i α 1 β k max {a β}. Then: 1 β k b giα A σγ=α 0 σ k 1 = A Cb α. σγ=α 0 σ k 1 a β b γ b γ = A b α b σ = Ab α max {a β } b γ. βγ=g i α βγ=g 1 β k i α 1 β k σγ=α 0 σ k 1 σγ=α 0 σ k 1 b σ b γ b σ A 1 b σ σγ=α 0 σ k 1 b σγ b σ 7

8 where C = A σ k 1 1 b σ. Thus, b α C b gi 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 7 Define f = a α X α by: 1 if α = 1, a α = 2 k if α = 12 k 1, 0 else. Let α = 2 k 1 and i = 1. Let s calculate b α b giα = b 2 k 1 = b 12k 1 βγ=2 k 1 βγ=12 k 1 a β b γ b α b gi α : (a 2 ) k a 1 = a β b γ 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 and therefore no C exists such that b α C b gi α. b α b C gi α We are now ready to complete the proof of Theorem 5. 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 : lim x m, δ α m f = lim m β F + n c m,β δ β, δ α f β F + n = 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). 8

9 Now we will show that (v) (iii): Notice the following: ε = sup = sup g b ε f g αδ α = sup δ α b g. α δ α g = sup b Thus, ε is bounded if and only if sup f α <, as desired. Thus, (i) b g α F + α n (ii) (iii) (iv) (v). 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 c β δ β = c β L α δ β = c β δ αβ. β F + n β F + n β F + n Let β 0 be such that β 0 0, but for all other β F + n follows that β = 0. Then: Xδ α = c β0 δ αβ0 + c β δ αβ. β β 0 β β 0 such that β < β 0 it Now δ αβ0 is orthogonal to δ αβ for every β β 0. Thus: Xδ α g = c β0 δ αβ0 + c β δ αβ β β 0 β β 0 g c β0 δ αβ0 g = c β0 δ αβ0 g = c β 0 b g αβ0. By applying lemma 6 iteratively, we can peel off the β 0 on the bottom of the fraction. This gives us: Xδ α g c β 0 K b g. α 1 Therefore, it follows that b g α Thus, Since α was arbitrary, it follows that C Xδ α g C X δ α f = C X 1. b g α C X. sup α F b g <. + α n 9

10 as desired. This leads immediately to the following theorem, which gives a simple method to calculate the norm of the ε map: Corollary 8 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: 3.2 Examples ε = sup b g α. 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 9 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 = 1, 1 j i }. In other words, b α 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. Proof. Let f = a α X α, with a α {0, 1} for all α F + n be given. The proof will be by strong induction. Base: Let α = g i. Now since a gi > 0 for all i n and a α {0, 1} for all α F + n, it follows that a gi = 1. Also recall that b 0 = 1. Now b gi = βγ=g i a β b γ = a gi b 0 = 1. But there is clearly only one way to decompose α = g i, so we re done. Induction: Assume the claim is true for all α with α m. Let α be given such that α = m + 1. Then b α = βγ=α a β b γ. By the induction hypothesis, b γ counts the number of ways γ can be decomposed into subwords γ = γ 1γ2 γ i such that a γ1 = a γ2 = = a γi = 1. Now if a β = 1, then any of the decompositions of γ will lead to proper decompositions of βγ = α, and so we should count them. This will lead to a β b γ = b γ new decompositions. However, if a β = 0, then none of the decompositions of γ will lead to proper decompositions of βγ = α, and so we should not count them. This will lead to a β b γ = 0 new decompositions. Finally, summing over all 10

11 βγ = α with β 1 gives us the final result. Since we ve counted all possible decompositions and only those possible decompositions, it follows that b α 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, as desired. ( Example 10 Let f = n X i + n X ij ). Then b α = F α +1, where F m is the mth Fibonacci number. i=1 j=1 Proof. Clearly b 0 = 1 = F 1 and b gi = 1 = F 2 for all i n. Assume b α = F α +1 for all α < m. Let α be given such that α = m. It then follows that b α = βγ=α a β b γ = a ρ b σ + a τ b ϕ, where ρσ = α = τϕ with ρ = 1 and τ = 2. Thus, b α = a ρ b σ + a τ b ϕ = F σ +1 + F τ +1 = F α + F α 1 = F α +1 as desired. While the finite case is quite pleasant, an example illustrates the potential difficulties encountered when transitioning to the infinite case: Example 11 Let n = 2 and look at f and g such that 1 if α = 1, 1 if α = 1, a f α = 2 k if α = 12 k 1, a g α = 4 k α = 12 if k 11, α = 12 0 else, k 12, 0 else. Look at ε : F 2 (f) F 2 (g). Is this bounded? ε 2 = sup α F b g bf 12 k 1 + α b g = n 12 k 1 = a f 1 ( ) k a f 2 a f 1 + af 12 k 1 a g 1 (ag 2 )k a g 1 γβ=12 k 1 γ 1 πσ=12 k 1 π 1 = 1 + 2k 1 a f γb f β a g πb g σ. So ε is unbounded. Now look at Y : F 2 (f) F 2 (g) by Y δ 0 = δ 1. Is this bounded? Let α be given. We can decompose α as α = α 1 α 2 α n such that α i { 1, 2, 12 k 1 }. This may yield several decompositions. However, it is easy to show that there exists a minimal decomposition in the sense that for every other decomposition α = α 1α 2 α m, m n. Next, note that for every i, if α i = 1, then 1 α 2 α i 1 α i α i+1 α n = 1 α 2 α i 1 α i+1 α n, and b g α 1 α 2 α i 1 α i α i+1 α n b g α 1α 2 α i 1α i+1 α n. Thus, we only need look at the case when α = α 1 α 2 α n 11

12 with α i = 12 k i 1. Thus: b g αg 1 = = = γβ=α γ 1 πσ=α1 π 1 a f γb f β a g πb g σ 1 + (a f α a f α n ) + (a f α 1 a f α 2 + a f α n 1 a f α n ) + + (a f α 1 a f α 2 a f α n ) 1 + (a g α a g α n ) + (a g α 1 a g α 3 + a g α 1 a g α 4 + a g α n 2 a g α n ) + + (a g α 1 a g α 3 a g α n ) 1 + (2 k kn ) + (2 k1 2 k kn 1 2 kn ) + + (2 k1 2 k2 2 kn ) 1 + (4 k kn ) + (4 k1 4 k3 + 4 k1 4 k4 + 4 kn 2 4 kn ) + + (4 k1 4 k3 4 kn ). It isn t difficult to notice that the bottom is always greater than or equal to the top. This is because if the top has n factors, the bottom can have at most n 2 factors. Thus, it follows that sup α F b g = bf 1 + α b g n 1 = 1 1 = 1. Thus, Y is bounded. This produces an example where f has infinitely many terms, ε is unbounded, but a nontrivial map is bounded. 4 Lifting Module Maps If f is a positive regular free holomorphic function, it induces {b α : α F + n } with the property that b α > 0 for every α F + n. We will now generalize this concept by looking at functions f : F + n (0, ) with f(α) = and a Hilbert space F 2 (f) with orthonormal basis δ α = 1. It is easy to see that f induces a set { } a f α : α F + n satisfying: b f γ = βα=γ b f γ = αβ=γ a f β bf α if γ 1, a f β bf α if γ 1. However, unlike positive regular free holomorphic functions, a f α need not be non-negative. In addition, the left and right creation operators may no longer be bounded, since the norm of the left and right creation operators will be given by: 12

13 L i 2 δ gi α 2 = sup α F + δ n α 2 = sup R i 2 δ αgi 2 = sup α F + δ n α 2 = sup b f g iα, g i. This leads immediately to the following definition: Definition 12 A function f : F + n (0, ) is called L-bounded if L i < for every i n. Similarly, A function f : F + n (0, ) is called R-bounded if R i < for every i n. We will look at the weighted Fock spaces generated by L-bounded and R- bounded functions. As with positive regular free holomorphic functions, the weighted Fock space generated by a L-bounded or R-bounded function f will be denoted F 2 (f). Due to the two different definitions we now have for F 2 (f), we will assume that f is a positive regular free holomorphic function unless explicitly stated. Using this more general idea of L-bounded and R-bounded maps, we are now in the position to modify the proof of the commutant lifting theorem for module maps from [12], which was based on [2] and [7]: Theorem 13 Let f : F + n (0, ) be an L-bounded map and g = a g αx α be a positive regular free holomorphic function with ε(f, g) 1. In addition, let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be -submodules of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and (F 2 (g) H 2 ; L 1 I H2, L 2 I H2,..., L n I H2 ), respectively. If X : M N is a module map, then there exists a module map X : F 2 (f) H 1 F 2 (g) H 2 such that X = X and XP M = P N X, i.e., the following diagram commutes: F 2 P M (f) H 1 M 0 X F 2 (g) H 2 P N N 0. X Proof. Let (M; V f 1, V f 2,..., V n f ) be a -L-submodule of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and (N ; V g 1, V g 2,..., V n g ) be a -submodule of (F 2 (g) H 2 ; L 1 I H2, L 2 I H2,..., L n I H2 ). Assume X : M N is a module map with X = 1. Let Y = XP M. Then Y : F 2 (f) H 1 N is also a module map with Y = X = 1. Now look at δ 0 H 2. If δ 0 H 2 N, let N 1 = δ 0 H 2 + N. If δ 0 H 2 N, find α F + n such that δ α H 2 N 13

14 but if α = βγ with β 1 then δ γ H 2 N. Then let N 1 = δ α H 2 + N. Note that in either case, (L α I H2 ) N 1 N for every α 1. (1) Let T g i = P N1 (L g i I H 2 ) P N1 for i n. Then (N 1 ; T g 1, T g 2,..., T g n) is a - L-submodule of (F 2 (g) H 2 ; L 1 I H2, L 2 I H2,..., L n I H2 ), and N is a -L-submodule of N 1. Now if α 1, it follows that: T g αp N1 N = 0. (2) Using this identity, a quick computation will reveal that for α 1 T g αβ P N = T g αv g β P N. (3) Now we need to find a module map Ŷ : F 2 (f) H 1 N 1 with Ŷ = X and P N Ŷ = Y. For every α F + n, define Y α : H 1 N by Y α (x) = Y (δ α x). Similarly, define Ŷα : H 1 N 1 by Ŷα(x) = Ŷ (δ α x). Now if Ŷ is to be a module map, each Ŷα, and thus Ŷ, will be determined by Ŷ0: Ŷ α (x) = Ŷ (δ α x) = Ŷ (( L f α I H1 ) (δ0 x) ) = T g αŷ (δ 0 x) = T g αŷ0(x). However, if α 1, then: Ŷ α (x) = T g αŷ0(x) = T g αp N1 N Ŷ0(x) + T g αp N Ŷ 0 (x) = T g αp N Ŷ 0 (x) = T g αy 0 (x). We can decompose F 2 (f) H 1 as F 2 (f) H 1 = [δ 0 H 1 ] [δ 0 H 1 ] and N 1 as N 1 = [N 1 N ] N and write Ŷ as a block matrix with respect to this decomposition: [ ] a b Ŷ = : [δ c d 0 H 1 ] [δ 0 H 1 ] [N 1 N ] [N ]. (4) Now Y = P N Ŷ = [ c d ], so the second row of Ŷ is already determined, and [ c d ] 1. Similarly, the second column of Ŷ is already determined, since [ ] b ) = (Ŷ [δ0 H1] : d N1 δ α H 1. is given by the following equation for x N 1 : ) (Ŷ [δ0 H1] (x) = δ α Ŷ α (x). 14

15 [ ] Next we need b 1. d ] b 2 ) 2 ) ) [ = (Ŷ [δ0 d H 1 ] = sup (Ŷ [δ0 H1] x, (Ŷ [δ0 H1] x x=1 = sup δ α Ŷ α (x), b f β δ β Ŷ β (x) x=1 ( ) = sup b f 2 α δα, δ α Ŷα (x), Ŷ α (x) x=1 = sup x=1 x, ŶαŶ α x. Thus it suffices to show that bf αŷαŷ α I. b g γvγ g Y 0 Y0 Vγ g b g γv g γ Vγ g Notice that I. γ F + n γ F + n b since Y is a contraction. In addition, since ε(f, g) is a contraction, sup f α b 1 g α F + α n by Corollary 8. Therefore, b g α for every α F + n. Thus: ŶαŶ α = TαY g 0 Y0 Tα g = = βγ=α γ F + n = γ F + n a g β bg γ = a g β T g β a g β T g β T g β T αy g 0 Y b g αtαy g 0 Y0 Tα g 0 Tα g a g β bg γt g βγ Y 0Y 0 T g βγ a g β bg γt g β V g γ Y 0 Y γ F + n 0 Vγ g b g γv g γ Y 0 Y a g β T g β 0 Vγ g T g β (L gβ I H 2 ) ( L g β I H 2 ) I. [ ] As desired. The last inequality follows from Proposition 4. Thus, b 1 d and [ c d ] 1. As in most commutant lifting theorems, we now apply 15

16 [ ] Parrott s Lemma parrott to find a such that Ŷ = a b = 1. Thus, c d Ŷ : F 2 (f) H 1 N 1 is a module map with Ŷ = X and P N Ŷ = Y, as desired. By iterating this process, it follows that we can find a module map X : F 2 (f) H 1 F 2 (g) H 2 such that X = X and XP M = P N X, completing the proof. We can also prove a similar result for the right creation operators R i, as shown in the following theorem: Theorem 14 Let f : F + n (0, ) be an R-bounded map and g = a g αx α be a positive regular free holomorphic function with ε(f, g) 1. In addition, let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be -submodules of (F 2 (f) H 1 ; R 1 I H1, R 2 I H1,..., R n I H1 ) and (F 2 (g) H 2 ; R 1 I H2, R 2 I H2,..., R n I H2 ), respectively. If X : M N is a module map, then there exists a module map X : F 2 (f) H 1 F 2 (g) H 2 such that X = X and XP M = P N X, i.e., the following diagram commutes: F 2 P M (f) H 1 M 0 X F 2 (g) H 2 P N N 0. X Proof. The proof of this theorem is very similar to the proof of the previous theorem. Thus, we will only provide a sketch of the proof. Let (M; V f 1, V f 2,..., V n f ) be a -R-submodule of (F 2 (f) H 1 ; R 1 I H1, R 2 I H1,..., R n I H1 ) and (N ; V g 1, V g 2,..., V n g ) be a -submodule of (F 2 (g) H 2 ; R 1 I H2, R 2 I H2,..., R n I H2 ). Assume X : M N is a module map with X = 1. Let Y = XP M. For every α F + n, define Y α : H 1 N by Y α (x) = Y (δ α x). Similarly, define Ŷα : H 1 N 1 by Ŷα(x) = Ŷ (δ α x). If δ 0 H 2 N, let N 1 = δ 0 H 2 + N. If δ 0 H 2 N, find α F + n such that δ α H 2 N but if α = βγ with β 1 then δ γ H 2 N. Then let N 1 = δ α H 2 + N. It follows that (1), (2), and (3) of theorem 13 hold. Now we need to find a module map Ŷ : F 2 (f) H 1 N 1 with Ŷ = X and P N Ŷ = Y. As before, Ŷ α, and thus Ŷ, will be determined by Ŷ 0, but this time we get that Ŷα(x) = T g αŷ0(x). Furthermore, if α 1, it follows that Ŷα(x) = T g α Y 0(x). Let T g i = P N1 (R g i I H 2 ) P N1 for i n. Then (N 1 ; T g 1, T g 2,..., T n) g is a -R-submodule of (F 2 (g) H 2 ; R 1 I H2, R 2 I H2,..., R n I H2 ), and N is a -R-submodule of N 1. We now decompose Ŷ into the same 2 2 matrix given by 4 in theorem

17 As in theorem 13, [ c d ] [ ] 1, and to prove that c 1, it suffices to d prove that bf αŷαŷ α = bf αŷ αŷ α I: b f αŷ αŷ α = b f α T αy g 0 Y0 Tα g = a = γ β= α β 1 γ F + n = γ F + n = a g βt g β g βb g γ T αy g 0 Y b g α T αy g 0 Y0 Tα g 0 Tα g a g βb g γ T g βγ Y 0Y 0 T g βγ a g βb g γ T g β V g γ Y 0 Y a g βt g β T g β γ F + n 0 Vγ g b g γ V g γ Y 0 Y a g β T g β 0 Vγ g T g β (R gβ I H 2 ) ( R g β I H 2 ) I. As desired. The last inequality follows from the second part of Proposition 4. Finally, iterating finishes the proof. We can actually say quite a bit more when F 2 (g) is the full Fock space. In this case, any module map between F 2 (f) and F 2 (g) ends up being a contraction, as shown in the following proposition: Proposition 15 Let f = a f αx α, where f has only finitely many terms, and g be the full Fock space. Let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be -submodules of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and (F 2 (g) H 2 ; L 1 I H2, L 2 I H2,..., L n I H2 ), respectively. If ε(f, g) is bounded, then ε(f, g) is a contraction, and therefore X can be lifted to X such that X = X and XP M = P N X. Proof. Popescu proved in [12] that for all α F + n b αα b α b α = b 2 α. Thus, if > 1 for any α, lim k bf lim α k k (bf α) k =. Therefore, if ε(f, g) is b bounded, sup f α b = sup b f g α F + α α < and so b α 1 for all α F + n. It follows n that ε = sup desired. b g α 1 and thus we can apply theorem 13 to obtain our X, as 17

18 Now if M is a submodule of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and ε : F 2 (f) F 2 (g) is bounded, then ε(m) is a submodule of F 2 (g). This follows immediately due to the nice property that L i δ α = δ gi α. However, it is important to note that the L i s do not behave nearly so nicely. It is easy to verify that L i δ g i α = bα b δ gi α. In particular, if (M; V f α 1, V f 2,..., V n f ) is a -submodule of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and ε(f, g) is bounded, (ϵ(m); V g 1, V g 2,..., V n g ) need not be a -submodule of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ). However, when F 2 (f) = F 2 (g), there is a nice correspondence between - submodules of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and those of (F 2 (g) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ), as shown in the following lemma: Lemma 16 Let f and g be positive regular free holomorphic functions with F 2 (f) = F 2 (g). Then if (M; V f 1, V f 2,..., V n f ) is a -submodule of F 2 (f), there exists a -submodule (N ; V g 1, V g 2,..., V n g ) of F 2 (g) such that M = N. Furthermore, there exists a module map ε(m, N ) : M N with ε ε such that εp M = P N ε. Proof. Let the -submodule (M; V f 1, V f 2,..., V n f ) be given. We can decompose F 2 (f) as F 2 (f) = M L. This means that M = F 2 (f)/l, where L is a submodule of F 2 (f). Now since F 2 (f) = F 2 (g), we can look at N = F 2 (g)/ε(l). Define V g i : N N by V g i (x + ε(l)) = L ix + ε(l). It is easy to see that V g i is well defined. By proposition 5.9 of Douglas and Paulsen [9], this induces a unique ε with ε ε such that the following diagram commutes: 0 L ε L 0 ε(l) F 2 (f) H 1 ε M F 2 (g) H 2 N 0. It immediately follows that (N ; V g 1, V g 2,..., V g n ) is a -submodule as desired. Now Theorems 13 and 14 only apply to cases where ε is a contraction. However, it turns out that as long as ε is bounded, we can lift all module maps with only a small penalty, as shown in the main theorem of this paper: Theorem 17 Let f and g be positive regular free holomorphic functions with ε(f, g) C. Let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be -submodules of (F 2 (f) H 1 ; L 1 I H1, L 2 I H1,..., L n I H1 ) and (F 2 (g) H 2 ; L 1 I H2, L 2 I H2,..., L n I H2 ), respectively. If X : M N is a module map, then there exists a module map X : F 2 (f) H 1 F 2 (g) H 2 such that X C X and XP M = P N X, i.e., the following diagram commutes: ε 0 18

19 0 L M F 2 P M (f) H 1 M 0 0 L N X LM X F 2 (g) H 2 P N N 0. X Proof. Assume without loss of generality that ε(f, g) = C > 1. Define an L-bounded function h : F + n (0, ) as follows: b h 0 = 1, b h α = bf α C 2 for every α F + n, α 1. Since f is a positive regular free holomorphic function, it follows immediately that h is indeed L-bounded. This gives us the weighted Fock space F 2 (h). In addition, F 2 (f) = F 2 (h). This is because for α 1: δ h α = 1 = b h α Note that since δ 0 h = 1 = 1 = 1 b h 0 C 2 b f 0 = = C δα f. δ f 0, F 2 (g) and F 2 (h) are not isometric. Now let ε(h, f) : F 2 (h) H 2 F 2 (f) H 2 be defined in the typical way. It follows that ε(h, f) = 1 and ε(h, f) 1 = C. Let M be the -submodule isomorphic to M given by Lemma 16. We get the following diagram: P M F 2 (f) H 1 M ε(h,f) 1 ε(h,f) 1 F 2 P M (h) H 1 M X ε(h,f) F 2 P N (g) H 2 N. X If we focus only on the bottom two rows, we obtain the following diagram: F 2 (h) H 1 P M M F 2 (g) H 2 P N N. X ε(h,f) 19

20 A brief calculation gives us ε(h, g) 2 b h α = sup α F b g = sup + α n = 1 C 2 = 1. sup α F b g + α n C 2 b g α = 1 ε(f, g)2 C2 We are thus in a position to apply Theorem 13. This gives us Ŷ : F 2 (h) H 1 F 2 (g) H 2 such that Ŷ X and (X ε(h, f))p M = P N Ŷ. The inequality is verified by a quick calculation: Ŷ = X ε(h, f) X ε(h, f) X ε(h, f) = X. Finally, define X := Ŷ ε(h, f) 1. Then X = Ŷ ε(h, f) 1 Ŷ ε(h, f) 1 C X. All that is left is to verify that XP M = P N X: as desired. XP M = XP M ε(h, f)ε(h, f) 1 = X ε(h, f)p Mε(h, f) 1 = P N Ŷ ε(h, f) 1 = P N X, Theorem 18 Let f and g be positive regular free holomorphic functions with ε(f, g) C. Let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be -submodules of (F 2 (f) H 1 ; R 1 I H1, R 2 I H1,..., R n I H1 ) and (F 2 (g) H 2 ; R 1 I H2, R 2 I H2,..., R n I H2 ), respectively. If X : M N is a module map, then there exists a module map X : F 2 (f) H 1 F 2 (g) H 2 such that X C X and XP M = P N X, i.e., the following diagram commutes: 20

21 F 2 P M (f) H 1 M 0 X F 2 (g) H 2 P N N 0. X Proof. Identical to the previous proof. Using theorems 17 and 18, we can build projective resolutions. To build such resolutions, we must first recall, with a slight reformulation, the construction of Poisson kernels developed by Popescu in [12] based on [5]: Lemma 19 (G. Popescu, [12]) Let f be a positive regular free holomorphic function, and let N F 2 (f) H be semi-invariant under the maps L i I H for i n, and let W i = P N (L i I H ) N for i n. Then: (i) The sequence N = a α W α Wα is non-negative and non-increasing, (ii) lim N γ >N αβ=γ α N α N b α a β W γ Wγ = 0, (iii) = lim N N exists, 0 I, and is not equal to zero, and, (iv) b α W α W α = I N. Theorem 20 (G. Popescu, [12] - Poisson kernels) Let f be a positive regular free holomorphic function and let N F 2 (f) H be semi-invariant under L i I H for i n, and let W i = P N (L i I H ) N for i n. Define D = 1 2, where is the non-negative operator given by the previous lemma. Define K : N F 2 (f) N by K(x) = b α δ α DW αx. Then K is an isometry and K is a module map. Moreover, N is isomorphic to K(N ) and K(N ) is a -invariant submodule of F 2 (f) H. We call K the Poisson Kernel of N. We use the Poisson kernel to construct resolutions. Proposition 21 Let f be a positive regular free holomorphic function and let M F 2 (f) H be semi-invariant under L i I H for i n. Then there exists Hilbert spaces H 1, H 2,..., partial isometric module maps Φ i : F 2 (f) H i+1 F 2 (f) H i, i = 1, 2,..., and a partial isometric module map Φ 0 : F 2 (f) H 1 M such that the following sequence is exact: 21

22 Φ 3 F 2 (f) H 3 Φ 2 F 2 (f) H 2 Φ 1 F 2 (f) H 1 Φ 0 M 0. Proof. Since M is a semi-invariant subspace of F 2 (f) H, it has a Poisson kernel K 0 : M F 2 (f) H 1, for some Hilbert space H 1. Set Φ 0 = K 0. Notice that ker Φ 0 is a submodule of F 2 (f) H 1, and hence it has a Poisson kernel K 1 : ker Φ 0 F 2 (f) H 2, for some Hilbert space H 2. Define Φ 1 : F 2 (f) H 2 F 2 (f) H 1 by Φ 1 = ι K 1, as illustrated in the following diagram: F 2 Φ 1 (f) H 2 F 2 Φ 0 (f) H 1 M K ι 1 ker Φ 0 0. The other maps are constructed similarly: ker Φ 1 is a submodule of F 2 (f) H 2, and hence it has a Poisson kernel K 2 : ker Φ 1 F 2 (f) H 3, for some Hilbert space H 3, Φ 2 = ι K 2. Let M F 2 (f) H and for i n let V i = P M (L i I H ) M. Recall that (M; V 1, V 2,..., V n ) is a subquotient of (F 2 (f) H; L 1 I H, L 2 I H,..., L n I H ) iff for every i, j n, P M (L i I H )P M (L j I H )P M = P M (L i I H )(L j I H )P M. Then the previous Proposition together with 17 and 18 give the following result: Theorem 22 Let f and g be positive regular free holomorphic functions. Let (M; V f 1, V f 2,..., V n f ) and (N ; V g 1, V g 2,..., V n g ) be subquotients of (F 2 (f) H; L 1 I H, L 2 I H,..., L n I H ) and (F 2 (g) Ĥ; L 1 IĤ, L 2 IĤ,..., L n IĤ), respectively. Assume there exist a module map T : M N. Then if ϵ(f, g) = C, there exists module maps T 1, T 2, T 3... with T i C i T, such that the following diagram commutes: Φ 3 F 2 Φ 2 (f) H 3 F 2 Φ 1 (f) H 2 F 2 Φ 0 (f) H 1 M 0 Ψ 3 T 3 F 2 (g) Ĥ3 Ψ 2 T 2 F 2 (g) Ĥ2 Ψ 1 T 1 F 2 (g) Ĥ1 T Ψ 0 N 0. Proof. Clearly we have that T 1 exists via theorems 17 and 18. Φ 1 = ι K 1 where K 1 : ker Φ 0 F 2 (f) H 2 is the Poisson kernel of ker Φ 0, and Ψ 1 = ι K 1 22

23 where K 1 : ker Ψ 0 F 2 (f) Ĥ2 is the Poisson kernel of ker Ψ 0, as shown below: Φ 1 F 2 K 1 (f) H 2 ι ker Φ 0 F 2 Φ 0 (f) H 1 M 0 F 2 (g) Ĥ2 K 1 T 1 T 1 ker Ψ 0 ι F 2 (g) Ĥ1 T Ψ 0 N 0. Ψ 1 Since K 1 and K 1 are isometries, by theorem 20, the Poisson kernels induce a module map T 1 : K 1 (ker Φ 0 ) K 1 (ker Ψ 0 ) with T 1 = T1. Therefore, theorems 17 and 18 give us T 2 with T 2 C T 1 such that the following diagram commutes: F 2 P (f) H 2 K 1 (ker Φ 0 ) T 2 F 2 (g) Ĥ2 Iterating this process gives us T 1 P K1 (ker Ψ 0 ). Φ 3 F 2 Φ 2 (f) H 3 F 2 Φ 1 (f) H 2 F 2 Φ 0 (f) H 1 M 0 Ψ 3 T 3 F 2 (g) Ĥ3 Ψ 2 T 2 F 2 (g) Ĥ2 Ψ 1 T 1 F 2 (g) Ĥ1 T Ψ 0 N 0, with T i C i T, as desired. References [1] J. Agler and J. McCarthy, Pick interpolation and Hilbert function spaces. Graduate Studies in Mathematics, 44. American Mathematical Society, Providence, RI, pp. 23

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