REFLEXIVITY OF THE SPACE OF MODULE HOMOMORPHISMS

REFLEXIVITY OF THE SPACE OF MODULE HOMOMORPHISMS JANKO BRAČIČ Abstract. Let B be a unital Banach algebra and X, Y be left Banach B-modules. We give a sufficient condition for reflexivity of the space of all bounded module homomorphisms from X to Y. The condition is closely connected with the notion of synthesis over a Banach module and is satisfied, for instance, by any triple (B, X, Y), where B is a unital C -algebra and X, Y are arbitrary left Banach B-modules. A corollary of our main result is a well known fact that every von Neumann algebra is reflexive. 1. Introduction Let X and Y be complex Banach spaces and S be a closed linear subspace of B(X,Y), the Banach space of all bounded linear operators from X to Y. The reflexive cover of S is the set Ref S := {T B(X,Y); Tx Sx for all x X}, where Sx = {Sx; S S} and Sx is the closure of Sx in Y. If S = Ref S, then S is said to be reflexive. This definition goes back to [4] and is closely connected with the theory of invariant subspaces in the following sense. Let X = Y. Define the lattice of invariant subspaces of S by Lat S = {M; M is a closed linear subspace of X such that TM M for all T S}. If L is a collection of closed subspaces of X, define Alg L = {T B(X); TM M for all M L}. Assume that S is a closed subalgebra of B(X) containing I, the identity operator on X. Then it is easily seen that Ref S = Alg Lat S (cf. [3], Proposition 22.3 (e)). Let B be a unital Banach algebra and X, Y be left Banach B-modules. Denote by B B (X,Y) the set of all bounded module homomorphisms, i.e. an operator T B(X,Y) is in B B (X,Y) if and only if T(b x) = b Tx, for all b B and all x X. We shall call T B(X,Y) a multiplier if it is in B B (X,Y). It is well known that B B (X,Y) is a closed linear subspace of B(X,Y). In particular, it is not hard to see that B B (X) is a subalgebra of 2000 Mathematics Subject Classification. Primary 46H25, 43A45. Key words and phrases. Synthesis over Banach module, reflexive subspace. 1

2 JANKO BRAČIČ B(X), closed according to the weak operator topology, and containing all multiplication operators that are induced by elements in the centre of B. In this paper we are concerned with the reflexivity of B B (X,Y). We give a sufficient condition on (B,X,Y) under which B B (X,Y) is reflexive. The condition is closely connected with the notion of synthesis over a Banach module (see [6]) and is satisfied, for instance, by any triple (B,X,Y), where B is a unital C -algebra and X, Y are arbitrary left Banach B-modules. A corollary of our main result (Theorem 4.3) is a well known fact that every von Neumann algebra is reflexive ([3], Proposition 56.6). 2. Preliminaries Let A be a unital commutative Banach algebra and Σ(A) be its character space, i.e. the set of all non-zero multiplicative linear functionals on A. It is well known that Σ(A) is a subset of the unit sphere of A, the topological dual of A. The relative weak- topology on Σ(A) is called the Gelfand topology. Denote by k(s) := {a A; ϕ(a) = 0 for all ϕ S} the kernel of a subset S Σ(A), and by h(u) := {ϕ Σ(A); ϕ(a) = 0 for all a U} the hull of U A. The algebra is semisimple if the kernel k(σ(a)) is trivial. The hull-kernel topology on Σ(A) is given through the closure operation S hk(s). This topology is weaker than the Gelfand topology; if both topologies coincide, the algebra is said to be regular. The Gelfand transform of a A is a function â : Σ(A) C that is defined by â(ϕ) = ϕ(a). The co-zero set of â is ω(a) := Σ(A) \ h(a), and the support of â is suppâ := ω(a), the closure of the co-zero set according to the Gelfand topology. A Banach space X is a left (respectively, right) Banach A-module, where A is a Banach algebra (not necessarily commutative or unital), if it is a left (respectively, right) A-module in the algebraic sense and a x a x holds for all a A and x X. If A has an identity, say 1, then it is assumed that 1 x = x for all x X. Let X be a left Banach A-module. The dual module of X is X, the topological dual of X, with a right module multiplication that is given by ξ a,x = ξ,a x (a A, ξ X, x X). In the case of a commutative algebra all modules can be considered as left or right modules, so the adjectives left and right will be omitted. Let A be a unital commutative Banach algebra and X be a Banach A- module. The annihilator of a subset M X is Ann A (M) := {a A; a x = 0 for all x M} and the spectrum of M is sp A (M) := h(ann A (M)). In particular, sp A (x) = {ϕ Σ(A); ϕ(a) = 0 for all a A such that a x = 0} is the spectrum of x X.

REFLEXIVITY OF THE SPACE OF MODULE HOMOMORPHISMS 3 Consider A as a Banach A-module through the usual multiplication in A. It is easily seen that suppâ sp A (a) for all a A. Moreover, if A is semisimple and regular, then suppâ = sp A (a) for all a A. The spectrum sp A (x) is always a compact subset of Σ(A) and it is empty if and only if x = 0. The most important properties of the spectra are the following: and hk(ω(a) sp A (x)) sp A (a x) sp A (a) sp A (x) (a A, x X) sp A (x 1 + x 2 ) sp A (x 1 ) sp A (x 2 ) Thus, for each F Σ(A), the set X (F) := {x X; sp A (x) F = } (x 1, x 2 X). is invariant for the action of the algebra. The closure of X (F) is denoted by X F and called the cospectral submodule of X corresponding to F. In particular, A F is the cospectral ideal of A corresponding to F. 3. Spectrally separable algebras and synthesis A unital commutative Banach algebra A is spectrally separable if, for any two distinct characters ϕ and ψ on A, there exist a and b in A such that ϕ(a) 0 ψ(b) and ab = 0. Every spectrally separable algebra is regular. Moreover, it can be proved that, for any pair of two distinct characters ϕ and ψ on A, there exist a and b in A such that ϕ(a) 0 ψ(b) and sp A (a) sp A (b) =. On the other hand, spectrally separable algebras are not necessarily semisimple; for an example see [1]. If A is a spectrally separable algebra and F Σ(A) is a closed set, then, by Proposition 4.1 in [2], the cospectral ideal A F is the smallest closed ideal of A with hull F. On the other hand, it is obvious that the kernel k(f) is the largest ideal of A with hull F. A closed subset F Σ(A) is a set of synthesis for A if A F = k(f). For instance, is a set of synthesis for every spectrally separable algebra. On the other hand, Σ(A) is a set of synthesis for a spectrally separable algebra A if and only if A is semisimple. Since the cross product of two spectrally separable algebras is a spectrally separable algebra ([1], Proposition 2.4 (ii)), it is not hard to give an example of a nonsemisimple spectrally separable algebra A and a closed subset F Σ(A) such that F is a set of synthesis for A. In Section 3 of [6], synthesis with respect to a Banach module over a semisimple, regular, commutative Banach algebra is defined. We extend that definition to the context of Banach modules over spectrally separable algebras in the following way. Let A be a spectrally separable algebra and X

4 JANKO BRAČIČ be a Banach A-module. A closed set F Σ(A) is a set of synthesis over X, if a x = 0 whenever x X and a A satisfy sp A (x) F h(a). Following the proof of Theorem 3.3 in [6] it is not hard to check the next proposition. Proposition 3.1. Let A be a spectrally separable algebra. For a closed set F Σ(A), the following assertions are equivalent: (a) F is a set of synthesis for A; (b) F is a set of synthesis over any A-module X; (c) F is a set of synthesis over the dual Banach A-module A. For a A, let T a be the multiplication operator induced by a on a Banach A-module X. If U A is a non-empty set, let ker U = a U ker T a. Let A be a spectrally separable algebra and X be a Banach A-module. The spectral submodule of X corresponding to a subset F Σ(A) is X(F) := {x X; sp A (x) F }. By [1], Proposition 2.7, X(F) is a closed submodule of X whenever F is a closed subset of Σ(A). For every closed set F Σ(A), we have ker k(f) X(F). Indeed, if x ker k(f), then k(f) Ann A (x) and consequently sp A (x) hk(f) = F. The opposite inclusion holds whenever F is a set of synthesis over X. More precisely, the following is true. Proposition 3.2. Let A be a spectrally separable algebra and X be a Banach A-module. A closed subset F Σ(A) is a set of synthesis over X if and only if X(F) ker k(f). Proof. Assume that F Σ(A) is a set of synthesis over X. If x X(F), i.e. sp A (x) F, then we have a x = 0, for each a k(f), which means x ker k(f). The opposite direction is obvious. The projective tensor product A A = { a i b i ; a i b i < }, where A is a spectrally separable algebra, is again a spectrally separable algebra (see [1], Proposition 2.4). It is well known that the character space of A A may be identified by the cross product Σ(A) Σ(A). The diagonal of A is the set D A := {(ϕ,ϕ); ϕ Σ(A)} Σ(A A). Proposition 3.3. Let A be a spectrally separable algebra. If the diagonal D A is a set of synthesis for A A, then A is semisimple.

REFLEXIVITY OF THE SPACE OF MODULE HOMOMORPHISMS 5 Proof. Consider the algebra A as a Banach A A-module through the multiplication ( ) a i b i x = a i xb i ( a i b i A A, x A). It is obvious that {a 1 1 a; a A} Ann A b A (x), for each x A. Thus, if (ϕ,ψ) Σ(A A) is in the spectrum sp A b A (x), then it annihilates each tensor of the form a 1 1 a. It follows ϕ(a) = ψ(a), for all a A. We conclude, that sp A b A (x) D A for all x A. Now assume that r A is in the radical of A. Then, of course, r 1 is in the radical of A A and therefore h(r 1) = Σ(A A). Thus sp A b A (x) D A h(r 1) for all x A. By Proposition 3.1, rx = r 1 x = 0 for all x A, which gives r = 0. A spectrally separable algebra A is in the class (SD) (cf. [5] and [2]) if the diagonal D A is a set of synthesis for the algebra A A. By the last proposition, each algebra in the class (SD) is semisimple (of course, it is also a regular unital commutative Banach algebra). For instance, the algebra C(K) of all complex-valued continuous functions on a compact Hausdorff space K is in the class (SD) (see [5]). In [2], we have extended the class (SD) in the following way. A unital Banach algebra B (not necessarily commutative) is locally in the class (SD) if there exists a set B 0 of generators of the algebra B such that, for each b B 0, there exists a closed subalgebra A b B that is in the class (SD) and contains b and 1, the identity of B. By Proposition 4.3 in [2], every unital C -algebra is locally in the class (SD). Proposition 3.4. If B is a unital commutative Banach algebra that is locally in the class (SD), then it is spectrally separable. Proof. Let B 0 be the set of generators of B that exists by the definition. Choose b and let A b B be a closed algebra which is in the class (SD) and contains b and 1. Then, A b is a spectrally separable algebra and B is a Banach A b -module through the usual multiplication. By Theorem 3.2 in [1], the multiplication operator T b that is induced by b on B is decomposable. Since the Gelfand transforms of the elements in B 0 separate points of Σ(B) we may conclude, by Theorem 3.5 in [1], that B is spectrally separable. 4. Reflexivity Let X and Y be Banach A-modules. Then B(X,Y) has a natural structure of a Banach A-bimodule. Namely, for S B(X,Y) and a, b A, the operators a S and S b are given by (a S)x = a Sx and (S b)x = S(b x) (x X).

6 JANKO BRAČIČ It follows that B(X,Y) is a Banach A A-module via the multiplication ( ) a i b i S = a i S b i. The following proposition extends a little bit Theorem 5 in [5]. Proposition 4.1. Let A be a spectrally separable algebra and X, Y be Banach A-modules. If D A is a set of synthesis over B(X,Y), then T B(X,Y) is a multiplier if and only if TX(F) Y(F), for every closed set F Σ(A). Proof. Assume that T B(X,Y) is a multiplier and that F Σ(A) is a closed set. If x X(F), i.e. sp A (x) F, then Ann A (x) Ann A (Tx), which gives sp A (Tx) sp A (x) F. Now, let T B(X,Y) be such that TX(F) Y(F), for every closed set F Σ(A). Consider the subsets M = {a 1 1 a; a A} and N = {a b; sp A (a) sp A (b) = } of A A. It is obvious that ker M = B A (X,Y). Let us show that T ker N. Choose an arbitrary a b N. Then ((a b) T)x = a (T(b x)) for all x X. Thus, sp A (((a b) T)x) = sp A (a (T(b x))) sp A (a) sp A (T(b x)) sp A (a) sp A (b x) sp A (a) sp A (b) sp A (x) =, for all x X, which means (a b) T = 0 and therefore T ker N. It remains to show that ker N ker M. For a A the hull of a 1 1 a M is that is h(a 1 1 a) = {(ϕ,ψ) Σ(A A); ϕ(a) = ψ(a)}, D A h(a 1 1 a) for all a A. If S is in ker N, then N Ann A b A (S). Since A is spectrally separable, for (ϕ,ψ) Σ(A A) \ D A, there exist a, b A such that which means that ϕ(a) 0 ψ(b) and sp A (a) sp A (b) =, a b N and (ϕ,ψ),a b 0. It follows that (ϕ,ψ) / sp A b A (S) and consequently sp Ab A (S) D A. We have seen that sp A b A (S) D A h(a 1 1 a) for all S ker N and for all a 1 1 a M. Since D A is a set of synthesis over B(X,Y) we conclude (a 1 1 a) S = 0 for all S ker N and for all a 1 1 a M, that is ker N ker M. We are ready to give our first result about reflexivity of the space of multipliers.

REFLEXIVITY OF THE SPACE OF MODULE HOMOMORPHISMS 7 Proposition 4.2. Let A be a spectrally separable algebra and X, Y be Banach A-modules. If the diagonal D A is a set of synthesis over B(X,Y), then Ref B A (X,Y) = B A (X,Y). Proof. We have to show that each T Ref B A (X,Y) is a multiplier. By Proposition 4.1, it is enough to see that TX(F) Y(F), for every closed set F Σ(A). Let F Σ(A) be closed and choose x X(F). For an arbitrary ε > 0 there exists S B A (X,Y) such that Tx Sx < ε. Thus, if a is in the annihilator of x, then a Tx = a Tx S(a x) a Tx Sx < ε a and consequently a Tx = 0. We have seen that Ann A (x) Ann A (Tx), which gives sp A (Tx) sp A (x) F, i.e. Tx Y(F). Let B be a unital Banach algebra (not necessarily commutative) and let X, Y be left Banach B-modules. We shall say, that (B,X,Y) is an (SD)- triple, if the following holds. There is a set of generators B 0 of B such that, for each b B 0, there exists a closed subalgebra A b B that is spectrally separable, contains b and 1, the identity of B, and the diagonal D Ab is a set of synthesis over B(X,Y). For instance, if B is locally in the class (SD), then, by Proposition 3.1, any triple (B,X,Y), where X and Y are arbitrary left Banach B-modules, is an (SD)-triple. In particular, this holds if B is a unital C -algebra. Theorem 4.3. Let B be a unital Banach algebra and let X, Y be left Banach B-modules. If (B,X,Y) is an (SD)-triple, then B B (X,Y) is reflexive. Proof. We have to see that each T Ref B B (X,Y) is in B B (X,Y). Choose b B 0 and let A b B be a spectrally separable algebra that exists by the definition of an (SD)-triple. Then X and Y are Banach A b -modules and D Ab is a set of synthesis over B(X,Y). Since B B (X,Y) B Ab (X,Y) we have Ref B B (X,Y) Ref B Ab (X,Y), in particular T Ref B Ab (X,Y). By Proposition 4.2, T is in B Ab (X,Y); hence T(b x) = b Tx for all x X. It follows that T(b 1...b n x) = b 1...b n Tx, for all x X and an arbitrary finite family {b 1,...,b n } B 0. Since T is linear and continuous we conclude that T B B (X,Y). It is well known that every von Neumann algebra of operators on a Hilbert space is reflexive. The following example shows how this can be deduced from our results. Example 4.4. Let H be a Hilbert space and C B(H) be a von Neumann algebra. Denote by B the commutant C. Of course, B is a unital C -algebra. Consider H as a left Banach B-module through the multiplication T h = Th for all T B and h H.

8 JANKO BRAČIČ Then, B B (H) = B = C and consequently, by the Double commutant theorem ([3]), B B (H) = C. Since B is a unital C -algebra the conditions of Theorem 4.3 are satisfied and we may conclude that C is reflexive. The last example will show that B B (X,Y) is not always reflexive. Example 4.5. Let X = C 2 and let {[ ] a b B = B(X); 0 a } a, b C. For the multiplication T x := Tx, where T B and x X are arbitrary, X is a left Banach B-module. It is easily seen that B B (X) = B and that Ref B B (X) is the algebra of all upper triangular matrices, which means that B B (X) is a proper subset of Ref B B (X). References [1] J. Bračič, Unital strongly harmonic commutative Banach algebras, Studia Math., 149 (2002), 253-266. [2] J. Bračič, Local operators on Banach modules, Math. Proc. Royal Irish Academy, 104A (2004). [3] J. B. Conway, A Course in Operator Theory, AMS, Graduate Studies in Mathematics, 21, Providence 2000. [4] A. I. Loginov and V. S. Shulman, Hereditary and intermediate reflexivity of W - algebras, Izv. Akad. Nauk. SSSR Ser. Mat., 39 (1975), 1260-1273 (Russian); Math. USSR Izv., 9 (1975), 1189-1201 (English transl.). [5] V. S. Shulman, Spectral Synthesis and the Fuglede-Putnam-Rosenblum Theorem, Teoriya Funktsii, Funktsional nyi Analiz i ikh Prilozheniya 54 (1990), 25-36 (Russian). [6] V. Shulman and L. Turowska, Operator synthesis II. Individual synthesis and linear operator equations, preprint, arxiv:math.fa/0401091 v2. University of Ljubljana, IMFM, Jadranska ul. 19, 1000 Ljubljana, Slovenia E-mail address: janko.bracic@fmf.uni-lj.si