Houston Journal of Mathematics. c 1999 University of Houston Volume 25, No. 1, 1999

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1 Houston Journal of Mathematics c 1999 University of Houston Volume 5, No. 1, 1999 MULTIPLIERS AND REPRESENTATIONS OF NONCOMMUTATIVE DISC ALGEBRAS ALVARO ARIAS Communicated by Vern I. Paulsen Abstract. The non-commutative disc algebra A n, n, is the norm closure of the non-selfadjoint algebra generated by the left regular representation of F + n, the free semigroup on n generators. We present examples of contractive representation of A n, n, which are not completely contractive. This answers a question from [30]. We characterize the (completely) contractive Schur multipliers of A n which are indexed by subsets of F + n, and we show that for n, B(F + n ) M 0 (F + n ) and M 0 (F + n ) \ B(F + n ). B(F + n ) is the space of coefficients of contractive representations of F + n and M 0 (F + n ) is the space of completely bounded Schur multipliers of A n. 1. Introduction In this paper we study some properties of the non-commutative disc algebras A n, n. These non-selfadjoint algebras were introduced in 1991 by Popescu. In a sequence of papers (see [5], [6], [7], [8], [9], and [30]) he established remarkable similarities between these algebras and classical spaces appearing in Harmonic Analysis. This line of study has been pursued recently by Davidson and Pitts (see [9] and [10]). We also refer to [1], [], and [11] for related results. For the moment, it suffices to say that A n is an algebra generated by n isometries with orthogonal ranges S 1, S,, S n. The finite products of the S i s can be indexed naturally by F + n, the free semigroup on n generators g 1, g,, g n. That is, if α = g i1 g i g ik F + n, then S α = S i1 S i S ik, and S 0 = I. Our study is focused on maps u : F + n C. We say that u is a bounded Schur multiplier of A n (respectively, completely bounded Schur multiplier of A n ) if the 1991 Mathematics Subject Classification. Primary: 47D5, Secondary: 47B49, 43A65. Supported in part by NSF DMS

2 100 ALVARO ARIAS operator Φ u : A n A n defined by Φ u (S α ) = u(α)s α is bounded (respectively, completely bounded). The set of bounded Schur multipliers of A n is denoted by M(F + n ), and the set of completely bounded Schur multiplier of A n is denoted by M 0 (F + n ). We say that u : F + n C is a coefficient of a contractive representation of F + n if there exist a Hilbert space H, a contractive representation π : F + n B(H), and two vectors ξ, η H such that for each α, u(α) = π(α)ξ, η. The set of coefficients of contractive representations is denoted by B(F + n ). In Section we set the notation and state the necessary background material. In Sections 3 and 4 we study Schur multipliers of A n indexed by subsets of F + n. If Λ F + n, define u Λ : F + n C by { 1 if α Λ, u Λ (α) = 0 if α Λ. In Section 3 we characterize the subsets Λ of F + n such that u Λ is a completely contractive multiplier, and in Section 4 we show that such multipliers are contractive if and only if they are completely contractive. Some of the results of these sections can be easily extended to other discrete semigroups. In Section 5 we show that B(F + n ) M 0 (F + n ) and that M 0 (F + n ) \ B(F + n ) for n. We also establish some properties of these spaces. Finally, in Section 6, we present an example of a contractive representation of A n, n, that is not completely contractive. This answers a question from [30]. We also give a simple example of a bounded representation of A n, n, that is not completely bounded. Several questions addressed in this paper are motivated by results on Herz- Schur multipliers of the Fourier algebra of locally compact groups (see [4], [5], [7], [1], [19]). If G is a discrete group, Cλ (G) is the C -algebra of the left regular representation of G, M(G) is the space of bounded Schur multipliers of Cλ (G), M 0 (G) is the space of completely bounded Schur multipliers of Cλ (G), and B(G) is the space of coefficients of unitary representations of G. We refer the reader to [13], [17], [0], and [1] for more information. It is well known that B(G) M 0 (G) M(G). Moreover, B(G) = M 0 (G) if and only if G is amenable (see [5], [19], and [33]). Since F, the free group on countably many generators, is not amenable, then M 0 (F ) \ B(F ). One can show that the function { 1, if α = 1, u : F C defined by u(α) = 0, otherwise,

3 MULTIPLIERS AND REPRESENTATIONS OF A n 101 belongs to M 0 (F ) \ B(F ) (see [33] and [, Lemma 3.]).. Preliminaries The non-commutative disc algebras A n, n, are subalgebras of B(l (F + n )), where F + n is the free semigroup on n generators g 1,, g n, and l (F + n ) is the Hilbert space with orthonormal basis {δ α : α F + n }. The left regular representation λ : F + n B(l (F + n )) is defined by λ(α)δ β = δ αβ. The non-commutative disc algebra A n is the norm closure of span{λ(α) : α F + n } in B(l (F + n )). An alternative description of A n uses non-commutative polynomials. Let P = P n be the set of polynomials in n non-commutative variables e 1,, e n, with product and identity e 0. A typical element of P looks like N n p = a 0 e 0 + a i1 i i k e i1 e i e ik. i=1 i 1,i, i k =1 The Hilbert space with orthonormal basis {e i1 e i e ik } is called the full Fock space of the n-dimensional Hilbert space H n with orthonormal basis e 1, e n. This space is usually denoted by F (H n ) = k=0 H k n, where Hn 0 = Ce 0 and Hn k is the tensor product of k copies of H n. The basis of F (H n ) can be indexed by F + n. If α = g i1 g i g ik F + n, let e α = e i1 e i e ik. Then {e α : α F + n } is the canonical orthonormal basis of F (H n ). We can view the elements of P as left multiplication operators on F (H n ). If p P, the map p : F (H n ) F (H n ), defined by ϕ p ϕ, has norm p = sup{ p ϕ : ϕ = 1, ϕ F (H n )}. The non-commutative disc algebra A n is the norm-closure of P in B(F (H n )). When n = 1, A n coincides with the disc algebra A. The two descriptions of A n are equivalent. To see this, identify δ β with e β, and λ(α) with e α. Then, l (F + n ) is canonically isomorphic to F (H n ), and the equality λ(α)δ β = δ αβ, which describes λ(α), is transformed into e α e β = e αβ, which describes e α as a left multiplication operator on F (H n ). Tuples of isometries S 1, S,, S n with orthogonal ranges have been studied for some time. Particularly in Cuntz algebras (see [8]) and in Dilation Theory (see [6], [15], [16], [4], [5], and [9]). In [5, Theorem 3.], Popescu gave a Beurling-Lax characterization of the invariant subspaces of S 1, S,, S n. In

4 10 ALVARO ARIAS 1989, he used this description to obtain a very general inner-outer factorization of operators (see [6, Theorem 5.]). In 1991, he introduced the algebras A n, n, and their wot-closure, F (H n ), n (see [7]). The factorization result from [6] is very transparent for these algebras. Popescu said that ϕ F (H n ) is inner if ϕ is an isometry, and ψ F (H n ) is outer if ψ has dense range (recall that ϕ and ψ are operators on l (F + n )). Then, Theorem 5. of [6] says that any η F (H n ) can be written as η = ϕ ψ were ϕ is inner and ψ is outer. Moreover, this factorization is essentially unique. Inner operators are useful to study A n and F (H n ). For example, they were used in [] to show that F (H n ) is a reflexive algebra and in [9] to show that F (H n ) is hyper-reflexive. In this paper we will use the following three examples, which appeared in []: (i) Inherited inner functions: Let ϕ(z) A and α F + n, then ϕ(e α ) A n and ϕ A = ϕ(e α ) An. If ϕ is inner in A, then ϕ(e α ) is inner in A n. (ii) Homogeneous polynomials: Let p A n be a homogeneous polynomial of degree k (i.e., p span{e α : α = k}, where α is the length of the word α), then p An = p. If p = 1, then p is inner. (iii) Let p(z) be a polynomial in the disc algebra A, then p(e 1 ) e An = p(e 1 ) e. If p = 1, then p(e 1 ) e is inner. We refer to [1] and [9] for related results. Two words α, β F + n are said to be orthogonal if αf + n βf + n =. This is equivalent to saying that λ(α) and λ(β) have orthogonal ranges. Or, that for every ϕ 1, ϕ l (F + n ), e α ϕ 1 is orthogonal to e β ϕ. The unitary flip Θ : l (F + n ) l (F + n ) is defined by Θ(e i1 e i e ik ) = e ik e i e i1. Its action is denoted by Θ(ϕ) = ϕ. This map is unbounded on A n. An operator space X is a subspace of B(H). The minimal (or spatial) tensor product of two operator spaces X and Y is denoted by X min Y. If X B(H) and Y B(K), this is the just the closure of the algebraic tensor product X Y in B(H K), where H K is the Hilbert-Schmidt tensor norm of H and K. Since M n min B(H) M n (B(H)), one identifies M n (X) with M n min X. Thus, an operator space X has a natural family of norms n on M n (X). The morphisms in the category of operator spaces are the completely bounded maps. A bounded linear map Φ : X Y between two operator spaces X and Y is completely bounded, or cb, if the map id Z Φ : Z min X Z min Y is bounded for every operator space Z. The completely bounded norm Φ cb of Φ is the supremum of the norms id Z Φ as Z ranges over all operator spaces. It

5 MULTIPLIERS AND REPRESENTATIONS OF A n 103 turns out that it is enough to look only at the M n s. That is, Φ cb = sup n Φ n, where Φ n = id n Φ : M n (X) M n (Y ). The fundamental theorem of completely bounded maps is due to Wittstock [3]. It was discovered independently by Haagerup and Paulsen, and the factorization was inspired by results of Stinespring [31] and Arveson [3]. Theorem.1. (Wittstock [3]) Let X B(H) be an operator space, Φ : X B(l ) a bounded linear map, and C > 0. Φ is completely bounded and Φ cb C if and only if there exist a -representation π : B(H) B(l ) and two bounded linear maps V 1, V : l H, V 1 V C, such that for every x X, Φ(x) = V π(x)v 1. If Φ u : A n A n is a completely bounded Schur multiplier of A n, then there exist a Hilbert space H, a -representation π : B(l (F + n )) B(H), and two bounded linear maps V 1, V : l (F + n ) H satisfying V 1 = V and V 1 V = Φ u cb, such that for each α F + n, Φ u (λ(α)) = V π(λ(α))v 1. For every β F + n, (1) u(α) = Φ u (λ(α))δ β, δ αβ = V π(λ(α))v 1 δ β, δ αβ = π(λ(α))v 1 δ β, V δ αβ. We say that a function u : Γ Γ C is a Schur multiplier of B(l (Γ)) if the map M u : B(l (Γ)) B(l (Γ)), which is defined by M u (T )δ s, δ t = u(s, t) T δ s, δ t, is bounded. These maps have a nice characterization. The following fundamental result is well known. It is essentially due to Grothendieck [18]. Theorem.. Let u : Γ Γ C be a function and C > 0. The following conditions are equivalent. 1. M u is a bounded Schur multiplier of B(l (Γ)) and M u C.. There exist a Hilbert space H and maps x, y : Γ H such that u(s, t) = x(s), y(t) and sup s x(s) sup t y(t) C. 3. M u is a bounded Schur multiplier of B(l (Γ)) and M u cb C. If G is a countable group, Bozejko and Fendler [5] proved that Φ is a completely bounded Schur multiplier of Cλ (G) if and only if Φ is the restriction of a Schur multiplier M of B(l (G)) with the same norm; i.e., M = Φ cb. A simple proof of this fact appears in [, Theorem 6.4]. G. Popescu observed that the same argument works in A n. The following proof is due to him Proposition.3. Let Φ be a completely bounded Schur multiplier of A n. Then Φ is the restriction of a Schur multiplier M of B(l (F + n )). Moreover, M = Φ cb.

6 104 ALVARO ARIAS Proof. Find H, π : B(l (F + n )) B(H), and V 1, V : l (F + n ) H satisfying (1). Assume that H is infinite dimensional and use the Dilation Theorem to find unitary maps U 1,, U n B(H H H) such that, for each i n, U i is upper triangular and the (,)-component of U i is S i = π(λ(g i )). Define V 1, V : l (F + n ) H by V i ϕ = (0, ϕ, 0), i = 1,. Then Φ(λ(α))δ α, δ αβ = S α V 1 δ β, V δ αβ = U α V1 δ β, V δ αβ. Since the U α s are unitary, we have that U α V1 δ β, V δ αβ = U β V 1 δ β, U αβ V δ αβ = x(β), y(αβ), where x, y : F + n H H H are defined by x(β) = U β V 1 δ β and y(α) = U α V δ α. Then Φ is the restriction of the Schur multiplier M of B(l (F + n )) defined by MT δ β, δ α = x(β), y(α) T δ β, δ α, and M V 1 V = Φ cb. 3. Completely Contractive Multipliers indexed by subsets of F + n In this section we consider multipliers indexed by subsets of F + n. Let Λ F + n and define the Λ-multiplier Φ Λ of A n by { λ(α), if α Λ, Φ Λ (λ(α)) = 0, otherwise. The main result is the following: Theorem 3.1. The Λ-multiplier Φ Λ is completely contractive (i.e., Φ Λ cb = 1) if and only if Λ = k γ kλ 1 for some semigroup Λ 1 with the left and right cancellation property, and some orthogonal words γ k with no final segment in Λ 1 ; i.e., if γ k = αβ and β 0, then β Λ 1. Suppose that Φ Λ is a completely contractive multiplier. Find a Hilbert space H, a -representation π : B(l (F + n )) B(H), and V 1, V : l (F + n ) H satisfying (1), and note that V 1 = V = 1. For every α Λ and β F + n, V π(λ(α))v 1 δ β = δ αβ. Since V π(λ(α)) 1, we see that V 1 δ β = 1. Hence, V 1 is an isometry. If in addition we assume that 0 Λ, we have that V 1 δ β = π(λ(0))v 1 δ β = V δ 0β = V δ β. Therefore, V 1 = V. Proposition 3.. If Φ Λ is completely contractive multiplier and 0 Λ, then Λ is a semigroup with the left and right cancellation property. Proof. From the comments before the statement of this Proposition, we see that the V 1 and V of (1) are equal to the isometry V : l (F + n ) H. This provides the following criterion to determine if an element belongs to Λ: () α Λ V δ α = π(λ(α))v δ 0.

7 MULTIPLIERS AND REPRESENTATIONS OF A n 105 We will check first that Λ is closed under products. Let α, β Λ. Since V π(λ(α))v δ β = Φ(λ(α))δ β = λ(α)δ β = δ αβ, then V δ αβ = π(λ(α))v δ β = π(λ(α))π(λ(β))v δ 0 = π(λ(αβ))v δ 0. Hence, it follows from () that αβ Λ. We will now verify that Λ has the left cancellation property. Suppose that αβ Λ and α Λ. Then V π(λ(α))v δ β = δ αβ. From here we see that π(λ(α))(v δ β ) = V δ αβ = π(λ(αβ))v δ 0 = π(λ(α))(π(λ(β))v δ 0 ). Since π(λ(α)) is an isometry, we conclude that V δ β = π(λ(β))v δ 0. Therefore, by (), β Λ. Finally, we will check that Λ has the right cancellation property. Suppose that αβ Λ and β Λ. Notice that (Φ(λ(α))δ β, δ αβ ) = (π(λ(α))v δ β, V δ αβ ) = (π(λ(α))π(λ(β))v δ 0, π(λ(αβ))v δ 0 ) = 1. Hence, Φ(λ(α)) 0, so α Λ. We will now prove that multipliers indexed by semigroups with the left and right cancellation property are completely bounded. More precisely, we will show that if Λ F + n is a semigroup with the left and right cancellation property, then Φ Λ is the restriction of a completely contractive Schur multiplier of B(l (F n)). We need the following Lemma 3.3. Let Λ be a semigroup with the left and right cancellation property. Define α β if there exist α 1, α Λ and γ F + n such that α = α 1 γ and β = α γ. Then the relation is an equivalence relation and αβ β if and only if α Λ. Proof. Clearly, is reflexive and symmetric. Suppose that α β and β γ. Then there exist α 1, α, α 3, α 4 Λ and η 1, η F + n such that α = α 1 η 1, β = α η 1, β = α 3 η, and γ = α 4 η. For simplicity, assume that the length of α is greater than or equal to the length of α 3. Then α = α 3 θ for some θ. This implies that θη 1 = η. Since α, α 3 Λ and Λ has the left cancellation property, θ Λ. Hence, γ = α 4 η = (α 4 θ)η 1. Since (α 4 θ) Λ, we get that α γ. Therefore, is an equivalence relation. We will now check that αβ β if and only if α Λ. If α Λ, then β αβ for each β. On the other hand, if β αβ, there exist α 1, α Λ and γ F + n such that β = α 1 γ and αβ = α γ. Hence, αα 1 = α. Since Λ has the right cancellation property, we conclude that α Λ.

8 106 ALVARO ARIAS Proposition 3.4. Let Λ F + n be a semigroup with the left and right cancellation property. Then there exists a completely contractive Schur multiplier M on B(l (F + n )) such that M restricted to A n is Φ Λ. Proof. The equivalence classes induced by the relation of Lemma 3.3 are of the form Λγ, for some γ F + n with no non-trivial initial segment in Λ. Partition F + n = k Λγ k and define x Λ : F + n l by x Λ (α) = e k if α Λγ k. Let M be the completely contractive Schur multiplier of B(l (F + n )) defined by MT δ α, δ β = x Λ (α), x Λ (β) T δ α, δ β. From Lemma 3.3, we see that for every β F + n, M(λ(α))δ β, δ αβ Therefore, Mλ(α) = Φ Λ (λ(α)). = x Λ (β), x Λ (αβ) { { 1, if β αβ, 1, if α Λ, = = 0, otherwise. 0, if α Λ. Notice that MT = k P kt P k, where P k is the orthogonal projection onto l (Λγ k ). Hence, M is completely positive and M(λ(α) ) = (M(λ(α)) for every α F + n. As an immediate consequence, we get Lemma 3.5. Let Λ be a semigroup with the left and right cancellation property, and β F + n. Let M = M Λ be the Schur multiplier of B(l (F + n )) described in Proposition 3.4. Then M(λ(β) ) 0 if and only if β Λ. We will now consider shifts of semigroups. We need the following results and examples. Example 3.6. It is well known that the multiplier p(z) p(z) p(0) has norm strictly greater than 1 in the disc algebra A. Notice that if γ, α F + n, then λ(β), if α = γβ, λ(γ) λ(α) = λ(β), if γ = αβ, 0, otherwise. Proposition 3.7. Let Λ 1 be a semigroup of F + n with the left and right cancellation property, γ Λ 1, and let Λ = γλ 1. Then Φ Λ is completely contractive if and only if no final segment of γ belongs to Λ 1 ; i.e., if γ = αβ and β 0, then β Λ 1.

9 MULTIPLIERS AND REPRESENTATIONS OF A n 107 Proof. Suppose that a non-trivial final segment of γ belongs to Λ 1. Then γ = αβ for some non-zero α, β with β Λ 1. This implies that β k Λ 1 for each k 0. Since γ Λ 1, then α Λ 1 either. Hence, αβ k γλ 1 = Λ iff k 1. Notice that for any polynomial p(e β ) on e β, ] Φ Λ (e α p(e β )) = e α [p(e β ) p(0)e 0. Using Example 3.6, we get that Φ Λ > 1. Suppose now that no final segment of γ belongs to Λ 1. Let M 1 be the completely contractive Schur multiplier of B(l (F + n )) associated to Λ 1 which was described in Proposition 3.4. Define Φ = λ(γ) M 1 (λ(γ)) ; i.e., Φ(T ) = λ(γ)m 1 (λ(γ) T ). Clearly, Φ is a completely contractive map on B(l (F + n )) and λ(α), if α γλ 1, Φ(λ(α)) = λ(γ)m 1 (λ(β) ), if γ = αβ, 0, otherwise. If γ = αβ and β 0 then, by assumption, β Λ 1. Hence, by Lemma 3.5, M 1 (λ(β) ) = 0. Therefore, Φ = Φ Λ. We use the tensor notation for the proof of the following statement. Recall that α, β F + n are orthogonal words if e α ϕ 1 is orthogonal to e β ϕ for every ϕ 1, ϕ in l (F + n ). Example 3.8. Let Λ = α β γ, where α, β are orthogonal words, γ 0, and γ is the semigroup generated by γ. Then Φ Λ > 1. Proof. Find polynomials p(z), q(z) A satisfying p = q = 1, p(0) 1, and p ± q 1 + ɛ, where ɛ > 0 is small enough (you can take p(z) = 1 z and q(z) = 1 N N k=1 zk, where N is large enough). Notice that It is easy to check that Φ Λ [ eα p(e γ ) + e β q(e γ ) ] = p(0)e α + e β q(e γ ). p(0)e α + e β q(e γ ) = p(0) + q On the other hand, since α and β are orthogonal, = 4. e α p(e γ ) +e β q(e γ ) = sup r 1 p(e γ ) r + q(e γ ) r = 1 sup ) r 1( (p + q)(eγ ) r + (p q)(e γ ) r ( ) p(eγ ) + q(e γ ) A n + p(e γ ) q(e γ ) A n 1 = 1 ( p + q + p q ) (1 + ɛ).

10 108 ALVARO ARIAS If ɛ < 1 9, we get that Φ Λ > 1. Proof of Theorem 3.1. Suppose that Φ Λ is completely contractive. If 0 Λ, it follows from Proposition 3. that Λ is a semigroup with the left and right cancellation property. Assume then that 0 Λ and let B be the set of those α Λ such that either α has no non-trivial initial segment in Λ, or that α has length one (i.e., α = g i for some i n). Then B = {γ 1, γ, } for some γ k Λ. It is easy to check that the γ k s are orthogonal words, and that every α Λ can be written as a product α = γ k θ for some k and some θ F + n. Let γ k B and consider the completely contractive map Φ k : A n B(l (F + n )) defined by Φ k = (λ(γ k )) Φ Λ λ(γ k ). Since { λ(α), if Φ k (λ(α)) = (λ(γ k )) γk α Λ, Φ Λ (λ(γ k α)) = 0, otherwise, we have that Φ k : A n A n is a completely contractive multiplier. Since Φ k (λ(0)) = λ(0), it follows from Proposition 3. that Φ k = Φ Λk for some semigroup Λ k with the left and right cancellation property. Hence, α Λ k if and only if γ k α Λ. This implies that Λ = k γ kλ k. We claim that all the Λ k s are equal to each other. Indeed, if this were not the case, we could find two indices k 1, k and γ F + n such that γ Λ k1 and γ Λ k. Using the notation of Example 3.8, ( we would get that Φ Λ eγk1 p(e γ )+e γk q(e γ ) ) = p(0)e γk1 +e γk q(e γ ), which contradicts that Φ Λ is contractive. Therefore, all the Λ k s are equal to each other and Λ = k γ kλ 1. Suppose now that Λ = k γ kλ 1, where Λ 1 is a semigroup with the left and right cancellation property, and the γ k s are orthogonal words with the property that no final segment of γ k belongs to Λ 1. Let x Λ1 : F + n l be the function associated to Λ 1 which was described in Proposition 3.4, and define x, y : F + n l as follows: { xλ1 (η), if β = γ k η for some k, x(α) = x Λ1 (α), and y(β) = 0, otherwise. Let M be the completely contractive Schur multiplier of B(l (F + n )) defined by MT δ α, δ β = x(α), y(β) T δ α, δ β. Then M(T ) = k [λ(γ k) M 1 λ(γ k ) ](T ) = k λ(γ k)m 1 (λ(γ k ) T ), where M 1 is the Schur multiplier associate to Λ 1 which was described in Proposition 3.4. Since no final segment of γ k belongs to Λ 1, it follows from the paragraph above

11 MULTIPLIERS AND REPRESENTATIONS OF A n 109 Lemma 3.5 that { [ λ(γk ) M 1 λ(γ k ) ] λ(α), if α γk Λ 1, λ(α) = 0, otherwise. Therefore the restriction of M to A n is equal to Φ Λ. 4. Contractive multipliers indexed by subsets of F + n In this Section we will show that a Schur multiplier indexed by a subset of F + n is contractive if and only if it is completely contractive. The main step is Theorem 4.1, which replaces Proposition 3.. Theorem 4.1. If Φ Λ is a contractive Schur multiplier and 0 Λ, then Λ is a semigroup with the left and right cancellation property. Following the proof of Theorem 3.1, where Theorem 4.1 replaces Proposition 3., we see that if Φ Λ is contractive, then Λ = k γ kλ k, where the Λ k s are semigroups with the left and right cancellation property, and the γ k s are orthogonal words with no final segment in Λ k. We use the argument of Example 3.8, which only assumes that Φ Λ is contractive, to conclude that all the Λ k s are equal to each other. Hence, we get, Theorem 4.. If Φ Λ is a contractive Schur multiplier, then there exist a semigroup Λ 1 with the left and right cancellation property, and some orthogonal words γ k with no final segment in Λ 1, such that Λ = k γ kλ 1. Consequently, Φ Λ is contractive if and only if Φ Λ is completely contractive. The proof of Theorem 4.1 requires several steps. We will present them separately, since they might be interesting on their own. For the remaining of this Section assume that 0 Λ and that Φ Λ is a contractive Schur multiplier. Step 1. (Powers) If α Λ, then α k Λ for every k 0. Proof. Define J α : A A n and Q α : A n A by J α z k = e k α, k 0, and Q α (e β ) = { z k, if β = α k, 0, otherwise. Let Ψ = Q α Φ Λ J α. Since J α and Q α are contractions, Ψ is a contractive multiplier of A. Notice that Ψ is an indicator multiplier indexed by σ = {k N : α k Λ}.

12 110 ALVARO ARIAS Since A is commutative, Ψ is also completely contractive. Since 0 σ, it follows from Proposition 3. that σ is a semigroup of N 0 = {0, 1,, } with the left and right cancellation property. Since 1 σ, then σ = N 0. Therefore, α k Λ for every k 0. Step. (More powers) If α, αβ Λ, then αβ k Λ for every k 0. Proof. Let Ψ = S αφ Λ S α and notice that Ψ(e γ ) = e γ if αγ Λ, and Ψ(e γ ) = 0 if αγ Λ. Since S α and S α are contractions, Ψ is a contractive multiplier indexed by the subset Λ = {γ : αγ Λ} F + n. By assumption, 0, β Λ. Hence, it follows from Step 1 applied to Λ that for every k 0, β k Λ and therefore αβ k Λ. For Steps 3, 4, 5, and 7 let p(z) = 1 z and q(z) = 1 N N k=1 zk. Then p A = q A = 1 and q = 1 N. Notice that q peaks around z = 1 and p(1) = 0. One can check that for every ɛ > 0, there exists N such that p ± q A 1 + ɛ. If β F + n, then p(e β ) ± q(e β ) An = p ± q A 1 + ɛ. Moreover, following the argument of Example 3.8, we see that if γ 1, γ are orthogonal words, then 5 e γ1 p(e β ) + e γ q(e β ) 1 + ɛ and e γ1 + e γ q(e β ) = 4. For Steps 3,4, 5 and 7 it suffices to take 1 + ɛ < 4. For Step 5 we need a more precise estimate, which is derived from the following elementary Lemma Lemma 4.3. Let p(z), q(z) A be such that p A = q A = 1 and p ± q A 1+ɛ. Let r be a unit vector in l (F + n ). If q(e β ) r 1 ɛ, then p(e β ) r ɛ. Proof. Since p(e β ) ±q(e β ) An = p ±q A 1 + ɛ, then p(e β ) r + q(e β ) r = 1 p(e β) r+q(e β ) r + 1 p(e β) r q(e β ) r 1 [ p+q + p q ] (1+ɛ). Since q(e β ) r 1 ɛ, then p(e β ) r (1+ɛ) (1 ɛ ) = 4ɛ. Step 3. (Products of orthogonal words) If α and β are orthogonal words and α, β Λ, then αβ Λ. Proof. Let α, β Λ be orthogonal words and suppose, contrary to the statement of Step 3, that αβ Λ. By Step 1, β k Λ for each k 0. Then ( Φ Λ e α e ) 0 e β + e β q(e β ) = e α + e β q(e β ). 5 Since e α p(e β ) + e β q(e β ) 1 + ɛ < 4 and e α + e β q(e β ) = Φ Λ > 1. This contradicts the assumption that Φ Λ is contractive. 5 4, then

13 MULTIPLIERS AND REPRESENTATIONS OF A n 111 Step 4. (Right cancellation property for orthogonal words) If α and β are orthogonal words and α, βα Λ, then β Λ. Proof. Let α, βα Λ and suppose, contrary to the statement of Step 4, that β Λ. Then ( Φ Λ e β 1 e ) α + e α q(e α ) = e βα + e α q(e α ). 5 Since e β p(e α ) + e α q(e α ) 1 + ɛ < 4 and e βα then Φ Λ > 1. This contradicts the assumption that Φ Λ is contractive. + e α q(e α ) = 5 4, Step 5. (Left cancellation property for orthogonal words) If α and β are orthogonal words and α, αβ Λ, then β Λ. Proof. Let α, αβ Λ and suppose, contrary to the statement of Step 5, that β Λ. Let λ = ɛ and consider ϕ = p(e β) + λe α q(e β ). It follows from Step that αβ k Λ for each k 0. Then Φ Λ (ϕ) = e 0 + λe α q(e β ). It is easy to check that ϕ and Φ Λ (ϕ) are normed by unit vectors of the form r = r(e α, e β ) with variables in e α and e β only. To see this, follow the argument of Proposition 1 of [1]. Moreover, we can assume that the constant term of r is zero. Indeed, if the constant term of r is not zero, take r e α (note that ψ A n, ψ r = ψ (r e α ) ). Hence, we will only consider vectors r of the form r = e α r 1 + e β r, where r 1 + r = 1. By Proposition 17 of [1], or by Example 6 of [AP], we have that q(e β ) e α An = q(e β ) = q. Hence, e α q(e β ) e α r 1 q r 1. Since we can choose q as small as we want, we see that e α q(e β ) e α r 1 0. Therefore, ϕ r e α [ ] r 1 + λq(e β ) e β r + eβ e 0 e β r, and Φ Λ (ϕ) r e α [ ] r 1 e + λq(e β ) e β r + β r. Clearly, Φ Λ (ϕ) An > λ. If r = e α r 1 + e β r satisfies r = 1 and ϕ r > λ, then λ q(e β ) e β r > λ. Therefore, q(e β ) e β r > 1 λ = 1 ɛ. By Lemma 3.3, we see that 1 (e 0 e β ) r ɛ. It is also clear that r 1 ɛ. Since r 1 + r = 1, then r 1 ɛ. Let µ = sup{ r 1 + λq(e β ) e β r : r 1 + r = 1}. Then, ϕ An µ + 4ɛ + q ɛ, and Φ Λ (ϕ) An µ + (1 ɛ) 4 q ɛ.

14 11 ALVARO ARIAS If we choose q and ɛ small enough, we get that Φ Λ > 1. Step 6. (Left Cancellation Property) If α, αβ Λ, then β Λ. Proof. If α and β are orthogonal, if follows from Step 5 that β Λ. If α and β are commutative, there exist γ F + n and k, l N such that α = γ k and β = γ l. An argument similar to the one used in Step 1 gives that σ γ = {m : γ m Λ} N 0 is closed under products and has the left and right cancellation property. Since we assumed that k, k + l belong to this set, we get that l σ γ. Therefore, β Λ. Assume that α and β are non-commutative and non-orthogonal. Find k 0, l 0 N such that α k 0 is orthogonal to β l 0 (take for instance k 0, l 0 so that α k 0 and β l 0 have the same length). Clearly, γα k is orthogonal to γβ l for every k k 0, l l 0, and γ F + n. In particular, αα k 0 = α k 0+1 is orthogonal to αβ l for every l l 0. By Step, αβ l Λ. Hence, by Step 3, (α k 0+1 )(αβ l ) = α k 0+ β l Λ. Since α k 0+ is orthogonal to β l, we use Step 5 to get β l Λ. Hence, β l Λ for every l l 0. We follow the argument used in Step 1 to conclude that σ β = {l : β l Λ} N 0 is closed under products and has the left and right cancellation property. Since l 0, l belong to this set, we see that 1 σ β also. Therefore, β Λ. Step 7. (Right Cancellation Property) If α, βα Λ, then β Λ. Proof. We assume, as we did in Step 6, that α and β are non-commutative and non-orthogonal. Find k, l N such that (βα) k is orthogonal to α l. Consider e (βα) k β e 0 e α + e α l q(e α ). If (βα) k β Λ, we would have that Φ Λ (e (βα) k β e 0 e α + e α l q(e α )) = 1 e 5 (βα) k β + e α l q(e α ) = 4, which would imply that Φ Λ > 1. Therefore, (βα) k β Λ. Since (βα) k Λ, we conclude from Step 6 that β Λ. Step 8. (Products) If α, β Λ, then αβ Λ. Proof. Assume again that α and β are non-commutative and non-orthogonal. Hence, β = αγ for some γ, which belongs to Λ by Step 6. Since we assumed that the length of β is greater that the length of α, we must consider αβ and βα. We will show first that βα = αγα Λ. Since β = αγ Λ, then β = αγαγ Λ. Since γ Λ, we get from Step 7 that αγα Λ. We will now show that αβ = ααγ Λ. Find k, l N such that α k is orthogonal to γ l. Then, by Step 3, α k γ l Λ. Since γ l 1 Λ, then, by Step 7, α k γ Λ. Since α k Λ, then, by Step 6, α γ Λ.

15 MULTIPLIERS AND REPRESENTATIONS OF A n Multipliers Associated with Representations In this section we show that coefficients of unitary representations of the free semigroup F + n induce completely bounded multipliers of A n. The proof is elementary. Theorem 5.1. Let σ : F + n B(H) be a unitary representation and let ξ, η H. Then the map Φ defined by Φ(λ(α)) = σ(α)ξ, η λ(α) is a completely bounded multiplier of A n satisfying Φ cb ξ η. Proof. We will show that Φ is the restriction of a Schur multiplier M on B(l (F n)) with norm M ξ η. Define x, y : F + n H by x(β) = (σ(β)) ξ and y(β) = (σ(β)) η, and define M on B(l (F + n )) by MT δ β, δ γ = x(β), x(η) T δ β, δ γ. It follows from Theorem. that M is a Schur multiplier with norm M cb ξ η. Since Mλ(α)δ β, δ αβ = x(β), y(αβ) = (σ(β)) ξ, (σ(αβ)) η = σ(αβ)(σ(β)) ξ, η = σ(α)ξ, η, we conclude that Mλ(α) = σ(α)ξ, η λ(α). Recall that B(F + n ) is the space of coefficients of contractive representations of F + n. That is, u : F + n C belongs to B(F + n ) if and only if there exist a Hilbert space H, a contractive representation π : F + n B(H), and two vectors ξ, η F + n such that (3) α F + n, u(α) = π(α)ξ, η. The norm of u is given by u B(F + n ) = inf{ ξ η : (3) holds }. One can check that the coefficients of contractive representations of F + n coincide with the coefficients of unitary representations of F + n ; and hence, Proposition 5.1 is also true for contractive representations. To see this, suppose that u(α) = π(α)ξ, η, where π : F + n B(H) is a contractive representation and ξ, η H. The unital representation π is determined by the contractions π(g i ) = T i B(H). Assume that H is infinitely dimensional and use the Dilation Theorem to find unitary operators U i B(H H H) such that the U i s are upper triangular and the (,)-component of U i is T i. Define the unitary representation σ : F + n B(H H H) by π(g i ) = U i, and let ξ = (0, ξ, 0), η = (0, η, 0) be vectors in H H H. Then, α F + n, π(α)ξ, η = σ(α) ξ, η.

16 114 ALVARO ARIAS It is not difficult to see that B(F + n ) is a Banach algebra and that it is the Banach space dual of the algebra generated by the universal isometric representation of F + n. This is analogous to the situation for discrete groups, where B(G) is the dual of the full C -algebra of G, C (G). Recall also that u : F + n C is a bounded Schur multiplier of A n if the operator Φ u on A n defined by Φ u (λ(α)) = u(α)λ(α) is bounded. M(F + n ) is the space of bounded Schur multipliers with norm u M(F + n ) = Φ u. And M 0 (F + n ) is the space of those u s such that Φ u is completely bounded. The norm on M 0 (F + n ) is given by u M0 (F + n ) = Φ u cb. Proposition 5.1 states that B(F + n ) M 0 (F + n ) and that the inclusion is contractive. We will show that M 0 (F + n ) \ B(F + n ). Example 5.. Let n = and Λ = g F +. Then Φ Λ is completely bounded but Φ Λ is not a coefficient of an isometric representation of F +. Proof. It is easy to see that Φ Λ cb. Suppose that there exists an isometric representation π : F + B(H) and ξ, η H such that { 1, if α Λ, π(α)ξ, η = 0, otherwise. Notice that π is determined by the isometries T 1 = π(g 1 ) and T = π(g ). Hence, for any α F +, T T α ξ, η = 1 and T 1 T α ξ, η = 0. Let C = conv{t T α ξ : α F + } and find z C satisfying (4) z = inf{ w : w C}. It is easy to see that such a z exists and is unique. Notice that for any α F +, T T α z C, T T α z, η = 1, and T 1 T α z, η = 0. Since T T α is an isometry, then T T α z = z. Moreover, since the minimum of (4) is unique, we conclude that α F +, T T α z = z. In particular, T z = T T 1 z = z. Hence, T (z T 1 z) = 0. Since T is one-to-one, we get that T 1 z = z. But this implies that 0 = T 1 z, η = z, η = 1. Therefore, Φ Λ is not the coefficient of a contractive representation. The space of coefficients of the left regular representation plays an interesting role. We say that u B (λ) (F + n ) if there exist ϕ, ψ l (F + n ) such that (5) α Λ, u(α) = λ(α)ϕ, ψ = e α ϕ, ψ, and u B (λ) (F + n ) = inf{ ϕ ψ : (5) holds }.

17 MULTIPLIERS AND REPRESENTATIONS OF A n 115 Davidson and Pitts [9] have recently proved that B (λ) (F + n ) is the predual of F (H n ), the wot-closure of A n. Davidson and Pitts called F (H n ) free semigroup algebras and denoted them by L n. If G is a countable group, B (λ) (G) is called the Fourier algebra of G and is denoted by A(G). It is known that A(G) is the predual of V N(G) (the von Neumann algebra generated by the left regular representation of G), and that A(G) is a closed ideal in B(G) (see [13]). We will show that B (λ) (F + n ) is an ideal in B(F + n ). However, unlike the situation for discrete groups, the norms do not coincide. Indeed, if u : F + n C is defined by u(α) = { 1, if α = 1, 0, otherwise, one can check that u B(F + n ) = 1 and u B (λ) (F + n ) = n. We need Fell s absortion principle. If π : F + n B(H) is a unitary representation, then λ π is unitarily equivalent to λ I. That is, λ π(α) = V (λ I)(α)V for some unitary operator on l (F + n ) H (see [14]). Proposition 5.3. B (λ) (F + n ) is an ideal in B(F + n ). Proof. Let u B (λ) (F + n ) and v B(F + n ). Find a Hilbert space with orthonormal basis {δ k : k 1}, a unitary representation π : F + n B(H), and two vectors ξ, η H such that for each α F + n, v(α) = π(α)ξ, η. Find also ϕ, ψ l (F + n ) such that u(α) = λ(α)ϕ, ψ. Use Fell s absortion principle to find a unitary map V on l (F + n ) H such that V (λ π)v = λ I. Let α F + n. Then, u(α)v(α) = λ(α)ϕ, ψ π(α)ξ, η = (λ π)(α)ϕ ξ, ψ η = (λ I)(α)V (ϕ ξ), V (ψ η). Write V (ϕ ξ) = k ϕ k δ k and V (ψ η) = k ψ k δ k for some ϕ k, ψ k l (F + n ). We employ the technique used by Davidson and Pitts to show that the dual of B (λ) (F + n ) is L n. This one consists in replacing the δ k s with orthogonal elements from l (F + n ). For each k, let θ k = e e k 1. Notice that for every φ 1, φ l (F + n ), φ 1, φ = φ 1 θ k, φ θ k. Moreover, if k j, then φ 1 θ k, φ θ j = 0. Then, it is easy to see that ϕ = k ϕ k θ k and ψ = k ψ k θ k satisfy ϕ = k ϕ k = k ϕ k δ k = V (ϕ ξ) = ϕ ξ and ψ = ψ η.

18 116 ALVARO ARIAS Moreover, u(α)v(α) = (λ(α) I)V (ϕ ξ), V (ψ η) = k=1 λ(α)ϕ k, ψ k = λ(α) ( k ϕ ) ( k θ k, j ψ ) j θ j = λ(α) ϕ, ψ. Since ϕ, ψ l (F + n ), uv B (λ) (F + n ) u B (λ) (F + n ) v B(F + n ). 6. Representations of A n Problem 6.3 of [30] asked if every contractive representation of A n, n, was also completely contractive. In this section we answer this question in the negative. Proposition 6.1. There exists a contractive representation of A n completely contractive. that is not We will also show that there are very simple bounded representations of A n, n, that are not completely bounded. This result is not new. The examples used recently by Pisier [3] to answer Halmos Similarity Problem work also in A n, n. However, they are more complicated. Recall that the unitary flip Θ is defined by Θ(e i1 e in ) = e in e i1 and is denoted by Θ(ϕ) = ϕ. Example 6.. Let n and 1 n < δ < 1. Let π : A n B(l n (F + n )) be the unital representation induced by π(s i ) = δs i, i n, where S i = λ(g i ). Then π is a bounded representation but it is not completely bounded. Proof. Let p P, p An = 1. Write p = p 0 + p 1 + p + + p N, where p k span{e α : α = k}. From Proposition 16 of [1], we have that p k An = p k. Notice that π(p) = p 0 + δp 1 (S1,, Sn) + δ p (S1,, Sn) + + δ N p N (S1,, Sn). Since (Si 1 Si Si k ) = S ik S i S i1, then if follows that p k (S1,, Sn) B(l (F + n )) = (p k(s1,, Sn)) B(l (F + n )) = p k(s 1,, S n ) An, where p k is the flip of p k. Using Proposition 16 of [1] again, we get that p k (S 1,, S n ) An = p k = p k 1. Therefore, π(p) N δ k p k (S1,, Sn) k=0 and π is a bounded representation. n k=0 δ k < 1 1 δ,

19 MULTIPLIERS AND REPRESENTATIONS OF A n 117 We will now check that π is not completely bounded. Fix k and let α 1,, α n k be an enumeration of the set {α : α = k}. Then [S α1 S α S αn k ] [S α 1 S α S α n k ] = n k i=1 S α i Sα 1 i = 1, and = n k 1 i=1 S α i S αi = n k. Since (1 π)[s α1 S αn k ] = δk [S α1 S αn k ], we conclude that π cb (δ n) k. To complete the proof of Proposition 6.1, we need to study the structure of some natural subspaces of A n. Let ϕ A n, and define l (F ϕ ) = span{e 0, ϕ, ϕ, } l (F + n ), A ϕ = span{e 0, ϕ, ϕ } A n. Let P ϕ be the orthogonal projection onto l (F ϕ ). Recall that ϕ A n is inner iff for every ψ 1, ψ l (F + n ), ϕ ψ 1, ϕ ψ = ψ 1, ψ ; and that ϕ A n is inner iff for every ψ 1, ψ l (F + n ), ψ 1 ϕ, ψ ϕ = ψ 1, ψ. Proposition 6.3. Let ϕ be inner. Then the map V ϕ : A A ϕ, defined by V ϕ z k = ϕ k, is an isometry onto. Moreover, if ϕ is also inner, then P ϕ : A n A ϕ is contractive. Proof. Suppose that ϕ is inner, and let p(z) = a 0 + a 1 z + a z + + a N z N be a polynomial. Since ϕ = 1, it follows from von Neumann s inequality that k a kϕ k k a kz k. Notice that l (F ϕ ) [ ϕ l (F ϕ ) ] = span {ψ} for some ψ l (F ϕ ), ψ = 1. Since ϕ is inner (i.e., an isometry), then {ϕ k ψ} k=0 is an orthonormal basis of l (F ϕ ). Let ɛ > 0, and find q = j b jz j satisfying q = 1 and p A(z) pq + ɛ. Since j b j[ϕ k ψ] = 1, we see that k a kϕ k ( k a kϕ k ) ( j b j[ϕ k ψ]) = k j a kb j [ϕ (k+j) ψ] = k j a kb j z k+j = pq p ɛ. Therefore, k a kϕ k An k a kz k A(z). Suppose that ϕ is inner too. Let p P, and write p = p 1 + p, where p 1 l (F ϕ ) and p (l (F ϕ )). From the first part of this Proposition, we see that p 1 = sup{ p 1 η : η = 1, η l (F ϕ )}. Let η l (F ϕ ). Clearly, p 1 η l (F ϕ ), and we claim that p η (l (F ϕ )). Indeed, it is enough to verify that for every k, m 0, ϕ k, p ϕ m = 0. Since

20 118 ALVARO ARIAS ϕ is inner, p (l (F ϕ )), and e 0, p = 0, we get that { ϕ ϕ k, p ϕ m (k m), p = 0, if k m, = e 0, p ϕ (m k) = 0, if k < m. Hence, p η = p 1 η + p η p 1 η and p 1 p. Proof of Proposition 6.1. Let T 1, T, T n B(H) be operators on H satisfying (6) T i T j = 0, i, j n, and n i=1 a it i n i=1 a i, a i C, i n. Let π : A n B(H) be the unital representation determined by π(s i ) = T i. We will check that π is contractive. Let p P, p = a 0 e 0 + (a 1 e a n e n ) + ( higher terms ). Notice that π(p) = p(t 1,, T n ) = a 0 I + (a 1 T 1 + a n T n ) = a 0 I + n a i T, where T = (a 1 T 1 +a T + a n T n )/ n i=1 a i. If follows from (6) that T 1. Let ϕ = (a 1 e 1 + a n e n )/ n i=1 a i and notice that ϕ and ϕ are inner, and that P ϕ (p) = a 0 e 0 + n a i ϕ + b ϕ + + b N ϕ N, i=1 for some b, b 3, b N C. Using the fact that T = = T N = 0, von Neumann s inequality, and Lemma 6.3, we get that π(p) i=1 = a 0 I + ai T = a 0 I + ai T + b T + + b N T N a 0 + ai z + b z + + b N z N = a 0 e 0 + ai ϕ + b ϕ + + b N ϕ N = P ϕ (p) p. To finish the proof, it is enough to find n operators satisfying (6) such that [T 1 T T n ] > 1. Choose T i to be basic elements of the row Hilbert space, shifted to the right (to insure that T i T j = 0). That is, T i = e 1 i+1 M n+1, where M n+1 is the set of (n + 1) (n + 1) matrices. Clearly, these operators satisfy (6), and hence, the unital representation π : A n M n+1 determined by π(λ(g i )) = T i is contractive. However, [λ(g 1 ) λ(g n )] = 1 and [T 1 T n ] = n. Since (1 π)[λ(g1 ) λ(g n )] = [T 1 T n ], we get that π cb n.

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