The Column and Row Hilbert Operator Spaces Roy M. Araiza Department of Mathematics Purdue University Abstract Given a Hilbert space H we present the construction and some properties of the column and row Hilbert operator spaces Hc and Hr, respectively. The main proof of the survey is showing that CB(Hc; Kc) is completely isometric to B(H ; K ); in other words, the completely bounded maps be- tween the corresponding column Hilbert operator spaces is completely isometric to the bounded operators between the Hilbert spaces. A similar theorem for the row Hilbert operator spaces follows via duality of the operator spaces. In particular there exists a natural duality between the column and row Hilbert operator spaces which we also show. The presentation and proofs are due to Effros and Ruan [1]. 1 The Column Hilbert Operator Space Given a Hilbert space H , we define the column isometry C : H −! B(C; H ) given by ξ 7! C(ξ)α := αξ; α 2 C: Then given any n 2 N; we define the nth-matrix norm on Mn(H ); by (n) kξkc;n = C (ξ) ; ξ 2 Mn(H ): These are indeed operator space matrix norms by Ruan's Representation theorem and thus we define the column Hilbert operator space as n o Hc = H ; k·kc;n : n2N It follows that given ξ 2 Mn(H ) we have that the adjoint operator of C(ξ) is given by ∗ C(ξ) : H −! C; ζ 7! (ζj ξ) : Suppose C(ξ)∗(ζ) = d; c 2 C: Then we see that cd = (cj C(ξ)∗(ζ)) = (cξj ζ) = c(ζj ξ) = (cj (ζj ξ)) : We are also able to compute the matrix norms on rectangular matrices. First note that for ξ; η 2 H , C(η)∗C(ξ)α = (αξj η) = α (ξj η) ; thus giving us the operator C(η)∗C(ξ) := (ξj η) : Now, for m; n 2 N; ξ 2 Mm;n(Hc); we have the induced amplification (m;n) ∼ n m C : Mm;n(Hc) −! Mm;n(B(C; H )) = B(C ; H ); and thus, we have 1 2 1 (m;n) (m;n) ∗ (m;n) ∗ 2 C (ξ) = C (ξ) C (ξ) = [C(ξij)] [C(ξij)] 1 1 1 " n # 2 " n # 2 X ∗ X = C(ξ`i) C(ξ`j) = (ξ`jj ξ`i) : `=1 `=1 Note that we have used the isometry ∼ n m ' : Mm;n(Hc) = B(C ; H ); and since both structures are indeed operator spaces, we actually have that ' is a complete isometry of these operator space structures. This is the case since for all p 2 N, ∼ pn pm n m ∼ Mp(Mm;n(Hc)) = Mpm;pn(Hc) =' B(C ; H ) = Mp(B(C ; H )) ='(p) Mp(Mm;n(Hc)): In particular we have ∼ m Mm;1(Hc) = Hc ; completely isometrically since ∼ ∼ m ∼ m Mm;1(Hc) =C Mm;1(B(C; H )) = B(C; H ) =C Hc : C denotes the column isometry on the corresponding spaces. We now wish to present another method (h) for computing the matrix norms in the column Hilbert operator space. Suppose that α 2 Mn; with (eh)1≤h≤p ⊂ H orthonormal vectors. We then have " # X (h) X (h) α ⊗ eh = αij eh h c h c " #! (n) X (h) = C αij eh h " !# X (h) = C αij eh h 1 " !#∗ " !# 2 X (h) X (h) = C αij eh C αij eh h h 1 " !# 2 X X (h) X (g) = α`j eh α`i eg ` h g 1 2 3 2 X (h) (h) = 4 α α 5 `j `i h;` 1 2 X (h)∗ (h) = α α h 2 (1)3 α 6 . 7 = 4 . 5 : α(p) The same formula will also work for rectangular matrices. 2 2 The Row Hilbert Operator Space We now wish to present the row Hilbert operator space. Recall that given a Hilbert space H we have the canonical isometry between the conjugate Hilbert space and its Banach dual given by ∗ θ : H −! H ; ξ 7! fξ; fξ(ζ) := (ζj ξ) ; and by the Riesz representation theorem we know that kfξk = kξk. Define the row isometry by ∗∗ ∗ R : H −! H = B(H ; C) = B(H ; C); ζ 7! R(ζ)(ξ) = θ(ξ)(ζ) = (ζj ξ) : Since B(H ; C) is an operator space, then we have an induced operator space structure on H . We define the row Hilbert operator space as n o Hr := H ; k·kr;n ; n where if given ξ 2 Mn(H ), then (n) (n) ∼ n n kξkr = R (ξ) ;R : Mn(H ) −! Mn(B(H ; C)) = B(H ; C ); is the induced amplification. The adjoint operator of the row isometry is given by ∗ R(ξ) : C −! H ; a 7! aξ: This is true since given ζ 2 H , ∗ (R(ξ) cj ζ = cR(ξ)ζ = cθ(ζ)ξ = c(ξj ζ) = c (ζj ξ) = cξ ζ : Thus, we have that for η; ξ 2 H ; c 2 C, then R(ξ): H −! C ;R(η)∗ : C −! H ; and R(ξ)R(η)∗c = R(ξ)(cη) = c (ξj η) ; and thus R(ξ)R(η)∗ := (ξj η) : As we did with the column isometry, let us compute the rectangular matrix norm for the row Hilbert operator space. Given ξ 2 Mm;n(Hr); we then have 1 1 " m # 2 " m # 2 1 (m;n) ∗ 2 X ∗ X kξkr = R (ξ) = [R(ξij)] [R(ξij)] = R(ξi`)R(ξj`) = (ξi`j ξj`) : `=1 `=1 We also see that ∼ ∼ n ∼ n M1;n(Hr) = M1;n(B(H ; C)) = B(H ; C) = Hr : As for the column Hilbert operator space, we have the very convenient method for computing the matrix 3 (h) norm. Letting α 2 Mm; and (eh)1≤h≤p ⊂ H orthonormal vectors, we have " # X (h) X (h) α ⊗ eh = αij eh h r h r " #! (m) X (h) = R αij eh h " !# X (h) = R αij eh h 1 " !# " !#∗ 2 X (h) X (h) = R αij eh R αij eh h h 1 " !# 2 X X (h) X (g) = αi` eh αj` eg ` h g 1 2 3 2 X (h) (h) = 4 α α 5 i` j` `;h 1 2 X (h) (h)∗ = α α h h i (1) (p) = α ··· α : We point out the dualities between the column and row isometries; C : H −! B(C; H ); ξ 7! C(ξ)α = αξ; ∗∗ R : H −! H = B(H ; C); ξ 7! R(ξ)(η) = (ξj η) : The corresponding adjoint mappings were defined for ξ 2 H as ∗ C(ξ) : H −! C; ζ 7! (ζj ξ) ; ∗ R(ξ) : C −! H ; α 7! αξ: This then gives us C(η)∗C(ξ) := (ξj η) R(ξ)R(η)∗ := (ξj η) : Suppose now that we once again have an orthonormal set of vectors (eh)1≤h≤p ⊂ H : We will calculate the column and row matrix norms on the row matrix k[e1 ··· ep]k 2 M1;n(H ): Using our calculations as before, we know that 1 p p 2 1 X X 2 k[e1 ··· ep]k = E1j ⊗ ej = Ej1E1j = kIk = 1: c j=1 j=1 c 4 In contrast we see that 1 2 p p X X p k[e1 ··· ep]k = E1j ⊗ ej = E1jEj1 = p: r j=1 j=1 r Thus, we do have that the induced operator space structures on the Hilbert space H by B(C; H ); and B(H ; C) are indeed different. Suppose that dim H = 1; thus H = C: We then have that given ξ 2 C; that C(ξ)ζ = ζξ (1) R(ξ)ζ = (ξj ζ) = ξζ: (2) In particular we see that both the column and row isometries are just the multiplication action of the vector ξ: This then tells us that for ξ 2 Mn; that kξkc = kξkr = k[ξij]k : Proposition 2.1. Given two Hilbert spaces H ; K , then we have the completely isometric identification ∼ B(H ; K ) = CB(Hc; Kc): ∼ n n Proof. Let T = [Tkl] 2 Mn(B(H ; K )) = B(H ; K ); and with this we define the induced operator ∼ Te 2 Mn(CB(Hc; Kc)) = CB(Hc;Mn(Kc)); defined by 3 ξ 7! [T (ξ)] 2 M ( ): Hc Te kl n Kc Pr r Now, given ξ 2 Mp(Hc) we may assume that ξ = j=1 αj ⊗ fj; where (fj)j=1 ⊂ H are orthonormal q vectors. Let Ho = spanj fj and let Ko = spank;l Tkl(Ho) and let (gi)i=1 ⊂ Ko be an orthonormal basis for Ko: We define the following complex numbers Tkl(i; j) as the coefficients given by X Tkl(fj) = Tkl(i; j)gi; i and we set To(i; j) = [Tkl(i; j)]k;l 2 Mn;To = [To(i; j)]i;j 2 Mnq;nr: We see that kTok ≤ kT k since To is n merely the restriction of T to Ho : Define the following pth-amplification (p) Te = IdMp ⊗ Te : Mp ⊗ Hc −! Mp ⊗ Mn ⊗ Kc: Then we compute (p) X X X Te (ξ) = αj ⊗ Te(fj) = αj ⊗ Ekl ⊗ Tkl(fj) j j k;l X X X = αj ⊗ Ekl ⊗ Tkl(i; j)gi = αj ⊗ Tkl(i; j)Ekl ⊗ gi = αj ⊗ To(i; j) ⊗ gi 2 Mp ⊗ Mn ⊗ Kc: i;j;k;l i;j;k;l i;j Thus, in computing the norm we see 2P α ⊗ T (1; j)3 2 3 2 3 j j o IdMp ⊗To(1; 1) ··· IdMp ⊗To(1; r) α1 ⊗ IdMn (p) 6 . 7 6 . 7 6 . 7 Te (ξ) = 4 . 5 = 4 . 5 4 . 5 Mpn(Kc) P j αj ⊗ To(q; j) IdMp ⊗To(q; 1) ··· IdMp ⊗To(q; r) αr ⊗ IdMn 2 3 α1 ⊗ IdMn .