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Operator Spaces Operator Spaces. Alonso Delf´ın University of Oregon. November 19, 2020 Abstract The field of operator spaces is an important branch of functional analysis, commonly used to generalize techniques from Banach space theory to algebras of operators on Hilbert spaces. A \concrete" operator space E is a closed subspace of L(H) for a Hilbert space H. Almost 35 years ago, Zhong-Jin Ruan gave an abstract characterization of operator spaces, which allows us to forget about the concrete Hilbert space H. Roughly speaking, an \abstract" operator space consists of a normed space E together with a family of matrix norms on Mn(E) satisfying two axioms. Ruan's Theorem states that any abstract operator space is completely isomorphic to a concrete one. On this document I will give a basic introduction to operator spaces, providing many examples. I will discuss completely bounded maps (the morphisms in the category of operator spaces) and try to give an idea of why abstract operator spaces are in fact concrete ones. Time permitting, I will explain how operator spaces are used to define a non-selfadjoint version of Hilbert modules and I'll say how this might be useful for my research. 1 Definitions and Examples Definition 1.1. Let H be a Hilbert space. An operator space E is a closed subspace of L(H). Example 1.2. 1. Any C∗-algebra A is an operator space. 2. Let H1 and H2 be Hilbert spaces. Then L(H1; H2) is regarded as an operator space by identifying L(H1; H2) in L(H1 ⊕ H2) by 0 0 L(H ; H ) 3 a 7! 2 L(H ⊕ H ) 1 2 a 0 1 2 3. Any Banach space E is an operator space. Indeed, it's well known that BE∗ is a compact space when equipped with the weak-∗ topology and E is identified with a subspace of C(BE∗ ) via the isometric mapping ξ 7! ξb, where ξb(') = '(ξ) ∗ for any ' 2 BE∗ . Since C(BE∗ ) is a C -algebra, it follows that E is an operator space. The main difference between the category of Banach spaces and that of operator spaces is in the morphisms. We will see below that we need to look at linear maps that behave well with respect to some natural matrix norms. 1 Matrix norms. Let E be an operator space and n 2 Z>0. Then, Mn(E), the space of n × n matrices with entries in E, is a subspace of Mn(L(H)). Thus, Mn(E) has a natural norm k · kn, which comes from the n n 2 identification of Mn(L(H)) as L(H ), where H is the ` direct sum of H with it self. More precisely, if ξ := (ξj;k) 2 Mn(E) 8 9 0 211=2 <> n n => X X kξkn = k(ξj;k)kn := sup @ ξj;khk A : h := (h1; : : : ; hn) 2 BHn :> j=1 k=1 ;> Of course the norm k · k1 coincides with the norm of E. The following lemma gives a useful way to compute kξkn: Lemma 1.3. Let E be an operator space and n 2 Z≥0. Then, 8 9 n n < X X = kξkn = k(ξj;k)kn = sup hξj;khk; fji : h; f 2 BHn : j=1 k=1 ; Proof. Let h; f 2 BHn . Recall that khk = supkfk=1 jhh; fij, from where we get 2 n n n n n n 2 X X X X X X 2 hξj;khk; fji ≤ h ξj;khk; fji ≤ ξj;khk ≤ kξk n j=1 k=1 j=1 k=1 j=1 k=1 For the reverse inequality, ... Completely bounded linear maps. Let E and F be operator spaces and u : E ! F a linear map. For each n 2 Z>0, u induces a linear map un : Mn(E) ! Mn(F ) in the obvious way un((ξj;k)) := (u(ξj;k)): Further, we set kunk := supfkun(ξ)kn : ξ 2 Mn(E); kξkn = 1g. We say that u is completely bounded (c.b.) if kukcb := sup kunk < 1: n2Z>0 We will denote by CB(E; F ) ⊂ L(E; F ) to the set of all c.b maps from E to F . Notice that when equipped with the norm k · kcb, CB(E; F ) is a Banach space and therefore also an operator space. Definition 1.4. Let E; F be operator spaces and u 2 CB(E; F ). 1. If kukcb ≤ 1 we say u is completely contractive. 2. If each un is an isometry, we say u is a complete isometry. 3. We say E and F are completely isomorphic if u is an isomorphism with u−1 2 CB(F; E). 4. We say E and F are completely isometrically isomorphic if u is a complete isomorphism that's also a complete isometry. Proposition 1.5. Let E; F and G be operator spaces and u 2 CB(E; F ), v 2 CB(F; G), then vu 2 CB(E; G) and kvukcb ≤ kvkcbkukcb. Proof. Since k(vu)nk = sup kvn(un(ξ))kn ≤ kvnkkunk; kξkn=1 it follows that kvuk = sup k(vu) k ≤ sup kv kku k = kvk kuk . n2Z>0 n n2Z>0 n n cb cb 2 Proposition 1.6. Let E; F be operator spaces and u 2 CB(E; F ) a rank one operator. That is, u(ξ) = '(ξ)η ∗ for ' 2 E and η 2 F . Then, kukcb = kuk. Proof. We always have kuk ≤ kukcb. For the reverse inequality, we recall first that kuk = k'kkηk. Then, notice that for any ξ = (ξj;k) 2 Mn(E) we have using the lemma above X X k'n(ξ)kn = k('(ξj;k))k = sup '(ξj;k)xkyj ≤ k'k sup ξj;kxkyj ≤ k'kkξkn x;y2B 2 x;y2B 2 `n j;k `n j;k Hence, kun(ξ)kn = kηkk'n(ξ)kn ≤ kηkk'kkξkn = kukkξkn from where it follows that kukcb ≤ kuk. Definition 1.7. Let n 2 Z>0. We write Mn instead of Mn(C). We use the natural norm on Mn, which 2 comes from the identification of Mn with L(` (f1; : : : ; ng)). If E is an operator space, the multiplication of elements in Mn(E) with Mn is done in the obvious way. Proposition 1.8. Let E ⊂ L(H) be an operator space and k · kn the norms on Mn(E) defined above. For ξ := (ξj;k) 2 Mn(E) and α := (αj;k); β := (βj;k) 2 Mn, we have (R1) kαξβkn ≤ kαkkξknkβk. (R2) If η := (ηj;k) 2 Mm(E), then ξ 0 = maxfkξk ; kηk g 0 η n m n+m Proof. To see (R1) we define for any α 2 Mn an element αe := (αj;kidH) 2 Mn(L(H)). Notice that kαk = kαekn. Furthermore, αξβ = αξe βe 2 Mn(E). Therefore kαξβkn = kαξe βekn ≤ kαeknkξknkβekn = kαkkξknkβk For (R2), notice that ξ ξ 0 idHn ξ 0 ξ 0 kξk = = ≤ kid n k = 0 0 η 0 0 η H 0 η n+m n+m and similarly ξ 0 kηk ≤ 0 η n+m This gives one inequality. For the reverse one, we take h 2 Hn and f 2 Hm and notice that 2 2 ξ 0 ξ ξh 2 2 2 2 = = kξhk + kηfk ≤ maxfkξkn; kηkmg(khk + kfk ): 0 η 0 ηf This finishes the proof. Theorem 1.9. (Ruan 1987) Suppose that E is a vector space, and that for each n 2 Z>0 we are given a norm k · kn on Mn(E) satisfying conditions (R1) and (R2) above. Then E is completely isometrically isomorphic to a subspace of L(H), for some Hilbert space H. 3 Sketch of Proof. Let U be the collection of all completely contractive maps u : E ! Mn for some n 2 Z≥0. For each u 2 U, we define nu 2 Z≥0 to be so that Mnu is the co-domain of u. Then, sup M M := Mnu u2U ∗ is a C -algebra and hence an operator space. We define v : E ! M by v(ξ) = (u(ξ))u2U . One checks v is a complete contraction. Furthermore, if ξ 2 Mn(E), by Hahn-Banach there is a complete contraction u : E ! Mn(E) such that kun(ξ)k = kξnk. Now consider the projection pu : M ! Mnu and notice that kvn(ξ)k = k(v(ξj;k))k ≥ k(pu(u(ξj;k)))k = kun(ξ)k = kξnk: Thus, vn is an isometry and therefore v is a complete isometry. \" 2 Column and Row Hilbert space. Let H be any Hilbert space. There are several (completely isometrically isomorphic) ways of giving H a canonical operator space structure which we call the column Hilbert space and denote by Hc. Informally, one should think of Hc as a \column in L(H)". We now give 3 equivalent descriptions of Hc for a general Hilbert space H. 1. Identify H with L(C; H) by regarding each h 2 H as a map th : C !H defined by th(λ) := λh. Notice ∗ ∗ that the operator th : H! C is such that th(f) = hf; hi and therefore ∗ (thtf )(1) = hf; hi c Using this we notice that the induced norm in Mn(H ) takes a nice form: 1=2 1=2 n ! n ! X ∗ X k(th )k = t th = hhi;k; hi;ji j;k hi;j i;k i=1 j;k i=1 j;k where we've used the C∗-identity. c 2. Fix a unit vector f 2 H. Look at the rank one operators θh;f (y) := hy; fih and identify H with fθh;f : h 2 Hg ⊂ L(H).
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