Pisier’s Self-Dual Hilbert Operator Space
Roy Araiza1
1Department of Mathematics, Purdue University, West Lafayette, IN, USA
Abstract Here we present Gilles Pisier’s remarkable result that given any Hilbert space ℋ there exists a unique operator space structure on ℋ such that the conjugate operator Hilbert space ℋ and the dual operator Hilbert space ℋ ∗ are completely isometric. We give the results as presented by Effros and Ruan [ER00], and the interested reader may reference Pisier’s original work [Pis96] for more details and properties of OH.
1 Pisier’s Self-Dual Operator Hilbert Space
As is well known, given any Hilbert space ℋ we may consider both the conjugate operator Hilbert space ℋ , and its ∗ ∗ dual operator Hilbert space ℋ . The map ∶ H ⟶ ℋ , ↦ f, f() ∶= (ð ) , is an isometry by the Riesz representation theorem. It was shown by Pisier that there is a unique operator space structure on ℋ so that becomes a complete isometry. Given two Hilbert spaces ℋ , K, define the operator
V ∶ ℋ ⊗2 K ⟶ (K, ℋ ), ⊗ ↦ x⊗, x⊗() = (ð ) .
Letting (es)s∈s be an orthonormal basis for K.
1 1 1 H I H I 2 2 2 ô ô ∗ 2 É ó É ó ó 2 x = trace x x = x es x es = esó = , ô ⊗ô2 ⊗ ⊗ ⊗ ó ⊗ ó ó ó ‖ ‖ ‖ ‖‖ ‖ ô ô s∈s ó s∈s ó ó
where the last equality is by Parseval’s equality. Thus, given any ∈ ℋ , ∈ K, we have x⊗ is a Hilbert-Schmidt operator from K to ℋ , and this has shown that the Hilbertian tensor product of the two Hilbert spaces is isometric
to the Hilbert-Schmidt operators between the respective spaces. By restricting the codomain of V to V (ℋ ⊗2 K) we have that V is a unitary, and thus have the unitary equivalence
−1 ∶ ℬ(ℋ ⊗2 K) ⟶ ℬ((K, ℋ )), u ↦ V uV .
∗ Given b ∈ ℬ(ℋ ), and a ∈ ℬ(K), we have (b ⊗ a)x = bxa . To see this suppose that x = x⊗. It then follows
−1 ∗ (b ⊗ a)x⊗ = V (b ⊗ a)V x⊗ = V (b ⊗ a)( ⊗ ) = V (b ⊗ a) = xb⊗a = bx⊗a .
To see this note that for any ∈ K we have x () = ( a) b = b( a∗ó ) = bx a∗. b⊗a ð ó ⊗ ∑ This then shows that for any u = bi ⊗ ai ∈ ℬ(ℋ ) ⊗ ℬ(K), $ % ôÉ ∗ô ‖u‖ = sup ô bixai ô ∶ ‖x‖2 ≤ 1 . ô ô2
1 Now, for Hilbert space ℋ with sesquilinear form (⋅ð ⋅), we have an induced matrix sesquilinear form given by M (ℋ ) × M (ℋ ) ⟶ M ⊗ M , (, ) ↦ (( )) = ó . m n m n ð kló ij i,j,k,l
Given , ∈ Mn(ℋ ), let
É (ℎ) = ⊗ eℎ É (ℎ) = ⊗ eℎ,
where (eℎ) ⊂ ℋ is an orthonormal basis. Then we have (( )) = ó ð kló ij É (ℎ) ó É (ℎ) = ⊗ eℎó ij ⊗ eℎ kl ó 4 0 15 É (ℎ) (ℎ) ó = e ó e ij kl ℎó ℎ ó É (ℎ) (ℎ) = ⊗ ∈ Mn ⊗ Mn.
At this point we are then able to prove an analogue to the Schwarz inequality.
Theorem 1.1 (Haagerup [Haa85]). Given a Hilbert space ℋ and n ∈ ℕ, then for any , ∈ Mn(ℋ ), we have
1 1 ‖((ð ))‖ ≤ ‖((ð ))‖ 2 ‖((ð ))‖ 2 .
Proof. Given , ∈ Mn(ℋ ),
É (ℎ) = ⊗ eℎ É (ℎ) = ⊗ eℎ then we may assume that ℋ = K = ℂn, since
É (ℎ) (ℎ) ((ð )) = ⊗ ∈ Mn ⊗ Mn.
Furthermore recall that if , ∈ Mn then HL MI ∗ É É É ó trace( ) = trace ik jk = ij ij = ijó ij . k i,j i,j ó
We then have by our above calculations that
ôÉ (ℎ) (ℎ)ô ‖((ð ))‖ = ô ⊗ ô ô $ ô % ôÉ (ℎ) (ℎ)∗ô = sup ô x ô ∶ ‖x‖2 ≤ 1 ô ô2 < = ó É ó = sup ótrace (ℎ)x (ℎ)∗y∗ ó ∶ x , y 1 . ó ó ‖ ‖2 ‖ ‖2 ≤ ó ó Thus, fix an x and y as above and let x = ðxð , y = w ðyð be the corresponding polar decompositions. Write x = 1 1 x x , x x 2 , x x 2 , y x x∗ x∗ , x∗x 1 2 1 = ð ð 2 = ð ð and take an analogous decomposition for . It then follows that 1 1 = ð ð 2 2 = ðxð , where the first equality follows since
óx∗ó2 = xx∗ = x x ∗ = x ∗ x ∗ = ( x ∗)2. ó ó ð ð ð ð ð ð ð ð ð ð
2 The analogous equalities also hold for y. We then have the following string of inequalities ó É ó É ótrace (ℎ)x (ℎ)∗y∗ ó ótrace (ℎ)x x (ℎ)∗y∗y∗ ó ó ó ≤ ó 1 2 2 1 ó ó ó ó ó É ó ∗ (ℎ) (ℎ) ∗ ∗ó = ótrace (y x1)(y2 x ) ó ó 1 2 ó 1 1 É ∗ (ℎ) ∗ (ℎ) ∗ 2 (ℎ) ∗ (ℎ) ∗∗ 2 ≤ trace y1 x1 y1 x1 trace y2 x2 y2 x2
The equality holds by decomposition of x and y into x1x2 and y1y2, respectively, and then using the tracial property, and the last inequality follows by applying the classical Cauchy-Schwarz inequality. Continuing we get
1 1 É ∗ (ℎ) ∗ (ℎ) ∗ 2 É (ℎ) ∗ (ℎ) ∗∗ 2 ≤ trace y1 x1 y1 x1 trace y2 x2 y2 x2 ôÉ ∗ (ℎ) ô ôÉ (ℎ) ∗ô = ô y x1ô ô y2 x ô ô 1 ô2 ô 2ô2 ôóÉ ∗ (ℎ) óô ôóÉ (ℎ) ∗óô = ôó y x1óô ôó y2 x óô ôó 1 óô2 ôó 2óô2 1 1 óÉ ∗ (ℎ) ∗ (ℎ)∗ ó 2 óÉ (ℎ) ∗ (ℎ)∗ ∗ó 2 = ó trace y x1x y1 ó ó trace y2 x x2 y ó ó 1 1 ó ó 2 2 ó 1 1 óÉ (ℎ) ∗ (ℎ)∗ ∗ ó 2 óÉ (ℎ) (ℎ)∗ ó 2 = ó trace óx ó óy ó ó ó trace ðxð ðyð ó ó ó ó ó ó ó ó ó 1 1 ó 1ó 2 ó 1ó 2 ó É (ℎ) (ℎ) ∗ ó ∗ ó ó É (ℎ) (ℎ) ó ó = ó ⊗ óx óó óy ó ó ó ⊗ ðxðó ðyð ó ó ó óó ó ó ó ó ó ó ó ó ó ó ó ó 1 1 ôÉ (ℎ) (ℎ)ô 2 ôÉ (ℎ) (ℎ)ô 2 ≤ ô ⊗ ô ô ⊗ ô . ô ô ô ô
Given a Hilbert space ℋ and n ∈ ℕ, we define the OH matrix norm ‖⋅‖o,n by
1 ‖‖o,n ∶= ‖((ð ))‖ 2 , ∈ Mn(ℋ ). ∑ (ℎ) Thus, we have that if (eℎ)ℎ ⊂ ℋ is an orthornormal basis and = ⊗ eℎ, that
1 1 0 < =1 2 É 2 ó É ó = ô (ℎ) ⊗ (ℎ)ô = sup ótrace (ℎ)x (ℎ)∗y∗ ó ∶ x, y ∈ B . ‖ ‖o,n ô ô ó ó (K,ℋ ) ô ô ó ó We will now prove that ‖⋅‖o satisfies Ruan’s axioms and furthermore that it induces the unique operator space structure on the Hilbert space such that its conjugate operator space and dual operator space are completely isometric. Theorem 1.2 (Pisier). Given a Hilbert space ℋ , then the OH matrix norm on ℋ satisfies Ruan’s axioms, and we let the , . ∗ corresponding operator space ℋ ‖⋅‖o,n n∈ℕ be denoted by ℋo Letting ℋ and ℋ have the induced conjugate and dual operator space structures, respectively, then ‖⋅‖o is the unique operator space matrix norm for which the corresponding mapping ∶ ℋ ⟶ ℋ ∗ is completely isometric.
Proof. We begin by proving that ‖⋅‖o satisfies Ruan’s axioms. Let (eℎ)ℎ ⊂ ℋ be an orthonormal basis for ℋ and let ∈ Mn(ℋ ), ∈ Mm(ℋ ). Then we have that for
É (ℎ) = ⊗ eℎ É (ℎ) = ⊗ eℎ,
3 that
É (ℎ) (ℎ) ℎ ℎ (( ⊕ ð ⊕ )) = ( ⊕ ) ⊗ ( ( ) ⊕ ( )) ℎ É = ( (ℎ) ⊗ (ℎ)) ⊕ ( (ℎ) ⊗ (ℎ)) ⊕ ( (ℎ) ⊗ (ℎ)) ⊕ ( (ℎ) ⊗ (ℎ)) ℎ = ((ð )) ⊕ ((ð )) ⊕ ((ð )) ⊕ ((ð )) and therefore
2 ‖ ⊕ ‖o = ‖(( ⊕ ð ⊕ ))‖ = ‖((ð )) ⊕ ((ð )) ⊕ ((ð )) ⊕ ((ð ))‖ 2 2 ≤ max ‖‖o , ‖‖o , ‖‖o , ‖‖o 2 2 = max ‖‖o , ‖‖o.
We therefore have that ‖ ⊕ ‖o ≤ max ‖‖o , ‖‖o . ∑ (ℎ) Let = ℎ ⊗ eℎ ∈ Mp(ℋ ), ∈ Mn,p and ∈ Mp,n. We first show that
(( ð )) = ( ⊗ ) ((ð )) ( ⊗ ).
First note that L M É (ℎ) = ikkl kj ∈ Mn(ℋ ), k,l i,j and thus we will we have
É (ℎ) (ℎ) (( ð )) = ⊗ ℎ L M É = (ℎ) (ℎ) kl ij ℎ i,j,k,l L H IH IM É É (ℎ) É (ℎ) = kaab bl iccd dj ℎ a,b c,d i,j,k,l L M É (ℎ) (ℎ) = kaab bl iccd dj a,b,c,d,ℎ i,j,k,l L M É (ℎ) (ℎ) = ka icab cd bl dj a,b,c,d,ℎ i,j,k,l L H I M É É (ℎ) (ℎ) = ka ic ab cd bl dj a,b,c,d ℎ i,j,k,l = ( ⊗ ) ((ð )) ( ⊗ ). Since we have that ô ⊗ ô = ô ô = 2 , ô ô ‖ ‖ ô ô ‖ ‖ ô ô ô ô 2 ô ⊗ ô = ‖ ‖ ô ô = ‖ ‖ , ô ô ô ô we then have the inequality 2 2 2 2 ‖ ‖o = ‖(( ð ))‖ ≤ ‖ ‖ ‖‖o ‖ ‖ ,
4 which gives us Ruan’s second axiom. Therefore, ‖⋅‖o is indeed an operator space matrix norm. ∗ We now prove that ∶ ℋo ⟶ ℋo is a complete isometry. Given n ∈ ℕ, and ∈ Mn(ℋo), we have
(n) ∗ () ∈ Mn(ℋo ) ≅ Cℬ(ℋo,Mn),
(n) and thus we set ' = (). Therefore for m ∈ M, and ∈ Mm(ℋo) we see
(m) (n) ' () = '(kl) = ()(kl) = (ij)(kl) = kló ij = (( )) . k,l k,l i,j,k,l ó i,j,k,l ð It then follows by applying our analogue of Cauchy-Schwarz that
ô (m) ô ô' ()ô ≤ ‖‖o ‖‖o . ô ô Thus, we have ô (n) ô ô ()ô ≤ ‖‖o . ô ôcb Conversely, letting = , implies ‖‖o
ô (n) ô 1 ô ()()ô = ‖((ð ))‖ = ‖‖o , ô ôcb ‖‖o implying that is a complete isometry.
¨ Suppose now that ‖⋅‖ is another operator space matrix norm such that is a complete isometry. Then by our above calculations we have ¨ ‖‖o = sup ‖((ð ))‖ ∶ ‖‖ ≤ 1 . Letting = ¨ implies that ‖‖ ¨ 1 ¨ ‖‖ ≥ ‖((ð ))‖ ¨ ⟹ ‖‖ ≥ ‖‖o . ‖‖ Conversely this also works for all vectors which implies
¨ ¨ ‖‖ = sup ‖((ð ))‖ ∶ ‖‖ ≤ 1
≤ sup ‖((ð ))‖ ∶ ‖‖o ≤ 1 1 = ‖((ð ))‖ 2 = ‖‖o .
Thus, ‖⋅‖o is the unique operator space matrix norm for which the conjugate Hilbert operator space and the dual Hilbert operator space are completely isometric.
References
[ER00] Edward G Effros and Zhong-Jin Ruan. “Operator spaces”. In: (2000). [Haa85] Uffe Haagerup. “Injectivity and decomposition of completely bounded maps”. In: Operator algebras and their connections with topology and ergodic theory. Springer, 1985, pp. 170–222. [Pis96] Gilles Pisier. The operator Hilbert space OH, complex interpolation and tensor norms. Vol. 122. American Mathematical Soc., 1996.
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