The Column and Row Hilbert Operator Spaces

Transcription

1 Te Column and Row Hilbert Operator Spaces Roy M Araiza Department of Matematics Purdue University Abstract Given a Hilbert space H we present te construction and some properties of te column and row Hilbert operator spaces H c and H r, respectively Te main proof of te survey is sowing tat C B(H c, K c) is completely isometric to B(H, K ), in oter words, te completely bounded maps between te corresponding column Hilbert operator spaces is completely isometric to te bounded operators between te Hilbert spaces A similar teorem for te row Hilbert operator spaces follows via duality of te operator spaces In particular tere exists a natural duality between te column and row Hilbert operator spaces wic we also sow Te presentation and proofs are due to Effros and Ruan [] Te Column Hilbert Operator Space Given a Hilbert space H, we define te column isometry C : H B(C, H ) given by ξ C(ξ)α : αξ, α C Ten given any n N, we define te nt-matrix norm on M n (H ), by ξ c,n C (n) (ξ), ξ M n (H ) Tese are indeed operator space matrix norms by Ruan s Representation teorem and tus we define te column Hilbert operator space as ( } ) H c H, { c,n n N It follows tat given ξ M n (H ) we ave tat te adjoint operator of C(ξ) is given by Suppose C(ξ) (ζ) d, c C Ten we see tat C(ξ) : H C, ζ (ζ ξ) cd (c C(ξ) (ζ)) (cξ ζ) c(ζ ξ) (c (ζ ξ)) We are also able to compute te matrix norms on rectangular matrices First note tat for ξ, η H, C(η) C(ξ)α (αξ η) α (ξ η), tus giving us te operator C(η) C(ξ) : (ξ η) Now, for m, n N, ξ M m,n (H c ), we ave te induced amplification C (m,n) : M m,n (H c ) M m,n (B(C, H )) B(C n, H m ), and tus, we ave C (m,n) (ξ) C (m,n) (ξ) C (m,n) (ξ) [C(ξij )] [C(ξ ij )]

2 [ n C(ξ li ) C(ξ lj )] l Note tat we ave used te isometry [ n (ξ lj ξ li )] l ϕ : M m,n (H c ) B(C n, H m ), and since bot structures are indeed operator spaces, we actually ave tat ϕ is a complete isometry of tese operator space structures Tis is te case since for all p N, M p (M m,n (H c )) M pm,pn (H c ) ϕ B(C pn, H pm ) M p (B(C n, H m )) ϕ (p) M p (M m,n (H c )) In particular we ave completely isometrically since M m, (H c ) H m c, M m, (H c ) C M m, (B(C, H )) B(C, H m ) C H m c C denotes te column isometry on te corresponding spaces We now wis to present anoter metod for computing te matrix norms in te column Hilbert operator space Suppose tat M n, wit (e ) p H ortonormal vectors We ten ave [ e ij ] c c ([ C(n) ij ]) e [ ( C ij )] [ ( )] [ ( C ij e C [ ( lj e l lj α() li,l α () α (p) Te same formula will also work for rectangular matrices g α (g) li g)] e ij )] e

3 Te Row Hilbert Operator Space We now wis to present te row Hilbert operator space Recall tat given a Hilbert space H we ave te canonical isometry between te conjugate Hilbert space and its Banac dual given by θ : H H, ξ f ξ, f ξ (ζ) : (ζ ξ), and by te Riesz representation teorem we know tat f ξ ξ Define te row isometry by R : H H B(H, C) B(H, C), ζ R(ζ)(ξ) θ(ξ)(ζ) (ζ ξ) Since B(H, C) is an operator space, ten we ave an induced operator space structure on H We define te row Hilbert operator space as ( } ) H r : H, { r,n, n were if given ξ M n (H ), ten ξ r R (n) (ξ), R (n) : M n (H ) M n (B(H, C)) B(H n, C n ), is te induced amplification Te adjoint operator of te row isometry is given by Tis is true since given ζ H, R(ξ) : C H, a aξ (R(ξ) c ζ ) cr(ξ)ζ cθ(ζ)ξ c(ξ ζ) c (ζ ξ) ( cξ ζ ) Tus, we ave tat for η, ξ H, c C, ten R(ξ) : H C, R(η) : C H, and R(ξ)R(η) c R(ξ)(cη) c (ξ η), and tus R(ξ)R(η) : (ξ η) As we did wit te column isometry, let us compute te rectangular matrix norm for te row Hilbert operator space Given ξ M m,n (H r ), we ten ave ξ r R (m,n) (ξ) [ [R(ξij )] [R(ξ ij )] m ] [ R(ξ il )R(ξ jl ) m (ξ il ξ jl )] We also see tat l M,n (H r ) M,n (B(H, C)) B(H n, C) H n r As for te column Hilbert operator space, we ave te very convenient metod for computing te matrix l 3

4 norm Letting M m, and (e ) p H ortonormal vectors, we ave [ e ij ] e r r ([ R(m) ij ]) e [ ( R ij )] e [ ( R [ ( l ij e il )] [ e il α() jl l, [ α () α (p)] We point out te dualities between te column and row isometries; ( R ij )] e α (g) jl g)] e g C : H B(C, H ), ξ C(ξ)α αξ, R : H H B(H, C), ξ R(ξ)(η) (ξ η) Te corresponding adjoint mappings were defined for ξ H as Tis ten gives us C(ξ) : H C, ζ (ζ ξ), R(ξ) : C H, α αξ C(η) C(ξ) : (ξ η) R(ξ)R(η) : (ξ η) Suppose now tat we once again ave an ortonormal set of vectors (e ) p H We will calculate te column and row matrix norms on te row matrix [e e p ] M,n (H ) Using our calculations as before, we know tat [e e p ] c E j e j E j E j c j j I 4

5 In contrast we see tat [e e p ] r E j e j E j E j r j j p Tus, we do ave tat te induced operator space structures on te Hilbert space H by B(C, H ), and B(H, C) are indeed different Suppose tat dim H, tus H C We ten ave tat given ξ C, tat C(ξ)ζ ζξ () R(ξ)ζ (ξ ζ) ξζ () In particular we see tat bot te column and row isometries are just te multiplication action of te vector ξ Tis ten tells us tat for ξ M n, tat ξ c ξ r [ξ ij ] Proposition Given two Hilbert spaces H, K, ten we ave te completely isometric identification B(H, K ) C B(H c, K c ) Proof Let T [T kl ] M n (B(H, K )) B(H n, K n ), and wit tis we define te induced operator defined by T M n (C B(H c, K c )) C B(H c, M n (K c )), H c ξ T [T kl (ξ)] M n (K c ) Now, given ξ M p (H c ) we may assume tat ξ r j α j f j, were (f j ) r j H are ortonormal vectors Let H o span j f j and let K o span k,l T kl (H o ) and let (g i ) q i K o be an ortonormal basis for K o We define te following complex numbers T kl (i, j) as te coefficients given by T kl (f j ) i T kl (i, j)g i, and we set T o (i, j) [T kl (i, j)] k,l M n, T o [T o (i, j)] i,j M nq,nr We see tat T o T since T o is merely te restriction of T to H n o Define te following pt-amplification T (p) Id Mp T : M p H c M p M n K c Ten we compute T (p) (ξ) j α j T (f j ) j α j k,l E kl T kl (f j ) i,j,k,l α j E kl T kl (i, j)g i i,j,k,l α j T kl (i, j)e kl g i i,j α j T o (i, j) g i M p M n K c Tus, in computing te norm we see T j α j T o (, j) Id Mp T o (, ) Id Mp T o (, r) α Id Mn (p) (ξ) Mpn(K c) j α j T o (q, j) Id Mp T o (q, ) Id Mp T o (q, r) α r Id Mn α Id Mn IdMp T o T ξ Mp(H c) α r Id Mn 5

6 We ten ave T T cb We need only sow now tat T T Begin by taking ξ (ξ l ) l H n Recall tat we ave te cb identification M n, (H c ) M n, (B(C, H )) B(C, H n ) Hc n Let (e j ) p j H be an ortonormal basis for span l ξ l, and terefore we know tat we may write for eac l, l n, ξ l (ξ e j ) e j, (ξ e j ) ξ j Now, T [T kl ] M n (B(H, K )), and tus we see tat T (ξ) T kl (e j ) (ξ l e j ) M n, (K c ) l,j Tus, at tis point we decompose te matrix as (ξ e ) T kl (e j ) (ξ l e j ) [ ] [T kl (e )] kl [T kl (e p )] kl (ξ l e j ), l,j (ξ n e p ) were (ξ e ) [ ] [Tkl (e )] kl [T kl (e p )] kl Mn,np (K c ), (ξ l e j ) M np, (ξ n e p ) We ave te induced map T : H c M n (K c ), ζ [T kl (ζ)], and terefore te first matrix becomes and if we write te coefficient matrix as T (,p) ([e e p ]), [(ξ l e j )] l,j, ten we ave te following inequalities; T (ξ) T (,p) ([e e p ]) [(ξ l e j )] T [e e p ] c ξ T ξ cb cb Tus, we ave te desired equality and ave sown tat bounded linear operators between two Hilbert spaces are completely isometric to te completely bounded maps between te corresponding column Hilbert operator spaces We now ave te following dualities; Letting K H, we ten ave (H c ) C B(Hc, C) B(H, C) B(H, C) B(H, C) (H ) r were tese dualites are complete isometries (K r ) H c H c (K ) c, 6

7 Proposition Given two Hilbert spaces H, K we ave te following completely isometric identification; B(K, H ) C B(H r, K r ) By our dualites already stated, we ten ave te completely isometric identifications (H c ) (H ) r (H )r Proposition 3 Given two Hilbert spaces H, K, we ave te completely isometric identification H c (H )c, H r (H )r References [] Edward G Effros and Zong-Jin Ruan Operator spaces 000 7