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http://ijmms.hindawi.com © Hindawi Publishing Corp.

COMPLETELY CONTRACTIVE MAPS BETWEEN

C

-ALGEBRAS

W. T. SULAIMAN

Received 15 November 2001

We give a simple proof that any completely contractive map betweenC∗-algebras is the top right hand corner of a two completely positive unital matrix operator. Some well-known results are deduced.

2000 Mathematics Subject Classification: 47B65.

1. Introduction. LetAandBbeC∗-algebras,S⊂Abe a subspace, andφ:S→Bbe a linear map. We defineφn:Mn(S)→Mn(B)by

φnaij=φaij. (1.1)

We said that φ is n-positive if φn is positive and thatφ is completely positive if

φn is positive for all n. The mapφ is said to ben-bounded (resp.,n-contractive) ifφn ≤c (resp.,φn ≤1). The mapφis said to be completely bounded (resp., completely contractive) ifφcb=supnφn<∞(resp.,φcc=supnφn ≤1).n -positivity (resp.,n-boundedness orn-contractivity) implies(n−1)-positivity (resp., (n−1)-boundedness or(n−1)-contractivity). The converse is not true in general.

For anyC∗-algebraA,Mn(Mp(A))is identified withMp(Mn(A))because there is a canonical isomorphism betweenMn(Mp(A))andMp(Mn(A))by the rearrangement of ann×nmatrix ofp×pblocks as ap×pmatrix ofn×nblocks with the(i,j)th entry of the(k,)-block becoming the(k,)th entry of the(i,j)th block. This rearrangement corresponds to a pre- and post-multiplying of a given matrix by a unitary and its adjoint.

2. Main results

Lemma2.1(see [1]). Let A be aC∗-algebra,R,S, andT∈AwithT being positive and invertible. Then,

T S

S∗ R

0⇐⇒R≥S∗T−1S. (2.1)

Proof. The lemma follows from the identity

0 0

−S∗T−1 I

T S

S∗ R

0 0

−S∗T−1 I

=

0 0

0 R−S∗T−1S

=

T S

S∗ R

T1/2 0

S∗T−1/2 0

T1/2 0

S∗T−1/2 0

.

(2)

Letφ:A→Bbe a linear map. We denote byφ∗:A→Bthe linear map defined by

φ∗(a)=φa∗∗. (2.3)

LetS(A)be the linear subspace ofM2(A)given by

S(A)=   

λ a b µ

:a,b∈A, λ,µ∈C

 

. (2.4)

LetΦ:S(A)→S(B)be defined by

Φ

λ a b µ

=

λ φ(a)

φ∗(b) µ

. (2.5)

Theorem2.2. The mapφisn-contractive which impliesΦisn-positive. Proof. Let

X=  

λij aij

bij µij

MnS(A)+

. (2.6)

We may identifyXwith

Y=

λij aij

bij µij

. (2.7)

Therefore, Y is positive which implies [λij] and [µij]are positive in Mn(C). Since

Y 0 if and only ifY+(1/m)I >0 for everym∈I, we may assume that[λij]and

[µij]are invertible. We have

λij aij

a∗ji µij

 0

⇐⇒   λij

aij

aij

µij

0, via identification

⇐⇒µij

≥aij

λij

1

aij

, byLemma 2.1

⇐⇒I≥µij

aij

λij

λij

aij

µij

, whereµij

=µij

1/2

, λij

=λij

1/2

⇐⇒I≥λij

aij

µij

λij

aij

µij

(3)

⇐⇒1≥λij

aij

µij

λij

aij

µij

⇐⇒10λij

aij

µij

2

⇐⇒1≥λijaijµ ij

⇐⇒10   n

s,t=1

λusastµ tr   u,r

⇐⇒1 φn

  n

s,t=1

λusastµ tr   u,r

⇐⇒1   n

s,t=1

λusφnastµ tr   u,r ⇐⇒1≥λij

φaij

µij

⇐⇒1≥λij

φaij

µij

2

⇐⇒1≥λij

φaij

µij

λij

φaij

µij

⇐⇒I≥λij

φaij

µij

λij

φaij

µij

⇐⇒µij

≥φaij

λij

1

φaij

⇐⇒

λij φaij

φaij∗ µij

 0

⇐⇒  

λij φaij

φ∗a∗ji

µij

0, via identification

⇐⇒φn

λij aij

a∗ji µij

 0.

(2.8)

This completes the proof of the theorem.

Theorem2.3. Letφ:E→Bbe a map from a selfadjoint subspaceEof aC∗-algebra

Ainto aC∗-algebraB. Define a mapΨ:S(E)→Bby

Ψ

λ a b µ

=(λ+µ)I+φ(a)+φ∗(b). (2.9)

(i) The mapφisn-contractive⇒Ψisn-positive. (ii) The mapΨis2n-positive⇒φisn-contractive.

(4)

Proof. (i) Define mapsΦ:S(E)→S(B),δ:M2(B)→Bby

Φ

λ a b µ

=

λ φ(a)

φ∗(b) µ

,

δ

a b c d

=a+b+c+d.

(2.10)

The mapΦisn-positive by Theorem 2.2. Asδisn-positive (in fact it is completely positive), thenΨ=δ◦Φisn-positive.

(ii) There are two methods to prove (ii).

Method1(see [4]). Via identification, we have

Ψn

 

λij

aij

aij

µij

=λij+µijI+φnaij+φ n

bij

,

λij,µij∈Mn(C); aij,bij∈Mn(A).

(2.11)

Let[aij] ≤1, we want to show thatφn[aij] ≤1. Now

aij1    In

aij

aij

In

  0

     

In 0 0

aij

0 0 0 0

0 0 0 0

aij

0 0

In

     0

        

Ψn

 In 0

0 0

Ψn

 0

aij

0 0

 

Ψn

0 0

aij

0

Ψn

 0 0

0 In

 

        

0

In φn

aij

φ∗n

aij∗ In

 0

⇒φn

aij≤1

⇒φn≤1.

(5)

Method2. SinceΨ2n2nΦ2nandΨ2n,δ2nare both positive, thenΦ2nis also positive. AsΦ2nis unital, thenΦ2n1. Hence

Ψ2nδ2nΦ2n=2·1=2. (2.13)

We identified

λij aij

bij µij

⊗Mn with

H A B K , (2.14) and write λij aij

bij µij

⊗Mn←→

H B A K , (2.15) where

H=λij, K=µij, A=aij, B=bij, H,K∈Mn, A,B∈Mn(A);

Ψ2n   H A B K ⊗M2

 =δ2n

 Φ2n

H A

B K

⊗M2

 =δ2n

 

H φ(A)

φ∗(B) K

⊗M2

 

←→δ2n

H⊗M2 φ(A)⊗M2

φ(B)⊗M2 K⊗M2

=

H φ(A)

φ∗(B) K

×4;

4φn(A)=

4

0 φn(A)

0 0 = Ψ2n   0 A 0 0 ⊗Mn

 

=Ψ2n

0 A 0 0 ⊗Mn =4A

⇒φn≤1.

(2.16) (iii) The proof of (iii) is obvious.

(iv) Via identification, we have

Ψ2n

H A

B K

=H+K+φn(A)+φ∗n(B)

φn(A)= Ψ2n 0 A 0 0 = δn◦Φn

0 A 0 0

=δnΦn

0 A 0 0 2A

⇒φn≤1.

(2.17)

3. Applications

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Proof. Define a mapΨ:S(E)→Bby

Ψ

λ a b µ

=(λ+µ)I+φ(a)+φ∗(b),

φis contractiveΨis positive (Theorem 2.3)

Ψis completely positive [1, Proposition 1.2.2] ⇒φis completely contractive (Theorem 2.3).

(3.1)

Theorem3.2(see [4, Theorem 1.1.13]). LetEbe a closed selfadjoint subspace of a C∗-algebra. LetBbe a commutativeC∗-algebra. LetΦ:E→Mn(B)be a linear map. If Φisn-positive then, it is completely positive.

The following theorem is a generalization ofTheorem 3.2.

Theorem3.3. LetEbe a selfadjoint subspace of aC∗-algebra. LetBbe a commu-tativeC∗-algebra. Letφ:E→Mn(B)be a linear map. Ifφisn-contractive then, it is completely contractive.

Proof. Define a mapΨ:S(E)→Mn(B)by

Ψ

λ a b µ

=(λ+µ)I+φ(a)+φ∗(b),

φisn-contractiveΨisn-positive (Theorem 2.3)

Ψis completely positive (Theorem 3.2) ⇒φis completely contractive (Theorem 2.3).

(3.2)

Theorem3.4(see [4, Theorem 1.1.12]). LetE be a closed selfadjoint subspace of a C∗-algebra, A containing the identity, and let φ:E →Mn =B(Cn) be n-positive map. Then,φpossesses a completely positive extensionΨ:A→Mnand therefore,φis completely positive.

In 1983 Smith [3] proved the following theorem.

Theorem3.5(see [3]). Letφ:A→Mnbe bounded. Thenφcb= φn. Here, we generalizeTheorem 3.5by giving the following theorem.

Theorem 3.6. LetE be a closed selfadjoint subspace of aC∗-algebra. If φ:E→

Mn(C)isn-contractive, thenφis completely contractive.

Proof. DefineΨ:S(E)→Mn(C)by

Ψ

λ a b µ

=(λ+µ)I+φ(a)+φ∗(b),

φisn-contractiveΨisn-positive (Theorem 2.3)

Ψis completely positive (Theorem 3.4) ⇒φis completely contractive (Theorem 2.3).

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References

[1] M. D. Choi,Some assorted inequalities for positive linear maps onC∗-algebras, J. Operator Theory4(1980), no. 2, 271–285.

[2] R. I. Loebl,Contractive linear maps onC∗-algebras, Michigan Math. J.22(1975), no. 4, 361–366.

[3] R. R. Smith,Completely bounded maps betweenC∗-algebras, J. London Math. Soc. (2)27 (1983), no. 1, 157–166.

[4] W. T. Sulaiman,Some classes of linear maps betweenC∗-algebras, Ph.D. thesis, Heriot-Watt University, Edinburgh, 1985.

W. T. Sulaiman: College of Education, Ajman University, P.O. Box346, Ajman, United Arab Emirates

References

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