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A Completely Positive Maps A Completely Positive Maps A.1. A linear map θ: A → B between C∗-algebrasiscalledcompletely positive if θ ⊗ id: A ⊗ Matn(C) → B ⊗ Matn(C) is positive for any n ∈ N. Such a map is automatically bounded since any ∗ positive linear functional on a C -algebra is bounded. ⊗ ∈ ⊗ C An operator T = i,j bij eij B Matn( ), where the eij’s are the usual C ∗ ≥ matrix units in Matn( ), is positive if and only if i,j bi bijbj 0 for any ∈ ⊗ C b1,...,bn B. Since every positive element in A Matn( )isthesumofn ∗ ⊗ → elements of the form i,j ai aj eij,weconcludethatθ: A B is completely positive if and only if n ∗ ∗ ≥ bi θ(ai aj)bj 0 i,j=1 for any n ∈ N, a1,...,an ∈ A and b1,...,bn ∈ B. A.2. If θ: A → B is completely positive and B ⊂ B(H), then there exist a Hilbert space K, a representation π: A → B(K) and a bounded operator V : H → K such that θ(a)=V ∗π(a)V and π(A)VH = K. The construction is as follows. Consider the algebraic tensor product AH, and define a sesquilinear form on it by (a ξ,c ζ)=(θ(c∗a)ξ,ζ). Complete positivity of θ means exactly that this form is positive definite. Thus taking the quotient by the space N = {x ∈ A H | (x, x)=0} we get a pre-Hilbert space. Denote its completion by K. Then define a representation π: A → B(K)by π(a)[c ζ]=[ac ζ], where [x] denotes the image of x in (A H)/N . Note that boundedness of π ∗ ∗ ⊗ ≤ follows from complete positivity of θ and the inequality i,j ci a acj eij 266 A Completely Positive Maps 2 ∗ ⊗ ∈ a i,j ci cj eij.IfA is unital, set Vξ=[1 ξ]forξ H. In the general case choose an approximate unit {ei}i in A,setViξ =[ei ξ], and consider any ≤ 2 1/2 ≤ weak operator limit point V of the Vi’s. Note that since Vi θ(ei ) θ1/2, such a point exists. Moreover, it is unique since it is determined by the condition (Vξ,[a ζ]) = (θ(a∗)ξ,ζ). Then V ∗([a ζ]) = θ(a)ζ and π(a)Vζ= [a ζ], whence V ∗π(a)V = θ(a). It follows from the construction that V ≤θ1/2,andifA is unital then V ≤θ(1)1/2. The equality θ(a)=V ∗π(a)V shows that the opposite inequalities hold. Hence V = θ1/2,andifA is unital then θ = θ(1). In particular, if θ is contractive then VV∗ ≤ 1, so that θ(a)∗θ(a)=V ∗π(a)∗VV∗π(a)V ≤ V ∗π(a∗a)V = θ(a∗a), that is, θ is a Schwarz map. The triple (K, π, V ) is called the Stinespring dilation of θ.Itiseasyto see that it is unique up to an isomorphism. Note also that any map of the form a → V ∗π(a)V ,whereπ is a representation and V is a bounded operator, is completely positive. A.3. If θ: A → B is a positive linear map, and either A or B is abelian, then θ is completely positive. Assume B is abelian. Then it suffices to prove that χ ◦ θ is completely positive for any character χ of B. In other words, we have to check that any positive linear functional on A is a completely positive mapping. But this is easy to see as the GNS-representation gives the Stinespring dilation of such a mapping. Assume now that A is abelian. It is obvious that any positive linear map C → B is completely positive. Hence any positive linear map Cn → B is completely positive. It follows that if A is AF, then θ is completely positive. In the general case consider the second transpose θ∗∗: A∗∗ → B∗∗ of θ.Since A∗∗ is an inductive limit of finite dimensional abelian C∗-algebras, we conclude ∗∗ that θ = θ |A is completely positive. A.4. Let θ: A → B be a contractive completely positive map. Let A∼ and B∼ be the C∗-algebras obtained by adjoining units to A and B, respectively. Then θ extends to a unital completely positive map θ˜: A∼ → B∼. To see this, consider a faithful unital representation of B∼ on a Hilbert space H,andlet(K, π, V ) be the corresponding Stinespring dilation of θ. By A.2 we have V ≤1. Consider the Hilbert space K˜ = K ⊕ H, the unital representationπ ˜: A∼ → B(K˜ )suchthat˜π(a)=π(a)⊕0fora ∈ A, and define V˜ : H → K,˜ Vξ˜ = Vξ⊕ (1 − V ∗V )1/2ξ. Then V˜ is an isometry, so that the completely positive map A∼ a → V˜ ∗π˜(a)V˜ ∈ B(H) is unital. It is easy to see that it coincides with θ on A,so that it is the required map θ˜. A Completely Positive Maps 267 ∼ Note that if B is unital, then since 1BB 1B = B,themapθ: A → B extends to a unital completely positive map A∼ → B. A.5. Let θ: A → B be a contractive completely positive map. Then for any x, y ∈ A we have θ(x∗y) − θ(x)∗θ(y)≤θ(x∗x) − θ(x)∗θ(x)1/2θ(y∗y) − θ(y)∗θ(y)1/2. By A.4 we may assume that A, B and θ are unital. Then the operator V in the Stinespring dilation of θ is an isometry. Thus, without loss of generality we may assume that θ is the compression map B(K) → pB(K)p, a → pap,where p ∈ B(K) is a projection. Set T (a)=(1− p)ap. Then for any a, c ∈ B(K) θ(a∗c) − θ(a)∗θ(c)=pa∗cp − pa∗pcp = T (a)∗T (c), so that θ(x∗y) − θ(x)∗θ(y)2 = T (x)∗T (y)2 = T (y)∗T (x)T (x)∗T (y) ≤T (x)T (x)∗T (y)∗T (y) = T (x)∗T (x)T (y)∗T (y) = θ(x∗x) − θ(x)∗θ(x)θ(y∗y) − θ(y)∗θ(y). → C A.6. Let θ: A Matn( ) be a linear map, θ(a)= i,j θij(a)eij. Consider ⊗ C ⊗ the linear functional ϕθ on A Matn( )definedbyϕθ( i,j aij eij)= i,j θij(aij). Then θ is completely positive if and only if ϕθ is positive. Assume ϕθ is positive. Consider the corresponding GNS-representation of ⊗ C ⊗ → ⊗ ∈ ⊗ n A Matn( ). It is necessarily of the form a x π(a) x B(H 2 ) → ⊗ ∈ ⊗ n for some representation π: A B(H). Let ξ = i ξi ei H 2 be the cyclic vector defining ϕθ,sothatθij(a)=(π(a)ξj,ξi). Consider the operator n → ∗ V : 2 H defined by Vei = ξi.ThenV ζ = i(ζ,ξi)ei, whence ∗ V π(a)Vej = (π(a)ξj,ξi)ei = θ(a)ej. i Thus θ is completely positive. Conversely, assume θ is completely positive. Let (H, π, V )beitsStine- n → spring dilation, where V : 2 H.Thenϕθ is defined by the representation ⊗ ⊗ C ⊗ n ⊗ π id of A Matn( )onH 2 and the vector ξ = i Vei ei,thatis, ϕθ(a ⊗ x)=((π(a) ⊗ x)ξ,ξ). Hence ϕθ is positive. Thus we get a one-to-one correspondence between completely positive maps A → Matn(C) and positive linear functionals on A ⊗ Matn(C). A.7. If M is a von Neumann algebra, then any completely positive map θ: M → Matn(C) can be approximated in the pointwise norm topology by normal completely positive maps. Moreover, if θ is unital then the approxi- mations can also be chosen unital. 268 A Completely Positive Maps The first assertion is an immediate consequence of A.6. The second one is true since if γ: M → Matn(C) is a completely positive map such that γ(1) is close to 1, then γ(1)−1/2γ(·)γ(1)−1/2 is a unital completely positive map, which is close to γ. A.8. If A is a C∗-subalgebra of a C∗-algebra B, then any completely positive map θ: A → B(H) can be extended to a completely positive map B → B(H). If H is finite dimensional, the result follows from A.6. In the general case first note that we may assume that B is unital and then by A.4 that A contains the unit of B. Choose a net {pi}i of finite rank projections in B(H)suchthat pi 1. For each i extend the map A a → piθ(a)pi ∈ piB(H)pi to B.Since A contains the unit of B, the norms of such extensions are not greater than the norm of θ. Then take any pointwise weak operator limit point of these extensions. This is a weak form of Arveson’s extension theorem. A more general form assumes only that B is a unital C∗-algebra and A ⊂ B is an operator system, that is, a self-adjoint subspace containing the unit. For a proof see e.g. [214, Theorem XV.1.1]. C → A.9. A linear map γ:Matn( ) A is completely positive if and only if the operator T = n γ(e ) ⊗ e ∈ A ⊗ Mat (C) is positive. γ i,j=1 ij ij n ⊗ ⊗ ∗ ⊗ Since i,j eij eij =( i e1i e1i) ( j e1j e1j) is positive, positivity of Tγ is a necessary condition for complete positivity of γ. Conversely, assume ∗ ⊗ Tγ is positive. Then Tγ is the sum of n elements of the form i,j ai aj eij.
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