This paper concerns classifying completely positive maps between certain C*-algebras. Several invariants for classifying completely positive maps are constructed. It is proved that one of them is isomorphic to the Ext-group of C*-algebra extensions in special circumstances. Furthermore, this invariant induces a functor from C*-algebras to abelian groups which is split-exact.
The theory of completely positive maps plays an important part in operator algebras, operator spaces, and extensions of C*-algebras. Many fundamental concepts and theorems are defined and proved via completely positive maps respectively, such as nuclearity, invertible extension, Stinespring’s Theorem ( [
On the other hand, as an effective tool to study the structure of C*-algebras and to classify C*-algebras, the theory extensions of C*-algebras originated from Busby’s work in 1960’s ( [
As we know, an extension of C*-algebras is determined by its Busby invariant with respect to unitary equivalence, so to an extent classifying extensions of C*-algebras is a sort of classifying homomorphisms between C*-algebras. It should be pointed out that the KK-groups were defined via homomorphisms in this way at the beginning ( [
This note is engaging in classifying completely positive maps between certain C*-algebras. Specifically, several invariants for classifying completely positive maps are introduced. As a main result, one of them is isomorphic to the Ext-group of C*-algebra extensions. In addition, this invariant induces a functor from C*-algebras to abelian groups which is split-exact.
In this section, we need to recall some notations and definitions for C*-algebras and extensions. One can also see [
Suppose that D is a C*-algebra. Recall that θ n : M n ( D ) → D is an inner isomorphism, if there are isometries S 1 , ⋯ , S n in D with ∑ i = 1 n S i S i * = 1 and S i * S j = 0 for i ≠ j , such that θ n = A d v , namely,
θ n ( [ x i j ] ) = v [ x i j ] v * = ∑ i , j S i x i j S j * ,
for [ x i j ] ∈ M n ( D ) , where v = ( S 1 , ⋯ , S n ) . Suppose that v 1 and v 2 are such elements. Then v 1 v 2 * ∈ D and v 1 v 2 * v 2 v 1 * = v 2 v 1 * v 1 v 2 * = 1 , and hence v 1 v 2 * is a unitary in D.
Let A and B be C*-algebras. An extension of A by B is a short exact sequence
e : 0 → B → α E → β A → 0.
Denote this extension by e or ( E , α , β ) .
The extension ( E , α , β ) is called trivial, if the above sequence splits, i.e. if there is a homomorphism γ : A → E such that β ∘ γ = i d A .
For an extension ( E , α , β ) , there is a unique homomorphism σ : E → M ( B ) such that σ ∘ α = ι , where M ( B ) is the multiplier algebra of B, and ι is the inclusion map from B into M ( B ) . The Busby invariant of ( E , α , β ) is a homomorphism τ from A into the corona algebra Q ( B ) = M ( B ) / B defined by τ ( a ) = π ( σ ( b ) ) for a ∈ A , where π : M ( B ) → Q ( B ) is the quotient map, and b ∈ E such that β ( b ) = a .
Two extensions e 1 and e 2 are called (strongly) unitarily equivalent, denoted by e 1 ~ s e 2 , if there exists a unitary u ∈ M ( B ) such that τ 2 ( a ) = π ( u ) τ 1 ( a ) π ( u ) * for all a ∈ A . Denote by E x t ( A , B ) or E x t s ( A , B ) the set of (strong) unitary equivalence classes of extensions of A by B.
Let H be a separable infinite-dimensional Hilbert space and K the ideal of compact operators in B ( H ) . If B is a stable C*-algebra (i.e. B ⊗ K ≅ B , where ⊗ is the tensor product operation), then the sum of two extensions τ 1 and τ 2 is defined to be the homomorphism τ 1 ⊕ τ 2 , where
τ 1 ⊕ τ 2 : A → Q ( B ) ⊕ Q ( B ) ⊆ M 2 ( Q ( B ) ) ≅ Q ( B )
and the isomorphism M 2 ( Q ( B ) ) ≅ Q ( B ) is induced by an inner isomorphism from M 2 ( M ( B ) ) onto M ( B ) , where ⊕ is the direct sum of C*-algebras.
The above sets of equivalence classes of extensions are commutative semigroups with respect to this addition when B is stable. One can similarly define these semigroups replacing B by B ⊗ K if B is not stable. Denote by E x t ( A , B ) the quotient of E x t s ( A , B ) by the subsemigroup of trivial extensions.
Suppose that D is a unital properly infinite C*-algebra, namely, there are two elements S 1 , S 2 ∈ D such that
S i * S i = 1 ( i = 1 , 2 ) , S i * S j = 0 ( i ≠ j ) , ∑ i = 1 2 S i S i * = 1.
For every C*-algebra A, we denote by C P ( A , D ) the set of all completely positive maps from A into D.
Definition 3.1. Two elements φ , ψ ∈ C P ( A , D ) are called (unitarily) equivalent, denoted by φ ≈ ψ , if there is a unitary u ∈ D such that A d u ∘ φ = ψ .
It is easy to check that ≈ is an equivalence relation on C P ( A , D ) . Denote by { φ } the equivalence class of φ .
Definition 3.2. C P 1 ( A , D ) is the equivalence classes in C P ( A , D ) under the equivalence relation ≈ , i.e. C P 1 ( A , D ) = C P ( A , D ) / ≈ .
Now we can define a diagonal addition in C P 1 ( A , D ) as follows:
{ φ } + { ψ } = { A d v ∘ ( φ ψ ) } = { ( S 1 S 2 ) ( φ ψ ) ( S 1 * S 2 * ) } ,
where A d v : M 2 ( D ) → D is the inner isomorphism with v = ( S 1 , S 2 ) .
Proposition 3.3. Equipped with the above addition, C P 1 ( A , D ) is an abelian semigroup.
Proof. The following is similar to the proof of ( [
Suppose that φ , φ ′ , ψ and ψ ′ are in C P ( A , D ) such that φ ≈ φ ′ and ψ ≈ ψ ′ . Then there are unitary elements u 1 , u 2 ∈ D such that φ ′ = A d u 1 ∘ φ and ψ ′ = A d u 2 ∘ ψ . Thus
A d v ∘ ( φ ′ ψ ′ ) = v ( u 1 u 2 ) ( φ ψ ) ( u 1 * u 2 * ) v * = v ( u 1 u 2 ) v * v ( φ ψ ) v * v ( u 1 * u 2 * ) v * .
Since
v ( u 1 u 2 ) v *
is a unitary in D, we have
{ A d v ∘ ( φ ′ ψ ′ ) } = { A d v ∘ ( φ ψ ) } .
It follows that the addition is well-defined.
Let θ 1 and θ 2 be two inner isomorphisms from M 2 ( D ) onto D with θ 1 = A d v 1 and θ 2 = A d v 2 . Then
A d v 1 ∘ ( φ ψ ) = v 1 v 2 * v 2 ( φ ψ ) v 2 * v 2 v 1 * = A d ( v 1 v 2 * ) ∘ A d v 2 ∘ ( φ ψ ) ,
and hence,
{ A d v 1 ∘ ( φ ψ ) } = { A d v 2 ∘ ( φ ψ ) } .
Therefore, the addition is independent of the choices of inner isomorphisms.
Suppose that φ , ψ ∈ C P ( A , D ) . Then
A d v ∘ ( φ ψ ) = v ( 0 1 1 0 ) ( 0 1 1 0 ) ( φ ψ ) ( 0 1 1 0 ) ( 0 1 1 0 ) v * = v ( 0 1 1 0 ) ( ψ φ ) ( 0 1 1 0 ) v * .
Let
v ′ = v ( 0 1 1 0 ) .
Then A d v ′ is an inner isomorphism from M 2 ( D ) onto D and hence
{ φ } + { ψ } = { ψ } + { φ } .
Suppose that φ 1 , φ 2 , φ 3 ∈ C P ( A , D ) and let S 1 , S 2 be isometries with S 1 * S 2 = 0 and S 1 S 1 * + S 2 S 2 * = 1 . One can check the following computation:
( { φ 1 } + { φ 2 } ) + { φ 3 } = { S 1 2 φ 1 S 1 * 2 + S 1 S 2 φ 2 S 2 * S 1 * + S 2 φ 3 S 2 * } = { ( S 1 2 , S 1 S 2 , S 2 ) ( φ 1 φ 2 φ 3 ) ( S 1 * 2 S 2 * S 1 * S 2 * ) }
and
{ φ 1 } + { { φ 2 } + { φ 3 } } = { S 1 φ 1 S 1 * + S 2 S 1 φ 2 S 1 * S 2 * + S 2 2 φ 3 S 2 * 2 } = { ( S 1 , S 2 S 1 , S 2 2 ) ( φ 1 φ 2 φ 3 ) ( S 1 * S 1 * S 2 * S 2 * 2 ) } .
Put v 1 = ( S 1 2 , S 1 S 2 , S 2 ) and v 2 = ( S 1 , S 2 S 1 , S 2 2 ) . Then A d v 1 and A d v 2 are two inner isomorphisms from M 3 ( D ) onto D. Note that
A d v 1 ∘ ( φ 1 φ 2 φ 3 ) = A d ( v 1 v 2 * ) ∘ A d v 2 ∘ ( φ 1 φ 2 φ 3 ) .
Since v 1 v 2 * is a unitary in D, it follows that
( { φ 1 } + { φ 2 } ) + { φ 3 } = { φ 1 } + ( { φ 2 } + { φ 3 } ) .
This completes the proof of associativity.
Therefore, C P 1 ( A , D ) is an abelian semigroup.
Remark 3.4. Suppose that φ , ψ ∈ C P ( A , D ) . We write
φ ⊕ ψ = ( S 1 S 2 ) ( φ ψ ) ( S 1 * S 2 * ) ,
or
θ 2 ∘ ( φ ψ ) = ( S 1 S 2 ) ( φ ψ ) ( S 1 * S 2 * ) .
Definition 3.5. Let H o m ( A , D ) be the set of homomorphisms from A into D. An element is called degenerate in C P ( A , D ) if it is also in H o m ( A , D ) .
Definition 3.6. Two elements { φ } , { ψ } ∈ C P 1 ( A , D ) are called equivalent, denoted by { φ } ~ 0 { ψ } , if there are φ ′ , ψ ′ ∈ H o m ( A , D ) such that { φ } + { φ ′ } = { ψ } + { ψ ′ } .
Then ~ 0 is an equivalence relation. The equivalence class of { φ } is denoted by [ { φ } ] 0 , or by [ φ ] 0 simply.
Definition 3.7. C P 2 ( A , D ) is the equivalence classes in C P 1 ( A , D ) under the equivalence relation ~ 0 , i.e. C P 2 ( A , D ) = C P 1 ( A , D ) / ~ 0 .
We define an addition + in C P 2 ( A , D ) by
[ φ ] 0 + [ ψ ] 0 = [ { φ } + { ψ } ] 0 , φ , ψ ∈ C P ( A , D ) .
To see the addition is well-defined, suppose that { φ ′ } ~ 0 { φ } and { ψ ′ } ~ 0 { ψ } . Then there exist φ 1 , φ ′ 1 , ψ 1 , ψ ′ 1 ∈ H o m ( A , D ) such that
{ φ } + { φ 1 } = { φ ′ } + { φ ′ 1 } , { ψ } + { ψ 1 } = { ψ ′ } + { ψ ′ 1 } ,
and hence
{ φ } + { ψ } + { φ 1 } + { ψ 1 } = { φ ′ } + { ψ ′ } + { φ ′ 1 } + { ψ ′ 1 } .
Since
{ φ 1 } + { ψ 1 } = { θ 2 ∘ ( φ 1 ψ 1 ) } ,
θ 2 ∘ ( φ 1 ψ 1 ) ∈ H o m ( A , D ) .
Similarly,
θ 2 ∘ ( φ ′ 1 ψ ′ 1 ) ∈ H o m ( A , D ) .
It follows that the addition is well-defined.
Remark 3.8. 1) Suppose that φ 1 , φ 2 ∈ C P ( A , D ) . Then [ φ 1 ] 0 = [ φ 2 ] 0 ∈ C P 2 ( A , D ) if and only if there exist σ 1 , σ 2 ∈ H o m ( A , D ) such that φ 1 ⊕ σ 1 is unitarily equivalent to φ 2 ⊕ σ 2 .
2) Suppose that η ∈ C P ( A , D ) . Then [ η ] 0 is the neutral element in C P 2 ( A , D ) if and only if for each φ ∈ C P ( A , D ) there exist σ 1 , σ 2 ∈ H o m ( A , D ) such that φ ⊕ η ⊕ σ 1 is unitarily equivalent to φ ⊕ σ 2 .
Theorem 3.9. C P 2 ( A , D ) is a unital abelian semigroup. An element [ φ ] 0 is the unit of C P 2 ( A , D ) if and only if φ ∈ H o m ( A , D ) .
Proof. Suppose that φ 1 , φ 2 , φ 3 ∈ C P ( A , D ) . Then
[ φ 1 ] 0 + ( [ φ 2 ] 0 + [ φ 3 ] 0 ) = [ φ 1 ] 0 + [ { φ 2 } + { φ 3 } ] 0 = [ { φ 1 } + ( { φ 2 } + { φ 3 } ) ] 0 = [ ( { φ 1 } + { φ 2 } ) + { φ 3 } ] 0 = ( [ φ 1 ] 0 + [ φ 2 ] 0 ) + [ φ 3 ] 0 .
It follows that C P 2 ( A , D ) is a semigroup. It is clear that C P 2 ( A , D ) is abelian.
Let η ∈ H o m ( A , D ) . For any φ ∈ C P ( A , D ) , take σ 1 ∈ H o m ( A , D ) and set σ 2 = η ⊕ σ 1 . Then
( φ ⊕ η ) ⊕ σ 1 ≈ φ ⊕ ( η ⊕ σ 1 ) ,
that is, ( φ ⊕ η ) ⊕ σ 1 ≈ φ ⊕ σ 2 . Since σ 1 , η ⊕ σ 1 ∈ H o m ( A , D ) and φ ⊕ η ~ 0 φ , we have [ φ ] 0 + [ η ] 0 = [ φ ] 0 by Remark 3.8. Hence [ η ] 0 is the unit of C P 2 ( A , D ) .
Suppose that ψ ∈ C P ( A , D ) such that [ ψ ] 0 is the unit of C P 2 ( A , D ) . For φ ∈ H o m ( A , D ) , [ φ ] 0 is also the unit of C P 2 ( A , D ) , and hence [ ψ ] 0 = [ φ ] 0 . Thus there exist φ 1 , ψ 1 ∈ H o m ( A , D ) such that { ψ } + { ψ 1 } = { φ } + { φ 1 } . Note that ψ ⊕ ψ 1 is unitarily equivalent to φ ⊕ φ 1 . Since φ and φ 1 are both homomorphisms,
θ 2 ∘ ( φ φ 1 )
is a homomorphism. Furthermore,
( ψ ψ 1 )
is in H o m ( A , M 2 ( D ) ) , and hence ψ is in H o m ( A , D ) .
Remark 3.10. The only invertible element in C P 2 ( A , D ) is the unit. In fact, suppose that [ φ ] 0 is an invertible element in C P 2 ( A , D ) with the inverse [ ψ ] 0 . Then [ φ ] 0 + [ ψ ] 0 is the unit and
θ 2 ∘ ( φ ψ )
is a homomorphism by Theorem 3.9. Thus,
( φ ψ )
is also a homomorphism. Therefore φ is in H o m ( A , D ) . It follows that [ φ ] 0 is the unit.
Definition 3.11. Let B be a closed ideal of D and π : D → D / B the quotient map. We define a relation ~ on C P 2 ( A , D ) as follows: for φ , ψ ∈ C P ( A , D ) , we write [ φ ] 0 ~ [ ψ ] 0 if there exist φ 1 , ψ 1 ∈ C P ( A , D ) such that [ φ 1 ] 0 = [ φ ] 0 , [ ψ 1 ] 0 = [ ψ ] 0 , and π ∘ φ 1 = π ∘ ψ 1 .
Suppose that φ ~ ψ , ψ ~ η . Then there exist φ 1 , ψ 1 , ψ 2 , η 2 such that
[ φ 1 ] 0 = [ φ ] 0 , [ ψ 1 ] 0 = [ ψ ] 0 , π ∘ φ 1 = π ∘ ψ 1 ,
[ ψ 2 ] 0 = [ ψ ] 0 , [ η 2 ] 0 = [ η ] 0 , π ∘ ψ 2 = π ∘ η 2 .
Since [ ψ 1 ] 0 = [ ψ 2 ] 0 , there exist ϕ 1 , ϕ 2 ∈ H o m ( A , D ) such that { ψ 1 } + { ϕ 1 } = { ψ 2 } + { ϕ 2 } . Thus there is a unitary u ∈ D such that
θ 2 ∘ ( ψ 1 ϕ 1 ) = A d u ∘ θ 2 ∘ ( φ 2 ϕ 2 ) .
Put
φ ′ 1 = θ 2 ∘ ( φ 1 ϕ 1 ) , η ′ 2 = A d u ∘ θ 2 ∘ ( η 2 ϕ 2 ) .
Then we have
[ φ ′ 1 ] 0 = [ φ 1 ] 0 + [ ϕ 1 ] 0 = [ φ 1 ] 0 = [ φ ] 0 ,
[ η ′ 2 ] 0 = [ η 2 ] 0 + [ ϕ 2 ] 0 = [ η 2 ] 0 = [ η ] 0 ,
and
π ∘ φ ′ 1 = π ∘ θ 2 ∘ ( φ 1 ϕ 1 ) = θ ′ 2 ∘ ( π ∘ φ 1 π ∘ ϕ 1 ) = θ ′ 2 ∘ ( π ∘ ψ 1 π ∘ ϕ 1 ) = π ∘ θ 2 ∘ ( ψ 1 ϕ 1 ) = π ∘ A d u ∘ θ 2 ∘ ( ψ 2 ϕ 2 ) = A d π ( u ) ∘ θ ′ 2 ∘ ( π ∘ ψ 2 π ∘ ϕ 2 ) = A d π ( u ) ∘ θ ′ 2 ∘ ( π ∘ η 2 π ∘ ϕ 2 ) = π ∘ A d u ∘ θ 2 ∘ ( η 2 ϕ 2 ) = π ∘ η ′ 2 ,
where θ ′ 2 is the inner isomorphism from M 2 ( D / B ) onto D / B induced by θ .
It follows that ~ is transitive, and hence ~ is an equivalence relation on C P 2 ( A , D ) . Denote the equivalence class of [ φ ] 0 by [ [ φ ] 0 ] , or by [ φ ] simply.
Let C P B ( A , D ) = C P 2 ( A , D ) / ~ . It is natural that we define an addition in C P B ( A , D ) as follows:
[ φ ] + [ ψ ] = [ [ φ ] 0 + [ ψ ] 0 ] .
Remark 3.12. The addition defined in Definition 3.11 is well-defined: for [ φ ] = [ φ ′ ] and [ ψ ] = [ ψ ′ ] , there exist φ 1 , φ ′ 1 , ψ 1 , ψ ′ 1 such that [ φ 1 ] 0 = [ φ ] 0 , [ φ ′ 1 ] 0 = [ φ ′ ] 0 , [ ψ 1 ] 0 = [ ψ ] 0 , [ ψ ′ 1 ] 0 = [ ψ ′ ] 0 , π ∘ φ ′ 1 = π ∘ φ 1 , and π ∘ ψ ′ 1 = π ∘ ψ 1 . Then
π ( S 1 φ ′ 1 S 1 * + S 2 ψ ′ 1 S 2 * ) = π ( S 1 φ 1 S 1 * + S 2 ψ 1 S 2 * ) ,
and hence
[ [ φ ] 0 + [ ψ ] 0 ] = [ [ φ 1 ] 0 + [ ψ 1 ] 0 ] = [ [ φ ′ 1 ] 0 + [ ψ ′ 1 ] 0 ] = [ [ φ ′ ] 0 + [ ψ ′ ] 0 ] .
It is easy to see that [ 0 ] is the unit of C P B ( A , D ) . Thus ( C P B ( A , D ) , + ) is a unital abelian semigroup. In particular, for B = { 0 } , we have ( C P B ( A , D ) , + ) = ( C P 2 ( A , D ) , + ) ; and for B = D , we have C P D ( A , D ) = { 0 } .
Definition 3.13. Let C P B − 1 ( A , D ) be the set of invertible elements in C P B ( A , D ) . Then C P B − 1 ( A , D ) is an abelian group.
Theorem 3.14. Let φ be in C P ( A , D ) . Then [ φ ] = 0 in C P B − 1 ( A , D ) if and only if there exist φ ′ , ϕ 1 , ϕ 2 ∈ H o m ( A , D ) and a unitary u ∈ M 2 ( D ) such that
( φ ϕ 1 ) = A d u ∘ ( φ ′ ϕ 2 ) .
Proof. Suppose that [ φ ] = 0 in C P B − 1 ( A , D ) . Since [ φ ] = [ 0 ] = 0 , there exist φ ′ , ϕ ′ ∈ C P ( A , D ) such that [ φ ′ ] 0 = [ φ ] 0 , [ ϕ ′ ] 0 = 0 and π ∘ φ ′ = π ∘ ϕ ′ . Hence, by Theorem 3.9, we have ϕ ′ ∈ H o m ( A , D ) . Since [ φ ′ ] 0 = [ φ ] 0 , there exist ϕ 1 , ϕ 2 ∈ H o m ( A , D ) such that { φ } + { ϕ 1 } = { φ ′ } + { ϕ 2 } . Then there is a unitary u ∈ M 2 ( D ) such that
( φ ϕ 1 ) = A d u ∘ ( φ ′ ϕ 2 ) .
Conversely, suppose that there exist φ ′ , ϕ 1 , ϕ 2 ∈ H o m ( A , D ) and a unitary u ∈ M 2 ( D ) such that
( φ ϕ 1 ) = A d u ∘ ( φ ′ ϕ 2 ) .
Set v 1 = ( S 1 , S 2 ) and v 2 = v u . Then A d v 1 , A d v 2 are both inner isomorphisms from M 2 ( D ) onto D. Therefore
A d v 1 ∘ ( φ ϕ 1 ) = A d v 2 ∘ ( φ ′ ϕ 2 ) .
Note that [ φ ′ ] = [ ϕ 1 ] = [ ϕ 2 ] = 0 . Thus [ φ ] = [ φ ] + [ ϕ 1 ] = [ φ ′ ] + [ ϕ 2 ] = 0 .
Remark 3.15. Suppose that [ φ ] = 0 in C P B − 1 ( A , D ) . By Theorem 3.14, we have
π ∘ ( φ ϕ 1 ) = π ∘ A d u ∘ ( φ ′ ϕ 2 ) = A d π ( u ) ∘ ( π ∘ φ ′ π ∘ ϕ 2 ) = A d π ( u ) ∘ ( π ∘ ϕ ′ π ∘ ϕ 2 ) = π ∘ A d u ∘ ( ϕ ′ ϕ 2 ) = π ∘ ϕ ,
where
ϕ = A d u ∘ ( ϕ ′ ϕ 2 ) ∈ H o m ( A , M 2 ( D ) ) ,
and π : M 2 ( D ) → M 2 ( D / B ) is induced by the quotient map π : D → D / B .
Set ϕ = ( ϕ i , j ) , we have π ∘ ϕ i , j = 0 ( i ≠ j ) and π ∘ φ = π ∘ ϕ 1 , 1 .
Theorem 3.16. Let φ ∈ C P ( A , D ) . Then [ φ ] ∈ C P B − 1 ( A , D ) if and only if there is ϕ = ( ϕ i , j ) ∈ H o m ( A , M 2 ( D ) ) and ψ ∈ C P ( A , D ) such that
π ∘ ( φ ψ ) = π ∘ ϕ .
Proof. Suppose that [ φ ] ∈ C P B − 1 ( A , D ) with the inverse [ φ ′ ] . Let v = ( S 1 , S 2 ) . Since [ φ ] + [ φ ′ ] = 0 , there exist ϕ ′ = ( ϕ ′ i , j ) in H o m ( A , M 2 ( D ) ) and a unitary u = ( u i , j ) in M 2 ( D ) such that
π ∘ ( φ φ ′ ) = A d π ( v * ) ∘ A d π ( v ) ∘ ( φ φ ′ ) = A d π ( v * ) ∘ ϕ ′ 1 , 1 = π ∘ ( S 1 * ϕ ′ 1 , 1 S 1 S 1 * ϕ ′ 1 , 1 S 2 S 2 * ϕ ′ 1 , 1 S 1 S 2 * ϕ ′ 1 , 1 S 2 ) .
Set
v 1 = ( S 1 * 0 S 2 * 0 0 I ) and v 2 = ( I 0 0 0 S 1 S 2 ) .
Then A d v 1 is an inner isomorphism from M 2 ( D ) onto M 3 ( D ) and A d v 2 is an inner isomorphism from M 3 ( D ) onto M 2 ( D ) . It follows that
π ∘ A d v 1 ∘ ϕ ′ = π ∘ ( S 1 * ϕ ′ 1 , 1 S 1 S 1 * ϕ ′ 1 , 1 S 2 S 1 * ϕ ′ 1 , 2 S 2 * ϕ ′ 1 , 1 S 1 S 2 * ϕ ′ 1 , 1 S 2 S 2 * ϕ ′ 1 , 2 ϕ ′ 2 , 1 S 1 ϕ ′ 2 , 1 S 2 ϕ ′ 2 , 2 ) = π ∘ ( φ φ ′ ϕ ′ 2 , 2 ) .
Set
ψ = ( S 1 S 2 ) ( φ ′ ϕ ′ 2 , 2 ) ( S 1 * S 2 * )
and
ϕ = A d v 2 ∘ A d v 1 ∘ ϕ ′ ∈ H o m ( A , M 2 ) ( D ) .
Then we have
π ∘ ϕ = π ∘ A d v 2 ∘ ( φ φ ′ ϕ ′ 2 , 2 ) = π ∘ ( φ ψ ) .
Conversely, since
π ∘ ( φ ψ ) = π ∘ ϕ ,
A d v ∘ π ∘ ( φ ψ ) = A d v ∘ π ∘ ϕ .
Then
π ∘ θ 2 ∘ ( φ ψ ) = π ∘ θ 2 ∘ ϕ .
Thus
[ φ ] + [ ψ ] = [ θ 2 ∘ ( φ ψ ) ] = [ θ 2 ∘ ϕ ] = 0.
Proposition 3.17. Suppose that φ ∈ C P ( A , D ) such that [ φ ] ∈ C P B − 1 ( A , D ) . Then π ∘ φ is a homomorphism.
Proof. Suppose that φ 1 ∈ C P B − 1 ( A , D ) and [ φ 2 ] is the inverse of [ φ 1 ] . Set
ψ = θ 2 ∘ ( φ 1 φ 2 ) = s 1 φ 1 s 1 * + s 2 φ 2 s 2 * .
By Theorem 3.14, there exist ϕ ∈ H o m ( A , M 2 ( D ) ) and ϕ 1 ∈ H o m ( A , D ) such that
π ∘ ( ψ ϕ 1 ) = π ∘ ϕ .
Hence π ∘ ψ is a homomorphism, and thus
π ( S 1 ) ( π ∘ φ 1 ( a b ) − π ( φ 1 ( a ) φ 1 ( b ) ) ) π ( S 1 * ) + π ( S 2 ) ( π ∘ φ 2 ( a b ) − π ( φ 2 ( a ) φ 2 ( b ) ) ) π ( S 2 * ) = 0.
Set x = π ∘ φ 1 ( a b ) − π ( φ 1 ( a ) φ 1 ( b ) ) and y = π ∘ φ 2 ( a b ) − π ( φ 2 ( a ) φ 2 ( b ) ) . Then
π ( S 1 ) x π ( S 1 * ) + π ( S 2 ) y π ( S 2 * ) = 0,
that is,
( π ( S 1 ) , π ( S 2 ) ) ( x y ) ( π ( S 1 * ) π ( S 2 * ) ) = 0.
Put v ′ = ( π ( S 1 ) , π ( S 2 ) ) . Then v ′ * v ′ = I ∈ M 2 ( D ) . Hence
( x y ) = 0.
This implies that x = y = 0 , and furthermore π ∘ φ 1 ( a b ) = π ( φ 1 ( a ) φ 1 ( b ) ) . It follows that π ∘ φ 1 is a homomorphism.
Lemma 3.18. ( [
1) [ ϕ ] is invertible in E x t ( A , B ) .
2) There exists ψ ∈ C P ( A , M ( B ) ) such that ϕ = π ∘ ψ .
3) There exists φ ∈ H o m ( A , M 2 ( M ( B ) ) ) such that
( ϕ 0 0 0 ) = π ( ( 1 0 0 0 ) φ ( 1 0 0 0 ) ) .
It is well known that M 2 ( M ( B ) ) and M ( B ) are innerly isomorphic if B is a stable C*-algebra. Then we have the following result.
Theorem 3.19. Let A and B be C*-algebras with B stable. Then
C P B − 1 ( A , M ( B ) ) ≅ E x t − 1 ( A , B ) .
Proof. Note that the condition that A is separable is not necessary in the proof of (1) ⇒ (2) in Lemma 3.18 ( [
Φ : E x t − 1 ( A , B ) → C P B ( A , M ( B ) )
by [ ϕ ] ↦ [ φ ] , where π ∘ φ = ϕ .
1) Prove that Φ is well-defined.
Suppose that ϕ 1 , ϕ 2 ∈ H o m ( A , M ( B ) / B ) such that [ ϕ 1 ] , [ ϕ 2 ] ∈ E x t − 1 ( A , B ) . Then there exist φ 1 , φ 2 ∈ C P B ( A , M ( B ) ) such that ϕ 1 = π ∘ φ 1 and ϕ 2 = π ∘ φ 2 . If [ ϕ 1 ] = [ ϕ 2 ] , there exist φ ′ 1 , φ ′ 2 ∈ H o m ( A , M ( B ) ) and u ∈ M 2 ( M ( B ) ) such that
θ ˜ B ∘ ( ϕ 1 π ∘ φ ′ 1 ) = A d π ( u ) ∘ θ ˜ B ∘ ( ϕ 2 π ∘ φ ′ 2 ) .
Hence,
π ( θ B ∘ ( φ 1 φ ′ 1 ) ) = π ( A d u ∘ θ B ∘ ( φ 2 φ ′ 2 ) ) .
Since θ B is an inner isomorphism,
[ φ 1 ] 0 = [ θ B ∘ ( φ 1 φ ′ 1 ) ] 0 and [ φ 2 ] 0 = [ A d u ∘ θ B ∘ ( φ 2 φ ′ 2 ) ] 0 .
Then [ φ 1 ] = [ φ 2 ] , and hence Φ is well-defined.
2) Prove that Φ is a homomorphism.
Note that
Φ ( [ ϕ 1 ] ) + Φ ( [ ϕ 2 ] ) = [ φ 1 ] + [ φ 2 ] = [ [ φ 1 ] 0 + [ φ 2 ] 0 ] .
Since
π ∘ θ B ∘ ( φ 1 φ 2 ) = θ ˜ B ∘ ( π ∘ φ 1 π ∘ φ 2 ) = θ ˜ B ∘ ( ϕ 1 ϕ 2 ) ,
we have
Φ ( [ ϕ 1 ] + [ ϕ 2 ] ) = Φ ( [ ϕ 1 ] ) + Φ ( [ ϕ 2 ] ) .
It follows that Φ is a homomorphism.
3) Prove that Φ ( E x t − 1 ( A , B ) ) ⊆ C P B − 1 ( A , M ( B ) ) .
Suppose that [ ϕ 1 ] is an invertible element with the inverse [ ϕ 2 ] . Then we have
Φ ( [ ϕ 1 ] ) + Φ ( [ ϕ 2 ] ) = Φ ( [ ϕ 1 ] + [ ϕ 2 ] ) = Φ ( 0 ) .
Therefore, Φ ( [ ϕ 1 ] ) is invertible.
4) Prove that Φ : E x t − 1 ( A , B ) → C P B − 1 ( A , M ( B ) ) is injective.
Suppose that Φ ( [ ϕ 1 ] ) = [ φ 1 ] and Φ ( [ ϕ 2 ] ) = [ φ 2 ] , where ϕ 1 = π ∘ φ 1 and ϕ 2 = π ∘ φ 2 .
If [ φ 1 ] = [ φ 2 ] in C P B − 1 ( A , M ( B ) ) , then there exist φ ′ 1 , φ ′ 2 ∈ C P ( A , M ( B ) ) such that [ φ 1 ] 0 = [ φ ′ 1 ] 0 , [ φ 2 ] 0 = [ φ ′ 2 ] 0 and π ∘ φ ′ 1 = π ∘ φ ′ 2 . Therefore there exist σ 1 , σ 2 , τ 1 , τ 2 ∈ H o m ( A , M ( B ) ) and unitary elements u 1 , u 2 ∈ M 2 ( M ( B ) ) such that
( φ 1 σ 1 ) = A d u 1 ∘ ( φ ′ 1 σ 2 ) , ( φ 2 τ 1 ) = A d u 2 ∘ ( φ ′ 2 τ 2 ) .
Put
X = ( φ ′ 1 σ 2 ) , Y = ( φ ′ 2 τ 2 ) , E = ( 1 0 0 0 0 1 0 1 0 ) .
Then we have
( φ 1 σ 1 τ 2 ) = ( u 1 X u 1 * 0 0 τ 2 ) = ( u 1 1 ) ( X 0 0 τ 2 ) ( u 1 * 1 ) ,
( φ 2 τ 1 σ 2 ) = ( u 2 Y u 2 * 0 0 σ 2 ) = ( u 2 1 ) ( Y 0 0 τ 2 ) ( u 2 * 1 ) ,
and
π ( ( u 1 1 ) ( X 0 0 τ 2 ) ( u 1 * 1 ) ) = ( π ( u 1 ) π ( 1 ) ) ( π ∘ X 0 0 π ∘ τ 2 ) ( π ( u 1 * ) π ( 1 ) ) = ( π ( u 1 ) π ( 1 ) ) π ( E ) ( π ∘ Y 0 0 π ∘ σ 2 ) π ( E ) ( π ( u 1 * ) π ( 1 ) ) = π ( ( u 1 1 ) E ( Y 0 0 σ 2 ) E ( u 1 * 1 ) ) .
Thus,
π ( ( φ 1 σ 1 τ 2 ) ) = π ( ( u 1 1 ) E ( u 2 * 1 ) ( φ 2 τ 1 σ 2 ) ( u 2 1 ) E ( u 1 * 1 ) ) .
Set
u 3 = ( u 1 1 ) E ( u 2 * 1 ) .
One can check that u 3 is a unitary in M 3 ( M ( B ) ) . Then we have
( ϕ 1 π ∘ σ 1 π ∘ τ 2 ) = A d π ( u 3 ) ∘ ( ϕ 2 π ∘ τ 1 π ∘ σ 2 ) .
It follows that
[ ϕ 1 ] = [ ϕ 1 ] + [ π ∘ σ 1 ] + [ π ∘ τ 2 ] = [ ϕ 2 ] + [ π ∘ τ 1 ] + [ π ∘ σ 2 ] = [ ϕ 2 ] .
Therefore, Φ is injective.
5) Prove that Φ : E x t − 1 ( A , B ) → C P B − 1 ( A , M ( B ) ) is surjective.
Suppose that [ φ 1 ] ∈ C P B − 1 ( A , M ( B ) ) . Then by Theorem 3.16 there exist [ φ 2 ] and an inner isomorphism ϕ ∈ H o m ( A , M 2 ( M ( B ) ) ) with ϕ = A d v and v = ( S 1 , S 2 ) , such that [ φ 1 ] + [ φ 2 ] = [ A d v ∘ ϕ ] . Since π ∘ φ 1 = ϕ 1 and π ∘ φ 2 = ϕ 2 , by Theorem 3.17, ϕ 1 and ϕ 2 are homomorphisms and
[ ϕ 1 ] + [ ϕ 2 ] = [ π ∘ A d v ∘ ϕ ] = [ 0 ] .
Thus [ ϕ 1 ] ∈ E x t − 1 ( A , B ) and Φ ( [ ϕ 1 ] ) = [ φ ] . This implies that Φ is surjective.
Similar to Lemma 3.18, we have the following result.
Corollary 3.20. Let A and B be C*-algebras with B stable and let ϕ ∈ H o m ( A , Q ( B ) ) . Consider the following three statements:
1) [ ϕ ] is invertible in E x t ( A , B ) .
2) There exists ψ ∈ C P ( A , M ( B ) ) such that ϕ = π ∘ ψ .
3) There exist φ ∈ H o m ( A , M 2 ( M ( B ) ) ) and ϕ ′ ∈ H o m ( A , M ( B ) ) such that
( ϕ ϕ ′ ) = π ∘ φ .
Then (1) ⇔ (3) ⇒ (2).
Proposition 3.21. Let A and C be C*-algebras and h , ϕ ∈ H o m ( A , C ) . Then
1) The map h * : C P 2 ( C , D ) → C P 2 ( A , D ) defined by [ φ ] 0 ↦ [ φ ∘ h ] 0 is a semigroup homomorphism.
2) The map ϕ * : C P B ( C , D ) → C P B ( A , D ) defined by [ φ ] ↦ [ φ ∘ ϕ ] is a unital semigroup homomorphism. Furthermore, it is a group homomorphism from C P B − 1 ( C , D ) into C P B − 1 ( A , D ) .
Theorem 3.22. Let C be the category of C*-algebras and S G the category of abelian semigroups. Define C P B ( ⋅ , D ) : C → S G by A ↦ C P B ( A , D ) and ϕ ↦ ϕ * for any A ∈ C and ϕ ∈ H o m ( A , C ) . Then C P B ( ⋅ , D ) is a contravariant functor from C to S G .
Proof. 1) For a C*-algebra A and [ φ ] ∈ C P B ( A , D ) , we have I * ( [ φ ] ) = [ φ ∘ I ] = [ φ ] . Then I * is the unit of C P B ( A , D ) .
2) Let φ 1 ∈ H o m ( A , E ) and φ 2 ∈ H o m ( E , C ) . Set F = C P B ( ⋅ , D ) . Then
F ( φ 2 ∘ φ 1 ) [ φ ] = [ φ ∘ φ 2 ∘ φ 1 ] = F ( φ 1 ) [ φ ∘ φ 2 ] = F ( φ 1 ) ∘ F ( φ 2 ) [ φ ] .
Thus C P B ( ⋅ , D ) is a contravariant functor.
Corollary 3.23. Let G be the category of abelian groups. Then C P B ( ⋅ , D ) induces a contravariant functor C P B − 1 ( ⋅ , D ) from C into G by A ↦ C P B − 1 ( A , D ) , and from H o m ( A , C ) into H o m ( C P B − 1 ( C , D ) , C P B − 1 ( A , D ) ) by ϕ ↦ ϕ * .
For a short exact sequence of C*-algebras 0 → C → φ 1 E → φ 2 A → 0 , the functor C P B ( ⋅ , D ) from C to S G is not exact, and it is even not split-exact. The following is a counterexample.
Example 3.24. Suppose that H is an infinite dimensional separable Hilbert space. Let A = C = K ( H ) , E = A ⊕ C , D = B ( H ) and B = 0 . Then C P B ( A , D ) = C P 2 ( A , D ) . Let f 1 : C → E be the inclusion map and let f 2 : E → A be the quotient map. Then the exact sequence
0 → C → f 1 E → f 2 A → 0
is split.
Take a nonzero element η ∈ C P ( A , ℂ I D ) . We define a map φ : E → D by φ | C = I C and φ | A = η . Then φ ∈ C P ( E , D ) and [ φ ∘ f 1 ] = 0 . If [ ψ ∘ f 2 ] 0 = [ φ ] 0 for some ψ ∈ C P 2 ( A , D ) , then there exist ϕ 1 , ϕ 2 ∈ H o m ( E , D ) and a unitary u ∈ U ( M 2 ( D ) ) such that
( ψ ∘ f 2 ϕ 1 ) = u ( φ ϕ 2 ) u * .
Put
u = ( u 1 u 2 u 3 u 4 ) .
Since ( ψ ∘ f 2 ) ( E ) = 0 , u 1 φ ( E ) u 1 * + u 2 ϕ 2 ( E ) u 2 * = 0 . Note that u 1 φ ( e ) u 1 * , u 2 ϕ 2 ( e ) u 2 * are positive if e is positive in E. It follows that
u 1 φ ( e ) u 1 * = u 2 ϕ 2 ( e ) u 2 * = 0.
Therefore u 1 K ( H ) u 1 * = 0 . Furthermore, u 1 u 1 * = 0 since there is a sequence in K ( H ) which is convergent to I in the strong operator topology on B ( H ) . Then u 1 = 0 . Hence u 4 = 0 and u 2 , u 3 ∈ U ( D ) . Therefore,
u ( φ ϕ 2 ) u * = ( u 2 ϕ 2 u 2 * u 3 φ u 3 * ) .
Since u 3 φ u 3 * = ϕ 1 is a homomorphism, φ is also a homomorphism. However, φ | A is not a homomorphism by the definition of φ . Otherwise, if φ | A is a homomorphism from K ( H ) to ℂ , then it follows that φ | A = 0 since a completely positive map preserves self-adjoint elements. This is in contradiction to the fact that φ | A ≠ 0 .
Theorem 3.25. Suppose that
0 → C → f 1 E → f 2 A → 0
is a split short exact sequence, then
0 → C P B − 1 ( A , D ) → ( f 2 ) * C P B − 1 ( E , D ) → ( f 1 ) * C P B − 1 ( C , D ) → 0
is also a split short exact sequence.
Proof. Since ( f 1 ) * ∘ ( f 2 ) * ( [ φ A ] ) = [ φ A ∘ f 2 ∘ f 1 ] = 0 , we have I m ( f 2 ) * ⊂ K e r ( f 1 ) * .
Assume that E = A ⊕ C . For any [ φ ] ∈ K e r ( ( f 1 ) * ) , let φ A = φ | A and φ C = φ | C . Then φ = φ A ⊕ φ C . Note that [ φ ] is invertible and
[ φ C ] = [ φ ∘ f 1 ] = ( f 1 ) * ( [ φ ] ) .
Hence, [ φ C ] ∈ C P B − 1 ( C , D ) . Similarly, [ φ A ] is also invertible.
Suppose that the inverses of [ φ A ] and [ φ C ] are [ φ ′ A ] and [ φ ′ C ] respectively. Let [ φ ′ ] = [ φ ′ A ⊕ φ ′ C ] . Now we show that [ φ ] + [ φ ′ ] is the unit. Suppose that [ φ ″ A ] 0 = [ φ A ] 0 + [ φ ′ A ] 0 and [ φ ″ C ] 0 = [ φ C ] 0 + [ φ ′ C ] 0 such that π ∘ φ ″ A = π ∘ ϕ A and π ∘ φ ″ C = π ∘ ϕ C , where ϕ A and ϕ C are homomorphisms. Then
[ φ ] 0 + [ φ ′ ] 0 = [ φ ″ A ⊕ φ ″ C ] 0 .
Since π ( φ ″ A ⊕ φ ″ C ) is a homomorphism, [ φ ] + [ φ ′ ] is the unit of C P B − 1 ( E , D ) . Since [ φ ] ∈ K e r ( ( f 1 ) * ) , [ φ ∘ f 1 ] = 0 . Then π ∘ φ C = π ∘ ϕ C and hence [ φ ′ C ] is the inverse of [ φ C ] . Therefore, [ φ ′ ] = [ φ ′ A ⊕ 0 ] is the inverse of [ φ ] . Since ( f 2 ) * ( [ φ ′ A ] ) = [ φ ′ ] , [ φ ] ∈ I m ( ( f 2 ) * ) . Thus,
I m ( ( f 2 ) * ) = K e r ( ( f 1 ) * ) .
Suppose that ( f 2 ) * ( [ φ A ] ) = 0 . Then [ φ A ⊕ 0 ] = 0 , and there exist ψ ∈ C P ( E , D ) and ϕ ∈ H o m ( E , D ) such that [ ψ ] 0 = [ φ A + 0 ] 0 and π ∘ ψ = π ∘ ϕ . Hence, π ∘ ψ | A = π ∘ ϕ | A . Note that ϕ | A ∈ H o m ( A , D ) and [ ϕ | A ] 0 = [ φ A ] 0 . It follows that [ φ A ] = 0 and ( f 2 ) * is an injective homomorphism.
Suppose that [ φ C ] ∈ C P B − 1 ( C , D ) . Then we have
( f 1 ) * ( [ 0 ⊕ φ C ] ) = [ ( 0 ⊕ φ C ) ∘ f 1 ] = [ φ C ] .
Therefore, ( f 1 ) * is surjective.
Define
f * : C P B − 1 ( C , D ) → C P B − 1 ( E , D ) , [ φ C ] ↦ [ 0 ⊕ φ C ] .
Then ( f 1 ) * ∘ f * = I . Finally,
0 → C P B − 1 ( A , D ) → ( f 2 ) * C P B − 1 ( E , D ) → ( f 1 ) * C P B − 1 ( C , D ) → 0
is a split short exact sequence.
Remark 3.26. For any C*-algebra B, we can define C P 2 ( A , B ) , C P I − 1 ( A , B ) , etc., to be C P 2 ( A , M ( K ⊗ B ) ) , C P I − 1 ( A , M ( K ⊗ B ) ) , respectively. Since for any stable C*-algebra its multiplier algebra is properly infinite, these invariants are well-defined.
This work was supported by the Shandong Provincial Natural Science Foundation (Grant No. ZR2020MA009).
The authors declare no conflicts of interest regarding the publication of this paper.
Wang, R.F., Liu, S.D. and Wei, C.G. (2022) A Classification of Completely Positive Maps. Journal of Applied Mathematics and Physics, 10, 3649-3664. https://doi.org/10.4236/jamp.2022.1012243