Completely positive order-zero map - Imported Knowlpedia Content

Definition

Let $A$ and $B$ be $C^*$ -algebras. A completely positive map $\phi:A\to B$ has order zero if it preserves orthogonality of positive elements:

a,b\in A_+,\quad ab=0 \quad\Longrightarrow\quad \phi(a)\phi(b)=0.

Because positive $a$ and $b$ with $ab=0$ also satisfy $ba=0$ , the condition is symmetric. Order zero does not mean that $\phi$ vanishes, has zero algebraic order, or is multiplicative. Contractivity is not included in the convention used here; a map that is both contractive and order zero is called a CPC order-zero map.

Structure theorem

Let $C=C^*(\phi(A))$ . The order-zero structure theorem supplies a positive element $h$ in the center of the multiplier algebra $M(C)$ and a $*$ -homomorphism $\pi_\phi:A\to M(C)$ such that

\phi(a)=\pi_\phi(a)h=h\pi_\phi(a).

For unital $A$ , one can take $h=\phi(1_A)$ . This factorization explains why order-zero maps retain much of the orthogonality behavior of $*$ -homomorphisms without themselves preserving products Winter–Zacharias, Theorem 2.3.

Cone correspondence

Contractive completely positive order-zero maps $A\to B$ correspond naturally to $*$ -homomorphisms

C_0((0,1])\otimes A\longrightarrow B,

with the coordinate function on $(0,1]$ sent, together with $a$ , to $\phi(a)$ . This converts a nonlinear-looking orthogonality condition into ordinary multiplicative data Winter–Zacharias, Corollary 3.1.

Examples and near-misses

Every $*$ -homomorphism has order zero, as does a positive scalar multiple of one. A compression $a\mapsto V^*aV$ is completely positive but generally not order zero: two orthogonal positive operators can acquire overlapping compressions. Thus complete positivity alone does not preserve orthogonality.

References

Wilhelm Winter and Joachim Zacharias, “Completely positive maps of order zero,” Münster Journal of Mathematics 2 (2009), 311–324. Author-institution record. Relevant: Definition 2.1, Theorem 2.3, and Corollary 3.1.