Take a unital cp map $f:B\to A$ between unital $C^*$ algebras. Given a state $\psi:B\to \mathbb{C}$ what conditions are necessary for there to exist a state $\phi:A\to \mathbb{C}$ so that $\phi\circ f=\psi$? I am sure that the answer must be known, and apologise for my ignorance.

In the case where $f$ is a unital $*$-algebra hom, and we consider the image of $f$ as a subalgebra, suitable conditions are given in -- Joel Anderson, Extensions, restrictions, and representations of states on $C^*$-algebras, Transactions of the American Mathematical Society 249(2):303-329, 1979.

This comes from considering a possibility of defining the degree of a cp map at a pure state. The classical theory requires taking the inverse image of points, which translates to the current question. The states may be assumed pure if it helps (as in the paper above).

As Nik points out below, the Hahn Banach theorem proves this if we have the inequality |ψ(b)|≤‖f(b)‖ for all b∈B.

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    $\begingroup$ Something's backwards ... do you mean $f: B \to A$? In that case, you need $f(x) = 0 \Rightarrow \psi(x) = 0$ so that the pushforward is well-defined, and you also need $f(x) \geq 0 \Rightarrow \psi(x) \geq 0$ so that the pushforward is positive on $f(B)$. Then you can just invoke Hahn-Banach to extend to a state on $A$. $\endgroup$ – Nik Weaver Jun 13 '17 at 13:37
  • $\begingroup$ Backwards - yes!! - will fix $\endgroup$ – Edwin Beggs Jun 13 '17 at 13:58
  • $\begingroup$ OK, had to do some checking - this uses the fact that on a unital $C^*$ algebra a unital norm 1 map to $\mathbb{C}$ has to be a state. And that can be made by the HB theorem, as long as $\phi$ on the image of $B$ (just as a normed space) is norm $\le 1$. Now why is that.... $\endgroup$ – Edwin Beggs Jun 13 '17 at 14:12
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    $\begingroup$ The fact you want is that if a linear functional on a unital C* algebra satisfies any two of the conditions (1) $\phi(1) = 1$, (2) $\phi \geq 0$, (3) $\|\phi\| = 1$, then it satisfies the third. $\endgroup$ – Nik Weaver Jun 13 '17 at 15:43
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    $\begingroup$ So you can go back and forth between $\phi$ being nonexpansive and being positive. $\endgroup$ – Nik Weaver Jun 13 '17 at 15:44

This is an expansion of Nik Weaver's comments. I'm using Effros+Ruan, "Operator Spaces" as a reference, but this is no doubt a little overkill.

A unital CP map is self-adjoint, so $f(B) \subseteq A$ is a (non-closed) operator system, namely a linear subspace which is unital and self-adjoint. (Most results in this area assume a closed subspace, but this is unnecessary). We have the following result:

A unital linear map between operator systems is completely positive if and only if it is completely contractive.

This implies that a unital functional from an operator system to $\mathbb C$ is positive if and only if it is contractive.

So give a state $\psi:B\rightarrow\mathbb C$, when does there exist a state $\phi:A\rightarrow\mathbb C$ with $\phi\circ f = \psi$?

Well-defined: If and only if $f(x)=0 \implies \psi(x)=0$.

Given this condition, there is a linear map $\phi:f(B)\rightarrow\mathbb C$ with $\phi f = \psi$. Notice that $\phi(1) = \phi(f(1)) = \psi(1) = 1$.

Positive: $f(x)\geq 0 \implies \psi(x)\geq 0$.

Then if $x\in f(B)$ and $x\geq 0$ there is $y$ with $f(y)=x$ so $\psi(y)\geq 0$ so $\phi(x)\geq 0$. Thus $\phi:f(B)\rightarrow\mathbb C$ is a positive unital functional, and so contractive. Hahn-Banach it to a contractive, unital functional on all of $A$, which must be a state, as required.

Equivalently (but not additionally) you could impose a different condition:

Contractive: $|\psi(x)| \leq \|f(x)\|$

Then with $y=f(x)$ we have $|\phi(y)| = |\psi(x)| \leq \|f(x)\| = \|y\|$ so $\phi$ is a unital contraction (hence positive). We may again apply Hahn-Banach.

  • $\begingroup$ OK, the new bit is the positive on operator systems implies contractive, giving the 2 conditions equivalent. I wonder what happens when the conditions are not satisfied... Anyway thanks to you and Nik. $\endgroup$ – Edwin Beggs Jun 16 '17 at 8:25

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