Completely positive maps as “positive operators”

Question

Let $A$ be a unital $C^{*}$-algebra and $\phi:A \rightarrow A$ be a completely positive map, i.e. $\phi^{(n)}:M_{n}(A) \rightarrow M_{n}(A)$ preserves positivity for any natural number $n$, where $\phi^{(n)}((A_{ij})_{ij})=(\phi(A_{ij}))_{ij}$. It is well-known that the norm $\|\phi\|=sup_{\|x\|\leq 1}\{\|\phi(x)\|\}=\|\phi(I)\|$.

Also recall that, given an element $x$ in a $C^{*}$-algebra $A$, the element $\|x\|-x$ is always a positive element.

It is often mighty tempting to want to treat a unital c.p. map like it is an element of a $C^{*}$-algebra and say that $\|\phi\|id-\phi$ is a positive map. I've heard that there are many circumstances where this is doable.

Question What are some natural circumstances where $\|\phi\|id-\phi$ is a positive map, with $\phi$ completely positive as above?

For example, if $A$ is a finite von Neumann algebra, then sometimes the map $\phi$ naturally gives rise to a bounded operator on $L^{2}(A)$. If this operator happens to be a positive operator then we get what we want. (I'd be interested in knowing natural circumstances where we know that precisely this happens.)

Answer 1

This answer deals with the case that $\phi$ is non-unital. In this case, $\phi$ must be of the form $\phi(a)=ha$, where $h$ is a positive element in the center of $A$.

Unfortunately, the solution I've got is somewhat long (hopefully correct): Let us assume that $\phi$ is contractive (otherwise one can rescale). Since $\phi(a)\leq a$ for any positive element $a$, $\phi$ preserves orthogonality of positive elements. There's a nice structural result for such maps proven in ``Completely positive maps of order zero", by Winter and zacharias. By Theorem 2.3 of that paper $\phi$ has the form $\phi(a)=h\pi(a)$, where $h=\phi(1)$, $$\pi:A\to M(C^*(\phi(A)))$$

is a unital homomorphism into the multiplier algebra of $C^*(\phi(A))$ and $h$ commutes with the image of $\pi$.

From $\phi(a)\leq a$ one gets $h\pi(a)\leq a$ for any positive $a$. Suppose that $\pi(b)=0$, with $b\in A_+$ positive contraction. Then $h=h\pi(1-b)\leq 1-b$. Similarly, $h\leq 1-b^{1/n}$ for all $n$. This implies that $hb=0$. So $h$ is orthogonal to $\ker \pi$. In particular, $\pi$ is injective on $\overline{hAh}$.

Let us show that $\pi$ is the identity on $\overline{hAh}$. Taking $n$-root in $h\pi(h^n)\leq h^n$ we get $h^{1/n}\pi(h)\leq h$. Since $h$ is a strictly positive element of $C^*(\phi(A))$ and $\pi(h)$ is a multiplier for that algebra, we have that $h^{1/n}\pi(h)\to \pi(h)$ in the strict topology. So, $\pi(h)\leq h$, and so $\pi$ maps $\overline{hAh}$ into itself. Passing to the limit in $h^{1/n}\pi(a)\leq a$, with $a\in \overline{hAh}$, one gets $\pi(a)\leq a$ for all such $a$. But if $\alpha$ and $\beta$ are homomorphisms such that $\alpha\leq \beta$ on all positive elements, then $\alpha+\tilde\alpha=\beta$, where $\tilde\alpha$ is a homomorphism with orthogonal range to $\alpha$. (There's probably a reference for this; I'll skip the argument to keep this answer short.) In our case, since $\pi$ is injective, we get $\pi(a)=a$ for $a\in \overline{hAh}$. In particular, $\pi(h)=h$.

We have that $$ \phi(a)=h\pi(a)=\pi(h^{1/2} a h^{1/2})=h^{1/2} a h^{1/2} $$for all $a\in A$. Finally, let us show that $h$ is in the center of $A$. We have $\pi(ah-ha)=\pi(a)\pi(h)-\pi(h)\pi(a)=0$. So $ha-ah\in \ker \pi$. It was argued above that this implies $h(ha-ah)=0$. So $h^2a=ah^2$. So $h$ is in the center.

Answer 2

If I am not mistaken, the map $\phi$ should be automatically equal to $id$ under such conditions. if $\psi:A\rightarrow B$ is a positive map between two $C^*$-algebras then it is bounded and $||\psi||\leq 2||\psi(1)||$. But $\psi(1)=id(1)-\phi(1)=0$, if we assume that $\phi$ is unital. therefore $\phi=id$.