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.)

  • $\begingroup$ Jon, I experienced similar problems; inserting a couple of symbols ` helps sometimes. See also the advice on the right side "How to write math." $\endgroup$ – Andreas Thom Mar 4 '11 at 13:17
  • $\begingroup$ I was hoping you'd tell me what magic you worked. I'll have a look. $\endgroup$ – Jon Bannon Mar 4 '11 at 13:19
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    $\begingroup$ Also, if anyone can think of a more appropriate title for this question, I'd be interested in seeing one...we really aren't looking at the cp map as a positive operator... $\endgroup$ – Jon Bannon Mar 4 '11 at 18:09

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.

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  • $\begingroup$ I haven't had a chance to check this yet, but it is the sort of thing one would expect in light of Kate's answer above. Thank you very much for taking the time to provide an answer to this!!! $\endgroup$ – Jon Bannon Mar 5 '11 at 22:39
  • $\begingroup$ And welcome to MathOverflow, Leonel! $\endgroup$ – Jon Bannon Mar 6 '11 at 12:35
  • $\begingroup$ Thanks Jon. I wanted to reply to your comment above about how to name the question. I didn't figure out how to do it so I'll do it here. In view of the answer I'm giving, one gets a minimality (extremality) property for homomorphism. That is, if a cpc map is bounded above by a homomorphism, from A to B, then the cpc map is the homomorphism times an element h in the center of the image of the homomorphism. In particular, if this center is trivial, the *-homomorphism is minimal among cpc maps. Along these lines there may be better approach to solving the question, or googling the answer. $\endgroup$ – Leonel Robert Mar 6 '11 at 15:27

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 $\phi=id$.

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  • $\begingroup$ Hi Kate! I had to give you the credit for answering what I asked. You also reminded me, though, that the case I am really interested in is the nonunital case. Can you help me with this case? Thanks again! $\endgroup$ – Jon Bannon Mar 4 '11 at 14:11
  • $\begingroup$ Hi, Jon! I've should posted this as a comment, since indeed the question that was posted before has only trivial answer. The question as it is now seems to be non-trivial, I'll try to think on examples of that. $\endgroup$ – Kate Juschenko Mar 4 '11 at 14:24
  • $\begingroup$ Absolutely not, Kate. I asked, you answered, and that is much appreciated! $\endgroup$ – Jon Bannon Mar 4 '11 at 15:42
  • $\begingroup$ I just found out I can switch accepted answers, and both of you will get credit...so I did. $\endgroup$ – Jon Bannon Feb 15 '12 at 15:14

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