Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

By results of Størmer and Woronowicz, every positive map $\Phi \colon \mathcal{M}_{d \times d} \rightarrow \mathcal{M}_{d' \times d'}$ for $dd' \leq 6$ can be decomposed as a convex combination

$$\Phi = p \phi + (1-p) ~ T \circ \psi$$

where $\phi$, $\psi$ are completely positive maps and $T$ is the transposition map.

For higher dimensions, this is in general false. Does there however (for fixed $d$, $d'$) exist a finite set of positive maps $(P_i)$ such that every general positive map $\Phi$ is a convex combination

$$\Phi = \sum p_i P_i \circ \phi_i$$

where the $\phi_i$ are suitably chosen completely positive maps?

share|improve this question
    
Perhaps you should give the definition of positive map, as well as of completely positive map. –  Denis Serre Feb 16 '11 at 9:29
    
Thank you for your suggestion. A linear map $\Phi$ of $C^*$-algebras is positive if sends the positive cone to the positive cone. It is $k$-positive if $\Phi$ tensored with the identity map on $k \times k$-matrices is positive. It is completely positive if is $k$-positive for all $k > 0$. –  Michael Feb 16 '11 at 10:00
1  
Maybe I'm worng, but I thought that is an open problem at the heart of quantum information theory (?) –  Stefan Waldmann Feb 16 '11 at 10:07
    
Michael, would you please include a reference for the first decomposition you mention, or at least to the specific results of Stormer and Woronowicz that lead to it? –  Jon Bannon Feb 16 '11 at 16:01
1  
Jon, the results can be found in the articles "Positive maps of low dimensional matrix algebras" (Woronowicz) and "Positive linear maps of operator algebras" (Størmer). –  Michael Feb 16 '11 at 17:37

1 Answer 1

up vote 3 down vote accepted

I'm quite late answering this question, but I figured it still deserves an answer. The answer to your question is "no": when $d,d^\prime \geq 3$ there does not exist such a finite (or even countable) set of positive maps.

In arXiv:1209.0437, we considered the following problem: given a set of positive maps $\mathcal{Q}$ from $\mathcal{M}_d$ to $\mathcal{M}_{d^\prime}$, what is the set

$$\mathcal{C}_\mathcal{Q} \stackrel{\text{def}}{=} \{\sum_i \psi_i \circ P_i \circ \phi_i : P_i \in \mathcal{Q}, \psi_i \text{ and } \phi_i \text{ are completely positive for all $i$}\}?$$

As you noted, Størmer and Woronowicz showed that if $\mathcal{Q} = \{T\}$ (where $T$ is the transpose map) and $dd^\prime \leq 6$, then $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps.

We didn't prove it in the paper, but as an offshoot Łukasz Skowronek proved in the $d,d^\prime\geq 3$ case that if $\mathcal{C}_\mathcal{Q}$ is the set of all positive maps then $\mathcal{Q}$ must be uncountable. I don't believe he ever did end up formally writing up the result, but he at least gave a talk about it (slides available here).

The slides go through the proof in pretty good detail: the rough idea is that there is a known uncountably infinite set of positive maps on $\mathcal{M}_3$ that are each exposed in the set of positive maps and are all indecomposable (i.e., can not be written in the form $\phi_1 + T \circ \phi_2$ for some completely positive $\phi_1,\phi_2$) due to Ha and Kye (see arXiv:1108.0130). Łukasz showed that there is no countable set $\mathcal{Q}$ such that $\mathcal{C}_\mathcal{Q}$ contains even just this particular (relatively small) set of positive maps.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.