Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this paper Man-Duen Choi gave a criteria for a completely positive map to be extreme. For convenience I am writing it below.

Let $\phi:\mathcal{M}_n\rightarrow\mathcal{M}_m$. Then $\phi$ is extreme in $CP[\mathcal{M}_n,\mathcal{M}_m]$, iff $\phi$ has an expression $\phi(x)=\sum_iV_i^*xV_i$ for all $x\in\mathcal{M}_n$ and $\lbrace V_i^*V_j\rbrace_{i,j}$ is a linearly independent set.

(I am assuming here unitality of the map. Otherwise, $\phi(I)=K\in\mathcal{M}_m$ for some fixed positive operator $K$.)

I have two questions.

1> Is there any extension of this theorem in any arbitrary $C^*$ algebra? (I mean extremal property, of course).

2> Even in finite dimension case, what can be the analogous result for (unital) completely bounded maps. Is it meaningful to ask this question for completely bounded maps?

I searched for the answers; but perhaps I am not giving the correct string; or did not understand some obvious points. Advanced thanks for any help.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

The only characterization I know of the extreme points of the CP maps on a C$^*$-algebra is Theorem 1.4.6 in Arveson's "Subalgebras of C$^*$-Algebras".

Arveson proves that if $\phi:A\subset B(H)\to B$ is CP with $\phi(I)=K$, and if $\phi=V^*\pi V$ is a Stinespring decomposition of $\phi$, then $\phi$ is extremal if and only if $\overline {VH}$ is faithful for $\pi(B)'$.

Here, faithful means that for every $x\in\pi(B)'$, $pxp=0\ \iff\ x=0$, where $p$ is the projection onto $\overline{VH}$.

I don't know enough about CB maps to say anything about your second question.

share|improve this answer
    
Thanks a lot. I have read that paper earlier. However, what I was hoping for, was something in the spirit of Choi. Perhaps it is not possible to make it simpler than the theorem of Arveson, except perhaps for some special cases, which can be intriguing. I did not get anything better in literature, and so decided to ask it to the community. From your answer, it seems, there is no better result known. Thanks once again. –  RSG Oct 15 '12 at 5:30
    
I'm no expert myself, but I did ask a colleague who is a C$^*$-convexity expert, and his answer was "Arveson I". The way I see it, there is no "Kraus formula" in a general C$^*$-algebra, so Stinespring is the best you have. In any case, you might want to ask this at Math Overflow. –  Martin Argerami Oct 15 '12 at 9:43
add comment

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.