Ordering of completely bounded maps - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T05:46:20Z http://mathoverflow.net/feeds/question/36998 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36998/ordering-of-completely-bounded-maps Ordering of completely bounded maps Matthew Daws 2010-08-28T20:54:12Z 2010-08-28T21:04:20Z <p>Let A be a C*-algebra, let H be a Hilbert space, and let $T:A\rightarrow B(H)$ be a completely bounded (cb) map (that is, the dilations to maps $M_n(A)\rightarrow M_n(B(H))$ are uniformly bounded). We can write T has $T_1-T_2+iT_3-iT_4$ where each $T_i$ is completely positive. If $T$ is <em>hermitian</em> in that <code>$T(x^*)^* = T(x)$</code> for all $x\in A$, then $T=T_1-T_2$. We can order the hermitian cb maps $A\rightarrow B(H)$ by saying that $T\geq S$ if $T-S$ is completely positive.</p> <p>I'm interested in criteria by which we can recognise that $T\geq S$. Even special cases would be good (for example, I'm happy to assume that $T$ is completely positive).</p> <p>An old paper of Arveson ("Subalgebras of C*-algebras") shows that if T and S are both completely positive, and T has the minimal Stinespring dilation <code>$T(x) = V^*\pi(x)V$</code>, then $T\geq S\geq 0$ if and only if <code>$S(x) = V^*\pi(x)AV$</code> where $0\leq A\leq1$ is a positive operator in the commutant of $\pi(A)$. This is nice, but suppose all I know is that <code>$T(x)=V^*\pi(x)V$</code> and <code>$S(x) = U^*\pi(x)U$</code> (notice that the representation $\pi$ is the same). Can I "see" if $T\geq S$ by looking at $U$ and $V$? What if S is only cb, so $S(x)=A\pi(x)B$? Maybe that's too much to hope for, but anything vaguely in this direction would be interesting.</p>