Tuesday, 6 September 2011

fa.functional analysis - Ordering of completely bounded maps

Let A be a C*-algebra, let H be a Hilbert space, and let $T:Arightarrow B(H)$ 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 hermitian in that $T(x^*)^* = T(x)$ for all $xin A$, then $T=T_1-T_2$. We can order the hermitian cb maps $Arightarrow B(H)$ by saying that $Tgeq S$ if $T-S$ is completely positive.



I'm interested in criteria by which we can recognise that $Tgeq S$. Even special cases would be good (for example, I'm happy to assume that $T$ is completely positive).



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 $T(x) = V^*pi(x)V$, then $Tgeq Sgeq 0$ if and only if $S(x) = V^*pi(x)AV$ where $0leq Aleq1$ is a positive operator in the commutant of $pi(A)$. This is nice, but suppose all I know is that $T(x)=V^*pi(x)V$ and $S(x) = U^*pi(x)U$ (notice that the representation $pi$ is the same). Can I "see" if $Tgeq S$ by looking at $U$ and $V$? What if S is only cb, so $S(x)=Api(x)B$? Maybe that's too much to hope for, but anything vaguely in this direction would be interesting.

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