# Ordering of completely bounded maps

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 hermitian in that $T(x^*)^* = T(x)$ 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.

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).

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 $T\geq S\geq 0$ if and only if $S(x) = V^*\pi(x)AV$ 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 $T(x)=V^*\pi(x)V$ and $S(x) = U^*\pi(x)U$ (notice that the representation $\pi$ is the same). Can I "see" if $T\geq S$ 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.

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I take it the criterion $U=\sqrt{A}V$ for some $0\leq A\leq1$ in the commutant of $\pi(A)$ is not what you're looking for. – Jonas Meyer Aug 28 '10 at 22:17
Ah, one should be a little bit careful. So if $U=A^{1/2}V$, then $T\geq S$. Conversely, if $T\geq S$, then S has the form $S(x)=W^*\pi(x)W$ with $W=A^{1/2}V$ for some suitable A. But if we also know that $S(x)=U^*\pi(x)U$, why does $U=W$? I'm certainly not going to claim that this dilation for S is minimal. Perhaps this CP question could be rescued, but I'm more interested in the CB question (that is, S is only known to be CB, but T is CP). – Matthew Daws Aug 29 '10 at 8:02