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Let $A$ be a $C^*$-algebra. The Stinespring construction shows that a completely positive contraction $T:A\rightarrow B(H)$ has the form $T(x) = U^* \pi(x) U$ where $U:H\rightarrow K$ is a contraction, and $\pi:A\rightarrow B(K)$ is a $*$-homomorphism. The Paulsen 2x2 matrix trick allows us to characterise complete contractions similarly, by replacing $U^*$ with $V$, another contraction.

Similarly, if $F$ is a Hilbert $C^*$-module over $B$ then a strict completely positive contraction $T:A\rightarrow L(F)$ is of the form $T(x) = U^* \pi(x) U$ where now $\pi:A\rightarrow L(G)$ is a $*$-homomorphism, $G$ being some Hilbert $C^*$-module over $B$, and $U\in L(F,G)$ is a contraction. (Recall, from e.g. Lance's book on Hilbert $C^*$-modules, that "strict" means that $(T(e_i))$ is a Cauchy net in the strict topology on $L(F)$, for some approximate unit $(e_i)$ for $A$. So, if you wish, you can just restrict to unital algebras, and then "strict" is a vacuous condition).

It seems to me that: (a) The Paulsen 2x2 trick should also work here (although I haven't thought very long about this at all, because...) (b) if so, surely someone has already written this down (although a bit of MathSciNet work didn't find anything). So my question is:

Is there a known characterisation of strict complete contractions? And what's a good reference?

Edit: Thinking more about, the Paulsen 2x2 matrix technique uses Arveson's extension theorem-- and this is likely to be somewhat more delicate in the Hilbert module / adjointable operator setting. Still, I'd be interested if anything is known...

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There must be a typo somewhere (you state a result on strict complete contractions and ask whether this result can be extended to strict complete contractions). By the way, could you recall the definition of a strict complete contraction? –  Mikael de la Salle Apr 6 '11 at 14:39
    
@Mikael, yes, sorry, fixed! –  Matthew Daws Apr 6 '11 at 14:55
    
Matthew, is it essential that $T$ is strict? Are there any cpc $T$ which is not strict and does not have the form $T(x)=U^*\pi(x)U$? –  Kate Juschenko Apr 6 '11 at 15:30
    
@Kate: Well, if you want $\pi$ to be non-degeneracy then you need strictness. So I guess that it's probably not really needed (but it seems to be a convenient extra assumption). –  Matthew Daws Apr 6 '11 at 16:00
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