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The set of completely bounded (CB) maps forms can be considered as a complex span of the set of completely positive (CP) maps. Can we find a basis for this complex linear space of CB maps such that every element of this is a CP map. What I am looking for is an explicit construction method for such a basis (if exists).

In same spirit we may consider the CB Hermiticity preserving maps which can be written as $\psi_1-\psi_2$, where $\psi_i\in$ CP maps and ask the similar question.

Even a partial answer is also helpful for me. (Example: Characterisations of all such maps between $\mathcal{B}(\mathbb{C}^n)\longrightarrow\mathcal{B}(\mathbb{C}^n)$, and similar partial cases). I hope, I explained my question correctly. I am ready to explain any ambiguity (if exists) and give further explanation if required (also re-edit my question to make it lucid and self-explanatory). Advanced thanks for all helps.

EDIT: As pointed by Tom Cooney, the above statement is not true for non-injective von Neumann algebras. Now the question is for injective $C^*$ algebras, does there exists a method by which we can actually construct a basis. For time being, we can consider only CB maps from $\mathcal{B(H)}$ to itself and ask for an explicit example of such basis.

Re-edit: Actually I tried to construct such basis for the maps from $\mathcal{B}(\mathbb{C}^n)$ to itself. However, structure of such maps, when CP, is well understood (Choi-Krauss representation), but does explicitly give a basis explicitly. At the best there is a result of Choi on the extremal points of such CP maps, but I failed to use it for the construction.

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This and this may help. – Norbert Jul 9 '12 at 9:04
Some comments: 1. What do you mean by convex span? Convex combinations of cp maps will again be cp and will not include cb maps which are not cp. 2. Yes, cb maps $B(\mathbb C^n) \to B(\mathbb C^n)$ are linear combinations of cp maps. However there are cb maps between operator algebras that cannot be decomposed as a linear combination of cp maps. See U. Haagerup's "Injectivity and decomposition of completely bounded maps" for further details. – Tom Cooney Jul 9 '12 at 10:38
@TomCooney Sorry, my stupid mistake. I wanted to write 'complex'. I have changed the question accordingly. I am trying to get a copy of the above mentioned paper. As per I know, by Wittstock’s decomposition theorem any cb map $\phi:\mathcal{A}\longrightarrow\mathcal{B(H)}$ can be written as $\psi_1\pm \psi_2\pm(\psi_3+\psi_4)$, where $\mathcal{A}$ is a $C^*$-algebra, $\psi_i$ are cp maps. Hence I am slightly uncomfortable about your statement, unless you want to say 'real linear'. That question I have asked in 2nd paragraph. Can you please explain it a bit more? – RSG Jul 9 '12 at 11:56
Yes, as you say, by Wittstock's theorem, any cb map from a $C^*$-algebra into an injective $C^*$-algebra (for example, $B(H)$) is a linear combination of completely positive maps. However if $N$ is any non-injective von Neumann algebra, then $CB(N,N) \neq \textrm{span} \ CP(N,N)$. This is shown in the paper of Haagerup's mentioned in my earlier comment. – Tom Cooney Jul 9 '12 at 13:14
One fundamental point that seems not to have been mentioned yet: in infinite dimensions, you need to be very careful about what you mean by a basis (Hamel? Schauder? Unconditional? Biorthogonal?) So if you are primarily interested in the finite-dimensional case, I suggest you clarify this in the question – user16299 Jul 10 '12 at 0:03

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