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In the paper titled "Nonlinear completely positive maps" M. D. Choi and T. Ando extended natural definition of completely positive maps ignoring the linearity condition (Aspects of positivity in functional analysis (Tübingen, 1985), 3–13, North-Holland Math. Stud., 122, North-Holland, Amsterdam, 1986). I do not have the access of the original paper, but I am reading some of the followup articles available in web. In this matter I have two question whose answer I did not get/understand from the materials I have in hand.

  1. Is it possible to have a Choi - Jamiołkowski isomorphism type theorem for such nonlinear CP maps? OK, it may not be an isomorphism; but may be a homomorphism? I know, it is nonlinear, but it would be nice if there is a method to check complete positivity of such maps.

  2. Does there exists nontrivial examples of nonlinear positive maps which are Hermiticity preserving and not CP? I have only few examples of maps which are nonlinear positive but not completely positive but they do not preserve Hermiticity.

    Examples:

    • $X\mapsto \|X\|^2$, which is $2$ positive not $3$ positive.

Further examples can be found in the book "Positive definite matrices" by Rajendra Bhatia. But I am yet to come across any example of nonlinear positive not CP map which preserves Hermiticity. Advanced thanks for any help/suggestions/references and etc.....

UPDATE: Actually I would prefer to have a matrix equation in the right hand side, not a scalar (or a scalar matrix).

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    $\begingroup$ Try for e.g., $X \mapsto \text{trace}(X^*X)^{1/2}$. This is 3-positive, but not more. Some more examples can be found in the paper Positive semidefinite 3 x 3 block matrices.... $\endgroup$
    – Suvrit
    Commented Jul 13, 2015 at 16:21
  • $\begingroup$ @Suvrit Actually I would like to have a matrix function at the right hand side, not a scalar (or scalar matrix). Sorry for not explicitly telling that in the question. $\endgroup$
    – RSG
    Commented Jul 13, 2015 at 16:25
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    $\begingroup$ The map $X\mapsto X \text{trace}(X^*X)^{1/2}$ is $3$-positive, which (I believe) it is not $4$-positive. $\endgroup$
    – M. Lin
    Commented Jul 15, 2015 at 17:12
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    $\begingroup$ @RSG: I contructed myself (perhaps it was known). As Suvrit mentioned $[\text{trace}(A_{i,j})^{1/2}I]_{i,j=1}^3$ is psd for whenever $[A_{i,j}]_{i,j=1}^3$ is. Then a Hadamard-like product of $[\text{trace}(A_{i,j})^{1/2}I]_{i,j=1}^3$ and $[A_{i,j}]_{i,j=1}^3$. $\endgroup$
    – M. Lin
    Commented Jul 15, 2015 at 17:38
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    $\begingroup$ @M.Lin this operation is a special case of Khatri-Rao tensor products --- Wikipedia also has an article about it :-) $\endgroup$
    – Suvrit
    Commented Jul 16, 2015 at 17:07

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