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Let $X_{AB}$ be an operator acting on the tensor-product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$. Suppose that $X_{AB}$ is block positive, meaning that (in Dirac notation)

$\langle \psi |_A \otimes \langle \varphi|_B X_{AB} | \psi \rangle_A \otimes | \varphi \rangle_B \geq 0 $

for all vectors $| \psi \rangle_A \in \mathcal{H}_A$ and $| \varphi \rangle_B \in \mathcal{H}_B$. Let us abbreviate this as

$X_{AB} \geq_{\text{BP}(A:B)} 0$.

Now let $Y_{CD}$ be an operator acting on the tensor-product Hilbert space $\mathcal{H}_C \otimes \mathcal{H}_D$. Suppose that

$Y_{CD} \geq_{\text{BP}(C:D)} 0.$

Is it then the case that

$X_{AB} \otimes Y_{CD} \geq_{\text{BP}(AC:BD)} 0$

or is there a counterexample?

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The answer to this question is "no," there is a simple counterexample. First, Fact 1: consider that a map $\mathcal{N}_{A\to B}$ is positive iff its Choi operator $J^{\mathcal{N}}_{AB}$ is block positive, where the Choi operator is defined as

$J^{\mathcal{N}}_{AB} = \sum_{i,j} | i\rangle \langle j|_{A} \otimes \mathcal{N}_{A\to B}(| i\rangle \langle j|_{A}),$

and $\{ |i\rangle \}_i$ is an orthonormal basis. This fact can be proven and is stated in the paper https://arxiv.org/pdf/1408.6981.pdf

So then our counterexample is given by

$F_{AB} \otimes I_{CD}$

where $F_{AB}$ is the swap operator, defined as

$F_{AB} = \sum_{i,j} | i\rangle \langle j|_{A} \otimes | j\rangle \langle i|_{B}.$

Consider that $F_{AB}$ is the Choi operator for the transpose map (corresponding to the basis $\{ |i\rangle \}_i$), whose action on an operator $X$ is defined as

$T(X) = \sum_{i,j} | i\rangle \langle j| X | i\rangle \langle j|$.

Since the transpose map is a positive map, it follows that $F_{AB}$ is block-positive by Fact 1.

Consider that $I_{CD}$ is the Choi operator for the identity map and of course the operator $I_{CD}$ and the identity map are positive.

However, $F_{AB} \otimes I_{CD}$ is the Choi operator for the map $T_B \otimes \text{id}_D$ which is not positive. So then, by appealing to Fact 1 again, this implies that $F_{AB} \otimes I_{CD}$ is not block-positive with respect to the cut $AC | BD$.

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