# tensor stability of block-positive matrices

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?

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 $$AC | BD$$.