Let $\pi\_1: A\_1 \to B\_1$ and $\pi\_2: A\_2 \to B\_2$ be positive linear maps between complex $\*$-algebras. Is the mapping
$$\pi\_1 \otimes \pi\_2: A\_1 \otimes A\_2 \to B\_1 \otimes B\_2$$
again positive?


I.e., if $\sum\_{i=1}^n x\_i \otimes y\_i \in A\_1 \otimes A\_2$, do we have
$$\sum\_{i,j=1}^n \pi\_1(x\_i^\*x\_j)\otimes \pi\_2(y\_i^\*y\_j) \ge 0$$


I know the proof for the case that $B\_1 = B\_2 = \mathbb{C}$. In that case, we can work with diagonalisation of matrices.



Tensor product of positive linear maps is positive
No. A standard example is given by $A\_1 = A\_2 = B\_1 = B\_2 = M\_2(\mathbb{C})$, where we choose $\pi\_1$ to be the identity map and $\pi\_2$ to be the transpose map. These maps are positive, but $\pi\_1 \otimes \pi\_2$ is not positive since, for example
$$
 (\pi\_1 \otimes \pi\_2)\left(\begin{bmatrix}
1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1
\end{bmatrix}\right) = \begin{bmatrix}
1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1
\end{bmatrix}
$$
has $-1$ as an eigenvalue. In other words, the transpose map is not [completely positive](https://en.wikipedia.org/wiki/Completely_positive_map).


