# Tensor product of positive linear maps is positive

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.

• See E. Stormer, Tensor products of positive maps of matrix algebras, Math. Scand. 111 (2012), 5-11. – Nik Weaver May 14 at 15:53

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.