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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.

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

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