positive elements in tensor products

Question

Let $A \otimes B$ be the algebraic tensor of two $C^{\ast}$ -algebras, and an element x in $A\otimes B$ is positive if $x=yy^{\ast}$. Then is it always possible to write x in the form $x=\sum a_i\otimes b_i$, where $a_i$ and $b_i$ are positive elements?

Answer 1

I think the answer is no. The matrix $$ a=\begin{bmatrix} 1&0&0&1\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&1 \end{bmatrix} $$ is positive in $M_4(\mathbb{C})$. When we see this algebra as $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, it cannot be obtained as a sum of elementary tensors with positive entries. .

(ok, several hours later, here is the argument)

First, $a$ is positive because it is selfadjoint and $a=\left(\frac1{\sqrt2}a\right)^2$. Now, if we have a sum of elementary tensors in $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, it will look like $$ \sum_j\begin{bmatrix} \alpha_j&\overline{\gamma_j}\\ \gamma_j&\beta_j\end{bmatrix} \otimes \begin{bmatrix}\alpha'_j&\overline{\gamma_j'}\\ \gamma_j'&\beta_j'\end{bmatrix} =\begin{bmatrix} \sum_j\alpha_j'\alpha_j& \sum_j \alpha_j'\overline{\gamma_j}& \sum_j\overline{\gamma_j'}\alpha_j&\sum_j\overline{\gamma_j'}\gamma_j\\ \sum_j\alpha_j'\gamma_j& \sum_j \alpha_j'\beta_j&*&*\\ *&*&*&*\\ *&*&*&* \end{bmatrix} $$ The assumption that each elementary tensor is made of the tensor of two positive matrices translates into $\alpha_j\geq0$, $\beta_j\geq0$, and $\alpha_j\beta_j\geq|\gamma_j|^2$ for all $j$ (and the "prime'' version too). Now if the matrix on the right is going to be our $a$ above, then the 2,2 entry forces the following: for each $j$, the product $\alpha_j'\beta_j=0$. If $\alpha_j'=0$, then $\gamma_j'=0$; and if $\beta_j=0$, then $\gamma_j=0$. That is, for each $j$, $\overline{\gamma_j'}\gamma_j=0$, and this forces the 1,4 entry to be $0$; but it is not zero in $a$.