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?



positive elements in tensor products
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 $a$. 


