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 $a$.

Answer 2

Further question: is the answer (to the original question) yes if we ask that all matrices are not only positive, but invertible?

Answer 3

It is nearly yes as you need to use linear combinations rather then sums. If $y=\sum_i e_i \otimes b_i$ then $x=yy^* =\sum_{i,j} e_ie_j^* \otimes b_ib_j^*$. This leaves you with two types of summands whose positivity is clear:

(1) $e_ie_i^*\otimes b_ib_i^*$

(2) $e_ie_j^* \otimes b_ib_j^* + e_je_i^* \otimes b_jb_i^*$ whose positivity is clear by the elementary calculation that boils down to multilinearization of $(\alpha e_i+\beta e_j)(\alpha e_i+\beta e_j)^* \otimes (\gamma b_i+\delta b_j)(\gamma b_i+\delta b_j)^*$. (Just write the second summand as a linear combination of these guys.)

Clearly, you need to use subtractions to clear up the summands of type (2).