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?

$\begingroup$ Further question: is the answer (to the original question) yes if we ask that all matrices are not only positive, but invertible? $\endgroup$ – Ruben A. MartinezAvendano Jan 10 '12 at 20:02

$\begingroup$ Only the element $x$ is in the question (the $a_i$ and $b_i$ would be part of the claimed conclusion). So are you asking: what if $x$ is invertible? $\endgroup$ – Matthew Daws Jan 10 '12 at 21:18
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 $a$.

$\begingroup$ Note that, at least in the case where both A and B are the C* algebras of square matrices of fixed size, this really is equivalent to the existence of entangled states in quantum physics. $\endgroup$ – Tsuyoshi Ito Oct 23 '10 at 13:03

$\begingroup$ I'm not really sure if it is the same. The existence of entangled states has to do with the fact that elements in the tensor product may fail to be elementary tensors (they are sums of elementary tensors, and these sums are the entangled states). The question here has to do with positivity: what the answer shows is that the positive part of the tensor product of two C$^*$algebras is not the tensor of the positive parts of the two C$^*$algebras. $\endgroup$ – Martin Argerami Oct 23 '10 at 16:58

$\begingroup$ This really does have everything to do with mixed state entanglement in quantum mechanics. A positive matrix is "separable" precisely when it can be written as a sum of tensor powers of positive matrices. These form a convex cone that is a strict subset of the cone of all positive matrices in the tensor product, and the complement of the separable operators in this cone are "entangled". The operators of unit trace correspond to (either separable or entangled) density matrices. There are close parallels to the classification of positive/completely positive linear maps on matrix algebras. $\endgroup$ – Jon Yard Oct 23 '10 at 17:30

$\begingroup$ My bad, then. I know little to nothing about that, and I always believed that entanglement was about the states (the vectors) and not the observables (the matrices). $\endgroup$ – Martin Argerami Oct 23 '10 at 19:16

1$\begingroup$ @Jon. Unbelievable! The four authors of this paper bear the same last name. What are their family links ? $\endgroup$ – Denis Serre Jan 11 '12 at 8:35
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).

$\begingroup$ I don't immediately see how you conclude, from an equality between two sums of elementary tensors, that the elementary tensors should be equal. $\endgroup$ – Martin Argerami Oct 22 '10 at 12:06

$\begingroup$ More to the point, say $e_i^*=e_i$ so $b_i^*=b_i$. But then why are $e_i$ and $b_i$ positive? $\endgroup$ – Matthew Daws Oct 22 '10 at 12:10


$\begingroup$ @Martin because the first components are linearly independent $\endgroup$ – Bugs Bunny Oct 22 '10 at 12:33

1$\begingroup$ Sorry, I'm being dumb. In part (2), I always get a $\alpha^2\gamma^2 e_ie_i^*\otimes b_ib_i^*$ term: how can I get rid of these by taking positive linear combinations? $\endgroup$ – Matthew Daws Oct 22 '10 at 15:02