Question

Let A, B be two C*-Algebras and $A\otimes B$ denote their minimal tensor product(I don't know whether C*-norm matters or not, but for simplicity we can assume that one of them is nuclear so all C*-norm coincide). Let x be a non-zero positive element in $A\otimes B$, can we always find a single tensor $0\neq x_1\otimes x_2$, where $x_1$ and $x_2$ are positive elements in A and B respectively, such that $x_1\otimes x_2\leq x$?

It's fairly easy to see that if both C*-algebras are communicative or one of them is a finite dimensional C*-Algebra(Sorry this is false), then the above assertion is true. So it's tempting to think that more general case should hold.

I asked a similar question before, where the stronger assertion that any positive element in tensor algebra is a sum of tensors of positive elements, is false. See the following link:

link text

Answer 1

The same answer as before, the matrix $$ a=\begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{bmatrix} $$ in $M_2(\mathbb{C})\otimes M_2(\mathbb{C})$, also works here since it is twice a rank one projection and so any smaller positive matrix must be a scalar multiple of $a$.

Answer 2

There is a result of Kirchberg that comes close to giving a positive answer to this question. Given $x\geq 0$ as in the question, there exists $z\neq 0$ such that $z^*z=x_1\otimes x_2$ and $zz^* \leq x$. See Lemma 4.1.9 of Rordam's book "Classification of nuclear C*-algebras".