Let $\mathcal M$ and $\mathcal N$ be two von Neumann algebras.


If $x$ is a positive element of $\mathcal M$ and $y$ is a positive element of $\mathcal N$, it is known that $x\otimes y$ is a positive element of $\mathcal M\overline\otimes\mathcal N$.


So I was wondering : *is every positive element $z$ of $\mathcal M\overline\otimes\mathcal N$ the supremum of the set of all the elements of $\mathcal M\overline\otimes\mathcal N$ of the form
$$\sum\_{n\in\mathbf N}x\_n\otimes y\_n$$ where $(x\_n)\_{n\in\mathbf N}$ and $(y\_n)\_{n\in\mathbf N}$ are finitely supported sequences of positive elements of $\mathcal M$ and $\mathcal N$ respectively such that $\sum\_{n\in\mathbf N}x\_n\otimes y\_n\leq z$ ?*


I wouldn't be surprised if this was not true in general and only worked when $z\geq \varepsilon\cdot 1\_{\mathcal M\overline\otimes\mathcal N}$ from some real numer $\varepsilon>0$ or under some condition of a similar fashion.



Approximation from below of positive elements in tensor product of von Neumann algebras
It fails for $M=N=M\_2(\mathbb C)$. Here $M\overline \otimes N = \mathcal B(\mathbb C^2 \otimes \mathbb C^2)$. Let $e\_1 =(1,0), e\_2 = (0,1) \in \mathbb C^2$. The rank 1 projection $z$ onto the span of $\tfrac{1}{\sqrt{2}} ( e\_1 \otimes e\_1 + e\_2 \otimes e\_2)$ gives a counter-example. In fact, let $x\otimes y \leq z$. Suppose $x\otimes y \neq 0$, then we may assume $x\otimes y = z$ (since $z$ has rank 1). Write $x\_{i,j} = \langle x e\_j, e\_i\rangle$ for the $(i,j)$'th matrix entry (similar for $y$). Then
\begin{equation}
x\_{i,j} y\_{k,l} = \langle (x\otimes y) (e\_j \otimes e\_l) , e\_i \otimes e\_k \rangle = \langle z (e\_j \otimes e\_l) , e\_i \otimes e\_k \rangle = \delta\_{i,k} \delta\_{j,l} /2.
\end{equation}
Hence $x\_{1,1}y\_{1,1} = 1/2$, $x\_{1,2}y\_{1,2} = 1/2$ and $x\_{1,1} y\_{1,2} = 0$, a contradiction.


So the only positive operator of the form $x\otimes y$ below $z$ is $0$.


