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This question seems elementary, but I have already asked an expert who does not know the answer, so I would like to post here.

Let $M$ and $N$ be von Neumann algebras, and let $M\bar{\otimes}N$ be their von Neumann algebra tensor product.

Question: Can every projection in $M\bar{\otimes}N$ be expressed as the supremum (join) of projections of the form $p\otimes q$, where $p$ and $q$ are projections in $M$ and $N$, respectively?

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The answer is no.

Proof. Let $\mathcal{H}$ and $\mathcal{K}$ be any infinite-dimensional Hilbert spaces, and let $\{\xi_n\}_{n=1}^\infty$ and $\{\eta_n\}_{n=1}^\infty$ be sequences of any orthogonal vectors of norm $1$ in $\mathcal{H}$ and $\mathcal{K}$, respectively. Then $\zeta:=\sum_{n=1}^\infty\frac{1}{2^n}\xi_n\otimes\eta_n$ is an element of $\mathcal{H}\otimes\mathcal{K}$ (the completion has been taken), but not an element of $\mathcal{H}\odot\mathcal{K}$ (the algebraic tensor product before taking completion). The rank-$1$ projection onto $\mathbb{C}\zeta$ is in $\mathbb{B}(\mathcal{H})\bar{\otimes}\mathbb{B}(\mathcal{K})$, but not in $\mathbb{B}(\mathcal{H})\otimes\mathbb{B}(\mathcal{K})$, and it cannnot be the supremum of projections of the form described in the question since it is of rank-$1$.

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    $\begingroup$ Nice. It would be interesting to see if the statement is true in the case of diffuse algebras. $\endgroup$
    – Adrián González Pérez
    Commented Jan 23, 2019 at 13:00
  • $\begingroup$ @Adrián González-Pérez: Thank you for a good question. I will think about it. $\endgroup$
    – Masayoshi Kaneda
    Commented Jan 23, 2019 at 19:02
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    $\begingroup$ Indeed, if $r$ is a projection in $M\bar{\otimes}N$ such that $p_\alpha\otimes q_\alpha\le r$ for all $\alpha$, then $p_\alpha\mathcal{H}\odot q_\alpha\mathcal{K}\subseteq(p_\alpha\otimes q_\alpha)(\mathcal{H}\otimes\mathcal{K})\subseteq r(\mathcal{H}\otimes\mathcal{K})$ for all $\alpha$. Thus $p\mathcal{H}\odot q\mathcal{K}\subseteq(p\otimes q)(\mathcal{H}\otimes\mathcal{K})\subseteq r(\mathcal{H}\otimes\mathcal{K})$, and hence $p\otimes q\le r$. $\endgroup$
    – Masayoshi Kaneda
    Commented Nov 19, 2023 at 3:27
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    $\begingroup$ After all, any projection in $M\bar{\otimes}N\setminus M\otimes N$ cannot be the supremum of projections of the form $p\otimes q$ with $p\in M$ and $q\in N$. This gives an alternate and more complete answer to my original question. $\endgroup$
    – Masayoshi Kaneda
    Commented Nov 19, 2023 at 3:32
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    $\begingroup$ @Adrián González Pérez: It was about your old question, but I answered in my comments above. Or, maybe you have already figured out. Thank you for asking, though. $\endgroup$
    – Masayoshi Kaneda
    Commented Nov 19, 2023 at 17:30

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