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Let $H$ and $K$ be Hilbert spaces. Recall that a closed subspace $V\subset B(H,K)$ is called a ternary ring of operators(TRO) if $xy^*z \in V$ for all $x,y,z \in V$. Obviously a TRO is also a operator space.

Let $V\subset B(H_1,K_1)$ and $W\subset B(H_2,K_2)$ be two TROs. We consider operator space Haagerup tensor product $V\otimes_{\rm h}W$ of operator spaces $V$ and $W$. It is well known that $V\otimes_{\rm h}W$ is operator space. Is it true that $V\otimes_{\rm h}W$ is also a TRO?

P.S: The same question was first posted on MSE but I dint get any answer there.

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  • $\begingroup$ No. Look at the row and the column Hilbert spaces. $\endgroup$
    – Narutaka OZAWA
    Commented Apr 5, 2021 at 0:58
  • $\begingroup$ @NarutakaOZAWA: sorry but I don't understand how this example works as Haagerup tensor product of the column Hilbert space with the row Hilbert space is the space of compact operators which is also a TRO. Can you please explain? $\endgroup$
    – Math Lover
    Commented Apr 9, 2021 at 6:46
  • $\begingroup$ Haagerup's tensor is not commutative. Row tensored with Column is the space of trace class operators. $\endgroup$
    – Narutaka OZAWA
    Commented Apr 9, 2021 at 6:57
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    $\begingroup$ @NarutakaOZAWA: But even trace class operators being a operator space is closed and also is an ideal so preserve ternary product? What am I missing? $\endgroup$
    – Math Lover
    Commented Apr 9, 2021 at 7:18
  • $\begingroup$ The space of trace class operators with the trace norm is not a TRO. $\endgroup$
    – Narutaka OZAWA
    Commented Apr 9, 2021 at 7:21

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