Let $M$ and $N$ be von Neumann algebras, and $\mathcal{H}$ a cyclic $MN$ correspondence with unit cyclic vector $\xi$. For which $\eta\in \mathcal{H}$ is the bimodule map extending $\xi\mapsto \eta$ a closable operator on $\mathcal{H}$? Is it closable for every $\eta$?

2$\begingroup$ No in general. First, if there is a closable $M$$N$ bimodule map $T$ with $T\xi = \eta$, then $\eta$ belongs to the closed subspace $K$ spanned by $(M\vee N^{\mathrm{op}})' \xi$. When $(M\vee N^{\mathrm{op}})'$ has a type III summand, not all vectors in $K$ correspond to measurable operators. $\endgroup$– Narutaka OZAWAJul 28, 2021 at 23:11

$\begingroup$ @NarutakaOZAWA: Thank you for the response. I think I have to go and look up what a measurable operator is in the type III case, since I'm only familiar with $\tau$measurable operators with $\tau$ a semifinite trace. I gather that if a vector in $K$ does not correspond to a measurable operator, this implies that the associated operator is not closable? $\endgroup$– Jon BannonJul 29, 2021 at 1:35

$\begingroup$ Well, I guess so, even though I'm not 100% sure because I'm not that knowledgeable of type III von Neumann algebras. $\endgroup$– Narutaka OZAWAJul 29, 2021 at 2:13
1 Answer
Here is a downtoEarth counterexample. Set $N=\mathbb C$ and $M=\mathcal B(K)$ with $H=K\otimes K$ and $M$ acting on the first tensor factor (should perhaps be $K\otimes\overline K$ but this will not affect the argument). Let $K$ have orthonormal basis $(e_n)$ and take for example $\xi = \sum_n n^{1} e_n\otimes e_n$.
Then $\xi$ is also separating, and so the module map sending $\xi$ to $\eta$ will exist for any $\eta$. (In general, this need not be the case I think, so already in complete generality the bimodule map sending $\xi$ to $\eta$ might not be welldefined.) I now make some choices: let $$ \eta = \Big( \sum_n n^{3/4} e_n \Big) \otimes e_1. $$ Consider the matrix unit $e_{m,n}$ which sends $e_n$ to $e_m$, so our operator is $$ T:e_{m,n}\cdot \xi = n^{1} e_m \otimes e_n \mapsto e_{m,n}\cdot\eta = n^{3/4} e_m \otimes e_1. $$ Hence $T:e_m \otimes e_n \mapsto n^{1/4} e_m \otimes e_1$. Consider the sequence $(e_1\otimes\alpha_k) = (k^{1/4} e_1\otimes e_k) \rightarrow 0$ in $H$, while $$ T(\alpha_k) = k^{1/4} k^{1/4} e_1\otimes e_1 = e_1\otimes e_1, $$ for all $k$. Thus $T$ is not closable.

$\begingroup$ Thanks a million, Matt! If you combine this with Taka's commutant comment, I think this is essentially complete now... $\endgroup$ Jul 30, 2021 at 11:14