Let $M$ and $N$ be von Neumann algebras, and $\mathcal{H}$ a cyclic $M-N$ 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 $\eta$?



Closability of a natural bimodule map between cyclic correspondences of von Neumann algebras
Here is a down-to-Earth counter-example. 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 well-defined.) 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.


