Closability of a natural bimodule map between cyclic correspondences of von Neumann algebras 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 $\eta$?
 A: 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.
