# 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 $$\eta$$?

• 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. Jul 28, 2021 at 23:11
• @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? Jul 29, 2021 at 1:35
• Well, I guess so, even though I'm not 100% sure because I'm not that knowledgeable of type III von Neumann algebras. Jul 29, 2021 at 2:13

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.