# Characterizing fullness of a von Neumann algebra by the topology of its bimodules

Let $\mathcal{M}$ be a $\mathrm{II}_1$ factor. Among other characterizations, it is said to be full iff the adjoint map: $$\mathrm{Ad}: U(\mathcal{M})/\mathbb{T} \longrightarrow \mathrm{Aut}(\mathcal{M}),$$ given by $\mathrm{Ad}([u])(x) = u x u^*$ is a homeomorphism onto the inner automorphisms $\mathrm{Inn}(\mathcal{M})$, where $U(\mathcal{M})$ represents the unitary group of $\mathcal{M}$, $\mathrm{Aut}(\mathcal{M})$ is the space of normal $\ast$-automorphisms and the corresponding topologies are given by pointwise norm convergence in the predual. I.e. $\alpha_n \to \alpha$ iff $\|(\alpha_n)_\ast(\varphi) - \alpha_\ast(\varphi) \|_{\mathcal{M}_\ast} \to 0$.

On the other hand, let us denote by $\mathrm{Bimod}(\mathcal{M})$ the space $$\mathrm{Bimod}(\mathcal{M}) = \Big\{ H : {}_{\mathcal{M}}H_{\mathcal{M}} \mbox{ is a Hilbert } \mathcal{M}\mbox{-}\mathcal{M}\mbox{-bimodule} \Big\}$$ By $\mathcal{M}$-bimodule we mean a Hilbert spaces $H$ with two commuting normal actions of $\mathcal{M}$, one right one and one left one. The natural equivalence relation between bimodules is given by the existence of a unitary isometry that respects the actions. The Automorpism group of the von Neumann algebra embeds into the space of bimodules by taking $H(\alpha)$ as $L^2(\mathcal{M})$ with actions given by $$x \cdot \xi \cdot y = \alpha(x) \xi y.$$ It is trivial to see that $H(\alpha) \cong H(\beta)$ iff $\alpha$ and $\beta$ are equal modulo the inner automorphism, see [AP, p 211].

Question: Is it possible to characterize fullness by a topological property of $\mathrm{Bimod}(\mathcal{M})$. I think of something along the lines of the equivalence classes of $\mathrm{Bimod}(\mathcal{M})$ being closed or perhaps by the quotient $\mathrm{Bimod}(\mathcal{M}) / \sim$ having some separability property.

There is a natural Fell's topology on $\mathrm{Bimod}(\mathcal{M})$ that behaves like the usual Fell topology of the space of representations of a group, see [AP, 13.3.2.]. But i am not sure whether that would be the adequate topology to use.

This question was posted before in math.stackexchange.com without receiving an answer.

• Is the argument given by Ozawa here (for separable II-1 factors) the kind of thing you're looking for? mathoverflow.net/a/323282 – Yemon Choi Feb 18 '19 at 13:21
• Yes. Although that is not what I have in mind originally. I thought that, since $H(\alpha)$ is weakly eq. with $H(id) = L_2(\mathcal M)$ when $\alpha$ is approximately inner, imposing that the Fell-topology closures of two bimodules are equal iff the bimodules are isomorphic will be equivalent to fullness. But that is too strong. Ozawa's answer gives the correct version of what I wanted. – Adrián González-Pérez Feb 22 '19 at 11:28