Let $\mathcal A$ be a conformal net, and let $\mathcal J$ be the set of all proper open sub-intervals of $S^1$.

We say that $\mathcal A$ satisfies von Neumann density, if for any representation $\pi$ of $\mathcal A$ and any $I\in\mathcal J$, $\pi(\mathcal A(I)\vee\mathcal A(I'))=\bigvee_{J\in\mathcal J}\pi(\mathcal A(J))$. ($I'$ is the open complement of $I$.) The importance of von Neumann density is that any representation of $\mathcal A$ is determined by the corresponding $\mathcal A(I)$-$\mathcal A(I')^{\mathrm{opp}}$ bimodule. So Connes fusions of representations of $\mathcal A$ cannot be well defined without the requirement of von Neumann density.

We say that $\mathcal A$ is strongly additive, if for any $I\in\mathcal J$, and any pair of disjoint open intervals $I_1,I_2$ obtained by removing a point on $I$, we have $\mathcal A(I)=\mathcal A(I_1)\vee\mathcal A(I_2)$.

It is easy to see that strong additivity implies von Neumann density. Question: are these two conditions equivalent?

  • $\begingroup$ When you say "von Neumann density", do you mean in the vacuum representation, or do you mean in every representation of the conformal net? $\endgroup$ Aug 19, 2017 at 21:05
  • $\begingroup$ Hi Andre. Thanks for asking, it should be on any representation. $\endgroup$
    – Bin Gui
    Aug 19, 2017 at 23:06


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