The following conjecture is the critical one for the proposed connection between self-constructing continua and AQFT models. It appears as a supposition in the main result of “Rigid boolean inclusions,” which is in turned used in an essential way by “AQFT as a possible source of self-constructing continua.” If the conjecture turned out to be false, the case for pursuing the Bergsonian Axioms project would be weakened, perhaps fatally.

**Conjecture**: If $R \subseteq S$ is a simple subfactor inclusion of injective von Neumann factors, then every partition of $R$’s projection lattice is also a partition of $S$’s projection lattice.

*Partition* here means maximal set of pairwise-disjoint projections (where “$P, Q$ are disjoint” means $P \wedge Q = 0$). For the definition of *simple subfactor* see R. Longo’s “Simple injective subfactors” (1987). A stronger version of the conjecture would replace “simple” with “irreducible” (meaning that $R$ has trivial relative commutant in $S$).

We have very little in the way of leads on proving (or disproving) this conjecture. The requirement that every partition of $R$’s projection lattice be a partition of $S$’s projection lattice is satisfied by an inclusion $\mathcal{R} \subseteq \mathcal{S}$ of separably-acting commutative type II von Neumann algebras (this is easy to show if we consider that in this case the lattices are measure algebras), but these are not factors and it is not obvious how this situation might apply to the factor case. On the other hand, if there existed a type I factor $\mathcal{N}$ “between” $\mathcal{R}$ and $\mathcal{S}$, i.e. $\mathcal{R} \subseteq \mathcal{N} \subseteq \mathcal{S}$, then the requirement would be violated. But if there exists such an $\mathcal{N}$ then $\mathcal{R} \subseteq \mathcal{S}$ is not irreducible.