# Approximately semifinite factors

For the sake of this question, lets call a factor $$M$$ approximately semifinite if there exists an increasing net of semifinite subfactors $$M_i$$, $$i\in J$$, with conditional expectations $$E_i:M\to M_i$$ such that

• $$E_i\circ E_j = E_i$$ if $$i\ge j$$,
• For all $$\varphi\in M_*$$ it holds that $$\lim_i ||\varphi-\varphi\circ E_i|| =0$$.

Clearly, all semifimite factors are approximately semifinite and I know of several results where it is shown that certain type $$III$$ factors are approximately semifinite. For instance, all type $$III_0$$ factors with separable predual are approximately semifinite, see [1, Prop. 8.3]. Also, every ITPFI factor $$M=\bigotimes_{n=1}^\infty (M_{k_n},\varphi_n)$$ is approximately semifinite (with $$E_i: M \to \bigotimes_{n=1}^i M_{k_n}$$, $$i\in J=\mathbb N$$, being the slice map $$E_i(x\otimes y)= (\otimes_{n=i+1}^\infty \varphi_n)(y)\cdot x$$ relative to $$M=(\bigotimes_{n=1}^i M_{k_n})\otimes (\bigotimes_{n=i+1}^\infty M_{k_n})$$. Since all non-ITPFI hyperfinite factors are of type $$III_0$$, this shows that every hyperfinite factor is approximately semifinite.

My question is: How large is the class of approximately semifinite factors really? Are factors (with separable predual) known that are not approximately semifinite?

[1] Haagerup, Uffe; Størmer, Erling, Equivalence of normal states on von Neumann algebras and the flow of weights, Adv. Math. 83, No. 2, 180-262 (1990).

However, there are type III$$_1$$ factors that are not approximately semifinite. For instance, you could take the free Araki-Woods factor associated to a measure without atoms. By Corollary C in https://doi.org/10.4171/JEMS/898 this gives you a nonamenable type III$$_1$$ factor $$M$$ with the property that for every faithful normal state $$\varphi$$ on $$M$$, the centralizer $$M^\varphi$$ is amenable.
It follows from this property that whenever $$P \subset M$$ is a semifinite von Neumann subalgebra with faithful normal conditional expectation $$E : M \to P$$, then $$P$$ must be amenable. Indeed, if $$P$$ is not amenable, one can choose a finite projection $$p \in P$$ such that $$pPp$$ is not amenable. Combining a faithful tracial state on $$pPp$$ with any faithful normal state on $$(1-p)P(1-p)$$, one obtains a faithful normal state $$\omega$$ on $$P$$ such that $$P^\omega \supset pPp$$ is nonamenable. Writing $$\varphi = \omega \circ E$$, we have that $$P^\omega \subset M^\varphi$$. So also $$M^\varphi$$ is nonamenable, which is a contradiction.
Since $$M$$ is itself nonamenable, it follows in particular that $$M$$ is not approximately semifinite.