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).


1 Answer 1


Since the class of approximately semifinite factors contains all injective factors and all semifinite factors and since it is closed under tensor products, it seems that this class is quite large and I would be surprised if this class could somehow be characterized.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.