We know that we can build an irreducible subfactor realizing a finite single chain lattice containing any finite index irreducible maximal subfactors, by using the free composition (see here).
Now about infinite single chain lattice:

Question: Is there an irreducible subfactor with an infinite homogeneous single chain lattice?


1 Answer 1


Yes, for example $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ the Prüfer $p$-group subfactor.

More precisely:
Let $p$ be a prime number and let the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \mathbb{Z}/p^3\mathbb{Z} \hookrightarrow \cdots$
given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is $\mathbb{Z}(p^{\infty})$ the Prüfer $p$-group.
(Note that $\mathbb{Z}(p^{\infty})$ is the Pontryagin dual of the group of $p$-adic integers $\mathbb{Z}_p$)

$\mathbb{Z}(p^{\infty})$ admits no extra subgroups so its lattice is an infinite homogeneous single chain.
Now it is a discrete group so by Galois correspondence for discrete group (theorem 3.13 p 48 of this paper of Izumi-Longo-Popa) for a given outer action of $\mathbb{Z}(p^{\infty})$ on the hyperfinite ${\rm II}_1$ factor $R$ then the subfactor $(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ admits an infinite homogeneous single chain lattice.

Remark: The planar subalgebra of $P(R \subset R \rtimes \mathbb{Z}(p^{\infty}))$ generated by all the biprojections "should be" $TLJ_{p^{1/2}}^{* \infty}$. So for any $\delta$ we should have the example $TLJ_{\delta}^{* \infty}$.
More generally, any infinite free composition of a maximal subfactor should be an example.

  • $\begingroup$ since this subfactor has infinite index, what do you mean by the "planar subalgebra" in the remark? $\endgroup$ Mar 6, 2015 at 16:32
  • $\begingroup$ @DavePenneys: yes it's not a subfactor planar algebra (because does not check the finite dim. axiom), but it's still a general planar algebra and the higher relative commutants are von Neumann algebras. What's the problem? $\endgroup$ Mar 7, 2015 at 3:13

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.