Theorem: $\mathcal{M}$ is of type ${\rm III}_{1}$.
Proof: By Corollaire 3.3.4 below, we need to show that the ratio set of the action $\alpha$ is $[0, \infty)$. By the classification below, it is enough to prove that the interval $[0,2)$ is included in the ratio set.
Consider $r_{\theta}$ and $s$ on $[0,1)$ directly, i.e., $r_{\theta}(x) = \lfloor x+\theta \rfloor$ and $s(x) = x^2$.
Let $r \in [0,2)$, $\epsilon > 0$ and $A \subseteq [0,1)$ be a Borel set with $\mu(A)>0$.
Let $x_0 \in A$ such that for any neighborhood $V$ of $x_0$ in $[0,1)$ we have $\mu(V \cap A)>0$.
Consider integers $n,m$ and let $f:=r^n_{\theta} \circ s \circ r^m_{\theta}$.
Then $f(x) = \lfloor \lfloor x + m\theta \rfloor^2 + n\theta \rfloor$ and $f'(x) = 2\lfloor x + m\theta \rfloor$. Note that $f$ depends on the integers $n$ and $m$, whereas $f'$ depends on $m$ only (and is well-defined up to a finite set).
By ergodicity, we can choose $m$ such that $|2r^m_{\theta}(x_0)-r|<\epsilon$, so that for a sufficiently small neighborhood $V$ of $x_0$, $|f'(x)-r|<\epsilon$ for all $x \in V$. Let $C:=V \cap A$ and $D:=s \circ r^m_{\theta}(C)$. Then $\mu(C)>0$ and $\mu(D)>0$, and by definition of ergodicity, we can choose $n$ such that $\mu(C \cap r^n_{\theta}(D))>0$. Thus, for $B= f^{-1}(C) \cap C = f^{-1}(C \cap r^n_{\theta}(D))$ we have:
- $\mu(B)>0$,
- $B \cup f(B) \subset A$,
- $|f'(x)-r|<\epsilon$, for all $x \in B$.
The result follows. $\square$
Conclusion: this action of $\mathbb{F}_2$ on $\mathbb{S}^1$ generates a non-hyperfinite ${\rm III}_1$ factor $\mathcal{M}$.
Let $\Omega$ be a cyclic-separating vector (i.e. $\mathcal{M}\Omega$ and $\mathcal{M}'\Omega$ dense in $H$) and $\sigma=(\sigma_t^{\Omega})$ be the one-parameter modular map. Then $\mathcal{M} \rtimes_{\sigma} \mathbb{R} $ is isomorphic to $\mathcal{N} \otimes B(H)$, with $\mathcal{N}$ a ${\rm II}_1$ factor (independent up to iso. of the choice of $\Omega$) of fundamental group $\mathbb{R}_{+}^*$.
Question: What is $\mathcal{N}$? Is it isomorphic to $\mathcal{L}(\mathbb{F}_2)$?
If so, it would solve the free group factors isomorphism problem, according to this paper of Florin Rădulescu.
Remark: If we replace $s:x \mapsto x^2$ by $s_n: x \mapsto x^n$ with $n \ge 2$, we generate a von Neumann algebra $\mathcal{M}_n$ which is also a non-hyperfinite ${\rm III}_1$ factor (the proof is similar). Are they isomorphic?
It should work as well if we replace $s$ by any bijection of $[0,1)$ polynomial of degree $\ge 2$.
An extract of the following paper of Alain Connes (page 88):
Connes, A. Structure theory for Type III factors. Proceedings of the International Congress of Mathematicians (Vancouver, B. C., 1974), Vol. 2, pp. 87--91. Canad. Math. Congress, Montreal, Que., 1975.
An extract of the following paper of Alain Connes (page 195):
Connes, Alain. Une classification des facteurs de type ${\rm III}$. (French) Ann. Sci. École Norm. Sup. (4) 6 (1973), 133--252.
