A non-hyperfinite type III factor from an action of the free group on the circle We define below a von Neumann algebra $\mathcal{M}$ from an action of the free group on the circle, and we prove that $\mathcal{M}$ is a non-hyperfinite type ${\rm III}$ factor.  

Question : Is $\mathcal{M}$ of type ${\rm III}_{0}$, ${\rm III}_{\lambda}$ or ${\rm III}_{1}$ ?


Definition : Let  $s, r_{\theta}:  \mathbb{R} / \mathbb{Z} \to  \mathbb{R} / \mathbb{Z} $, defined by $s( x) = x^{2}$ (choosing representatives in $[0,1[$) and $r_{\theta} (x) = x+\theta$. 
Now,  identifying $  \mathbb{R} / \mathbb{Z}$ and $\mathbb{S}^{1}$, we define the action $\alpha$  of  $\mathbb{F}_{2}  = \langle a, b \vert \  \rangle$,  generated by $\alpha (a) = s$ and $\alpha (b) = r_{\theta}$ in Homeo($\mathbb{S}^{1}$).  
Lemma: If $\theta$ is transcendental, the action $\alpha$ is faithful.
Proof: A relation $s^{n_{1}}r_{\theta}^{m_{1}}...s^{n_{k}}r_{\theta}^{m_{k}} = e $ can be translated into an algebraic equation in $x$ and $\theta$, where $\theta$ has to be a root $\forall x$. Then, if $\theta$ is transcendental, we are sure that there is no relation. $\square$  
Remark: For a fixed transcendental $\theta$, each non-trivial relations  can be realized for at most finitely many  $x \in \mathbb{R} / \mathbb{Z}$, i.e. roots of the related algebraic equation.  
Theorem:  $\mathcal{M} = L^{\infty}(\mathbb{S}^{1}, Leb) \rtimes_{\alpha} \mathbb{F}_{2}  $ is a non-hyperfinite type III factor.
Proof :  The action $\alpha$ of $\mathbb{F}_{2}$ on $\mathbb{S}^{1}$ is: 


*

*(a) Measure class preserving:  the set of null measure subspaces is invariant.

*(b)  Essentially free:  a fixed point set for $\gamma \ne e$ is at most finite, so with null measure.

*(c) Properly ergodic:  ergodicity comes from irrational rotation, next, every $\mathbb{F}_{2}$-orbit have null measure.

*(d) Non-amenable (Edit, Aug. 2018):  for any $\eta > 0$, there is $n \in \mathbb{Z}_{>0}$ such that $\lfloor n\theta \rfloor < \eta$. Now, for $\lambda=Leb$ and $g = r_{\theta}^n$, $\partial (\lambda g)/\partial \lambda = 1$ because $\lambda$ is $g$-invariant. It follows that the action $\alpha$ is indiscrete, and then by the proposition below, it is non-amenable.  

*(e)  Non equivalent measure preserving: by ergodicity, an equivalent invariant measure $m$ is proportional to $Leb$. Then $m([1/4 , 1/2]) = 2m([1/16 , 1/4])$, and by $\alpha(a)$ invariance, $m([1/4 , 1/2]) = m([1/16 , 1/4])$. In fact, the only invariant measure are $0$ or $\infty$.
(a), (b), (c) give a factor, (d) gives non-hyperfinite, (e) gives a type ${\rm III}$.  $\square$   

Here are two extracts of the following recent paper (April 2018):   
Bartholdi, Laurent. Amenability of groups and $G$-sets. Sequences, groups, and number theory, 433--544, Trends Math., Birkhäuser/Springer, Cham, 2018. 

 A: 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. $\square$  
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.


Conclusion: this action $\alpha$ 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 group. Then $\widetilde{\mathcal{M}}:=\mathcal{M} \rtimes_{\sigma} \mathbb{R} $ is a ${\rm II}_{\infty}$ factor called the core of $\mathcal{M}$.
It is independent (up to isomorphism) of the choice of $\Omega$, and is isomorphic to $\mathcal{N} \otimes B(H)$, with $\mathcal{N}$ a non-hyperfinite ${\rm II}_1$ factor of fundamental group $\mathbb{R}_{+}^*$.
By Theorem 2.23 below, $$\widetilde{\mathcal{M}} = L^{\infty}(\mathbb{S}^1 \times \mathbb{R}_{+}^*, Leb) \rtimes_{\widetilde{\alpha}} \mathbb{F}_2$$ where the action $\widetilde{\alpha}$ of $\mathbb{F}_2$ on $\mathbb{S}^1 \times \mathbb{R}_{+}^*$ is given by $$\gamma(x,\lambda) = (\gamma(x), [\gamma'(x)]^{-1}\lambda),$$
with $x \in \mathbb{S}^1$,  $\lambda \in \mathbb{R}_{+}^*$ and $\gamma \in \mathbb{F}_2$ identified with $\alpha(\gamma)$ or $\widetilde{\alpha}(\gamma)$ when appropriate.
Remarks and questions:
Note that $\widetilde{\mathcal{M}}$ has a Cartan subalgebra. Can we deduce that $\mathcal{N}$ has also a Cartan subalgebra?
 If so, $\mathcal{N}$ cannot be isomorphic to $L(\mathbb{F}_2)$, because this last has no Cartan subalgebra, by this paper of Dan Voiculescu. Anyway, can we deduce that $L(\mathbb{F}_2)$ has also for fundamental group $\mathbb{R}_{+}^*$?
If so, it would solve the free group factors isomorphism problem, according to this paper of Florin Rădulescu.  

Extracts of the following book of Masamichi Takesaki (pages 28 and 17):   
Takesaki, M. Theory of operator algebras. III. Encyclopaedia of Mathematical Sciences, 127. Operator Algebras and Non-commutative Geometry, 8. Springer-Verlag, Berlin, 2003. xxii+548 pp. 
The result below provides the core $\widetilde{\mathcal{R}}$ of the crossed-product $\mathcal{R}$.  

From Corollary 2.5 below, $\delta(x,sx) = \delta(sx,x)^{-1}$, whereas Lemma 2.4 provides $\delta(sx,x)$.  
 
