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):

essentially freeif $\lambda$-almost every point has a trivial stabilizer, namely $\lambda(\{ x \in X \ | \ G_x \neq 1 \}) = 0$. $\endgroup$1more comment