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 III factor.

Question: Is $\mathcal{M}$ of type III$_{0}$, III$_{\lambda}$ or 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 $\theta$, which $\theta$ 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)
*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*: by Connes-Feldman-Weiss, if such an action is amenable, there exist a transformation $T$ of $\mathbb{S}^{1}$, such that $\forall x \in \mathbb{S}^{1}$ up to a null set, $\mathbb{F}_{2}.x = T^{\mathbb{Z}}.x$. Then, it exists $n \in \mathbb{Z}$ and $\gamma \in \mathbb{F}_{2}$, such that $a.x = T^{n}.x$ and $T.x = \gamma.x$. So, $a.x = \gamma^{n}.x$ and $x$ is in the null set of algebraics with $\theta$.

[*This argument is incorrect and needs to be completed*(see comments below)]. (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 III. $\square$