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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. $\square$


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I'm pretty sure that it's type $III_1$, because the following set is dense in $\mathbb R_{>0}$: $\{f'(x) : x\in \mathbb S^1, f\in F_2, f(x)=x\}$. –  André Henriques Sep 8 '13 at 15:09
    
@AndréHenriques : Thank you for your comment. Is it a (sketch of) proof ? Is there a theorem characterizing the $III_{1}$ factors in such a way ? –  Sébastien Palcoux Sep 8 '13 at 15:16
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I don't follow your logic in part (d), why does $a.x = \gamma^n.x$? –  Jesse Peterson Sep 8 '13 at 16:23
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I don't know the literature, so I can't point to a reference. But here's how things go: given an (let's say a.e. smooth) action of a group $\Gamma$ on a manifold $M$, you can form the bundle of densities $\Omega^{top}_{>0}M$, which is a principal bundle with structure group $\mathbb R_{>0}$. The action of $\Gamma$ on $M$ induces an action on $\Omega^{top}_{>0}M$, and the vN algebra $L^\infty(M)\rtimes \Gamma$ is a type $III_1$ factor iff the action of $\Gamma$ on $\Omega^{top}_{>0}M$ is ergodic. If that action is not ergodic, the vN algebra $L^\infty(\Omega^{top}_{\>0}M)^\Gamma$... –  André Henriques Sep 8 '13 at 16:58
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... is equipped with an action of $\mathbb R_{>0}$ (coming from the action on $\Omega^{top}_{>0}(M)$). This corresponds to a action of $\mathbb R_{>0}$ on some measure space $X$. If that action is transitive, it is equivalent to $\mathbb R_{>0}$ acting on $\mathbb R_{>0}/\mathbb Z^\lambda$ for some $\lambda\in(0,1)$, and the factor $L^\infty(M)\rtimes\Gamma$ is of type $III_\lambda$. Otherwise, $L^\infty(M)\rtimes\Gamma$ is of type $III_0$. –  André Henriques Sep 8 '13 at 17:05
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