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Let $\mathcal{A} , \mathcal{B} \subset B(H)$ be ${\rm II}_1$-factors such that $\mathcal{A}', \mathcal{B}' $ are also a ${\rm II}_1$-factors.

Question: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \, \, \Rightarrow \, \, \mathcal{A}' \cap \mathcal{B}' $ hyperfinite?
Else, what are counterexamples? (see examples in this post and its answers).

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The answer is no. If $N_1$ and $N_2$ are both finite index subfactors of a nonamenable ${\rm II}_1$ factor $M \subset \mathcal B(L^2M)$ such that $N_1 \cap N_2 = \mathbb C$, then $N_1'$ and $N_2'$ are both finite and $N_1' \cap N_2'$ is nonamenable since it contains $M'$.

For an example of such a situation consider non-trivial finite groups $G$ and $H$, set $\Gamma = G * H$, and $M = *_{\gamma \in \Gamma} \mathbb M_2(\mathbb C)$. Then the action of $\Gamma$ on itself by left multiplication induces a properly outer action on $M$ and we can set $N_1 = M^G$ and $N_2 = M^H$, so that $N_1 \cap N_2 = M^\Gamma = \mathbb C$.

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  • $\begingroup$ Thanks to this paper of Ken Dykema, $\mathbb M_2(\mathbb C)^{\star n} \simeq L(\mathbb{F}_{3n/4})$ for $n>1$, and $M \simeq L(\mathbb{F}_{\infty})$, isn't it? $\endgroup$
    – Sebastien Palcoux
    Aug 4, 2014 at 20:23
  • $\begingroup$ There is something unclear for me: let $G$ be a non-trivial finite group and $N = *_{\gamma \in G} \mathbb M_2(\mathbb C)$, then $N \simeq L(\mathbb{F}_s)$, $N^G = \mathbb{C}$ but $N^G$ is finite index in $N$. Where is the error? $\endgroup$
    – Sebastien Palcoux
    Aug 4, 2014 at 20:42
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    $\begingroup$ @SébastienPalcoux: Yes, $M \cong L(\mathbb F_\infty)$. The reason that $M^\Gamma = \mathbb C$ is not because the action of $\Gamma$ on itself is transitive, but rather because the action has no finite orbits. $\endgroup$
    – Jesse Peterson
    Aug 4, 2014 at 20:58
  • $\begingroup$ I see, thank you! But then, what's the difference with $\bigotimes_{\gamma \in \Gamma} \mathbb M_2(\mathbb C)$ generating the hyperfinite ${\rm II}_1$-factor $\mathcal R$, what's about $\mathcal R^Γ$? $\endgroup$
    – Sebastien Palcoux
    Aug 4, 2014 at 21:53
  • $\begingroup$ The situation is similar, for the shift action $\mathcal R^\Gamma = \mathbb C$ if and only if $| \Gamma | = \infty$. $\endgroup$
    – Jesse Peterson
    Aug 4, 2014 at 22:41

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