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](http://en.wikipedia.org/wiki/Von_Neumann_algebra#Amenable_von_Neumann_algebras)?  

 Else, what are counterexamples? (see examples in [this post](https://mathoverflow.net/q/177597/34538) *and* its answers). 



${\rm II}_1$-factors with finite commutant: $\mathcal{A} \cap \mathcal{B} = \mathbb{C} \Rightarrow \mathcal{A}' \cap \mathcal{B}'$ hyperfinite?
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 $N\_1 \cap N\_2 = M^\Gamma = \mathbb C$.


