Let $H$ be an $\infty$-dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.  

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


*Examples*:  

(1) Take $\mathcal{B} = \mathcal{A}'$ then $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$ by definition of a factor.  

(2) Take $(\mathcal{A}' \subset \mathcal{B})$ an irreducible subfactor, then $\mathcal{A} \cap \mathcal{B} = \mathbb{C}$ by definition of irreducibility.


Obviously $\langle \mathcal{A}' , \mathcal{B}' \rangle = \mathbb{C}' = B(H)$, with the *notation* $\langle S \rangle := (S \cup S^\* \cup \mathbb{C}) ''$. 


**Question**: Is it also true that $\langle \mathcal{A} , \mathcal{B} \rangle = B(H)$, or equivalently, that $\mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$ ?  

Else, what are counterexamples? 


*Remark*: It's true for the examples (1) and (2).



${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?
No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II\_1$ factors and their intersection is still trivial, but their commutants $M\_2(\mathcal{A}')$ and $M\_2(\mathcal{B}')$ now intersect in the scalar matrices. *The answer is now in its true, correct, and final form. I will notify the Cyber Police of any additional unauthorized edits.*


