# ${\rm II}_1$-factors with finite commutant and trivial intersection generate $B(H)$?

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).

• @ougao: I don't understand where you use $N$ for building $\mathcal{A}$ and $\mathcal{B}$, perhaps there is a typo. – Sebastien Palcoux Jul 30 '14 at 21:04

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.

• I rolled back Sebastien's edit, which seemed to change completely the meaning of what Nik Weaver wrote – Yemon Choi Aug 1 '14 at 17:54
• However I agree with part of Sebastien's edit, namely that you probably mean M2 of A', etc – Yemon Choi Aug 1 '14 at 17:56
• @SébastienPalcoux you changed "now intersect" to "don't intersect". That is not fixing a typo. – Yemon Choi Aug 1 '14 at 18:17
• Sorry, I pressed enter before I was finished ... I was saying, now replace with $\mathcal{A}\otimes\mathcal{M}$ and $\mathcal{B}\otimes\mathcal{N}$ where $\mathcal{M}$ and $\mathcal{N}$ are $II_1$ factors with trivial intersection whose commutants have nontrivial intersection. – Nik Weaver Aug 1 '14 at 18:38
• Honestly? I think you should spend a couple of days working on it yourself before opening a new post. Otherwise it feels like you are abusing the community's goodwill to have people do your thinking for you. Just my opinion. – Nik Weaver Aug 1 '14 at 19:28

Transformation of the answer of Nik, for an hyperfinite infinite dimensional intersection:

He gave the counterexample $(\mathcal{A} \otimes I_2)' \cap (\mathcal{B} \otimes I_2)' \simeq M_2(\mathbb{C})$, for $\mathcal{A}$, $\mathcal{A'}$, $\mathcal{B}$, $\mathcal{B'}$ ${\rm II}_1$-factors and $\mathcal{A} \cap \mathcal{B} = \mathcal{A'} \cap \mathcal{B'} = \mathbb{C}$.
Let two states on $(\mathcal{A} \otimes I_2)^{\otimes \infty}$ and $(\mathcal{B} \otimes I_2)^{\otimes \infty}$ generating ${\rm II}_1$-factors $\mathcal{M}=\overline{(\mathcal{A} \otimes I_2)^{\otimes \infty}}$ and $\mathcal{N}=\overline{(\mathcal{B} \otimes I_2)^{\otimes \infty}}$ such that $\mathcal{M'}=\overline{(\mathcal{A'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ and $\mathcal{N'}=\overline{(\mathcal{B'} \otimes M_2(\mathbb{C}))^{\otimes \infty}}$ are ${\rm II}_1$ factors, $\mathcal{M} \cap \mathcal{N} = \mathbb{C}$ and $\mathcal{M'} \cap \mathcal{N'} \simeq \overline{M_2(\mathbb{C})^{\otimes \infty}} \simeq \mathcal{R}$ the hyperfinite ${\rm II}_1$-factor.
For more details about infinite tensor product of von Neumann algebras, see this answer of Nik.

Generalized question (posted here):
Let $\mathcal{A}_1 \dots \mathcal{A}_n \subset B(H)$ be ${\rm II}_1$-factors such that $\forall i \, \, \mathcal{A}_i'$ is also a ${\rm II}_1$-factor and $\bigcap_i \mathcal{A}_i = \mathbb{C}$.
Is it true that $\bigcap_i \mathcal{A}'_i$ is hyperfinite? Else, what are counterexamples?

• Seems like some justification is needed, but I think this works. – Nik Weaver Aug 3 '14 at 18:18