# Is $vN(M_1,M_2) \cap M_3= vN(M_1,M3) \cap vN(M_2,M3)$?

Let $M_1,M_2,M_3$ be von Neumann algebras (i.e. weakly closed subalgebras of $B(H)$ where $H$ is a Hilbert space).

Let $vN(M_1,M_2)$ denote the von Neumann algebra generated by $M_1$ and $M_2$ inside $B(H)$.

Is it true that $vN(M_1,M_2) \cap M_3= vN(M_1,M3) \cap vN(M_2,M3)$?

• @JonasMeyer: Note that $vN(M_i,M_3),i=1,2$ is contained in LHS, as $M_i,i=1,2,3$ is contained in LHS. Since LHS and RHS are von Neumann algebras, so the RHS is trivially is contained in LHS. Isn't this argument correct? – voldemort Aug 26 '14 at 4:12
• @JonasMeyer: Thanks. I realize my error. – voldemort Aug 26 '14 at 4:16
• OK, I deleted the previous comments. Note that LHS$\subseteq M_3\subseteq$ RHS, and either containment can be strict. – Jonas Meyer Aug 26 '14 at 4:19
• @JonasMeyer: Thanks:). Please let this comment stay- as it is certainly useful for me. – voldemort Aug 26 '14 at 4:20

Let $M_1=M_2=M_2(\mathbb C)$, and $M_3=\mathbb C\,I_2$. Then $$vN(M_1,M_2)\cap M_3=\mathbb C\,I_2,$$ $$vN(M_1,M_3)\cap vN(M_2,M_3)=M_1\cap M_2=M_2(\mathbb C).$$