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

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  • $\begingroup$ @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? $\endgroup$
    – voldemort
    Aug 26, 2014 at 4:12
  • $\begingroup$ @JonasMeyer: Thanks. I realize my error. $\endgroup$
    – voldemort
    Aug 26, 2014 at 4:16
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    $\begingroup$ OK, I deleted the previous comments. Note that LHS$\subseteq M_3\subseteq$ RHS, and either containment can be strict. $\endgroup$ Aug 26, 2014 at 4:19
  • $\begingroup$ @JonasMeyer: Thanks:). Please let this comment stay- as it is certainly useful for me. $\endgroup$
    – voldemort
    Aug 26, 2014 at 4:20

1 Answer 1

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

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