Do we have a counterexample of injective von Neumann algebras $\mathcal{M}$ and $\mathcal{N}$ acting on the Hilbert space $H$ such that $\mathcal{M}\cap \mathcal{N}$ is not injective ?
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1$\begingroup$ If the answer below is satisfactory, please consider marking it as "accepted" so that the question does not stay in the "unanswered" queue. $\endgroup$– Yemon ChoiCommented Oct 17 at 0:25
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Since a von Neumann subalgebra of $B(H)$ is injective if and only if its commutant is injective, it suffices to find two injective von Neumann subalgebras $M_1,M_2\subset B(H)$ such that the von Neumann algebra generated by $M_1$ and $M_2$ is not injective.
This naturally suggests that we take $H=\ell^2(F_2)$ and, writing $a$ and $b$ for generators of $F_2$, define $M_a$ to be the (abelian, hence injective) von Neumann algebra generated inside $B(H)$ by $\lambda_a$; define $M_b$ analogously. Then the von Neumann algebra generated by $M_a$ and $M_b$ is the free group factor $L({\bf F}_2)$.
Just to make the link to your original question explicit: put $N_a=M_a'$ and $N_b=M_b'$. Then $N_a$ and $N_b$ are injective, but their intersection is $L({\bf F}_2)'$; this is the von Neumann algebra generated by the right translations, hence isomorphic to $L({\bf F}_2)$, hence not injective.
