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Consider a von Neumann algebra $A_0$ and an injective *-isomorphism $\pi : A_0 \to B(H)$.

Then we have a *-subalgebra $\pi(A) \subset B(H)$, which is abstractly *-isomorphic to the von Neumann algebra $A_0$, but which might not be a "von Neumann subalgebra" of $B(H)$, i.e. we are not guaranteed $\pi(A)''=\pi(A)$. (E.g. this answer.)

If $A_0$ is injective, can we conclude $\pi(A)''$ injective? If it helps, I'm interested in the case $A_0$ is the hyperfinite $II_1$ factor.

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No, it isn't. An irreducible representation of a II$_1$ factor represents faithfully as a dense sot-subalgebra of a non-separable $B(H)$, which then cannot be injective.

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