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Proposition 1.16.2 of Sakai's "$\mathcal{C}^\ast$-Algebras and $\mathcal{W}^\ast$-Algebras" states

Let $\phi$ be a weakly-$\ast$ continuous $\ast$-homomorphism of a $\mathcal{W}^*$-algebra $\mathcal{M}$ (i.e. a $\mathcal{C}^*$-algebra that is isometrically isomorphic to the topological dual of a Banach space) into another $\mathcal{W}^*$-algebra $\mathcal{N}$. Then the image $\phi(\mathcal{M})$ is closed with respect to the weak-* topology on $\mathcal{N}$.

However, consider the following: Let $\mathcal{M}$ be a $\mathcal{W}^*$-algebra, and let $(\pi,\mathcal{H})$ be its universal representation. Then $\pi: \mathcal{M} \to \mathcal{B}(\mathcal{H})$ is an injective $\ast$-homomorphism. In particular, $\pi(\mathcal{M})$ is also a $\mathcal{W}^*$-algebra. On the other hand, all $\ast$-isomorphisms between $\mathcal{W}^*$-algebras are normal (i.e. suprema-preserving) and therefore weakly-$\ast$ continuous by $\mathcal{W}^\ast$-theory. Using the above proposition, this would imply that $\pi(\mathcal{M})$ is already weakly-$\ast$ closed, i.e. it coincides with the enveloping von Neumann algebra. However, this clearly contradicts the Sherman–Takeda theorem, which states that the enveloping von Neumann algebra can identified with the double dual of $\mathcal{M}$, where the latter is strictly larger than $\mathcal{M}$ if $\mathcal{M}$ has infinite-dimension. Where is the mistake in my reasoning?

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The mistake in the reasoning is that the normality occurs on the image and not in the codomain. So you cannot conclude that it is continuous with the ambient weak$^*$-topology.

For example, consider a non-type I von Neumann algebra $R$, say the hyperfinite II$_1$ for instance. As any other C$^*$-algebra, $R$ has an irreducible representation $\rho$. Because $R$ is simple, $\rho$ is faithful. So $\rho(R)\subset B(H_\rho)$ is weak$^*$-dense, which implies that $\rho(R)$ is not closed.

Precisely because of this is that people in the past used to make the distinction between a $W^*$-algebra (as you defined), which is abstract, and a von Neumann algebra. These days (almost?) no one cares because if you start with a $W^*$-algebra you take the normal universal representation (i.e., the sum of the GNS of the normal states) and you get a bona-fide von Neumann algebra.

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  • $\begingroup$ Glad I could help :) $\endgroup$ – Martin Argerami Nov 24 '17 at 4:05

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