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?


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.

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

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