# Examples of isomorphic W* algebra with non-homeomorphic weak topology

Due to the uniqueness of the predual, a W* algebra, when realized as a von Neumann algebra in any way, always has a unique, well-defined ultraweak (or $$\sigma$$-weak) topology. The same can be said about the ultrastrong (or $$\sigma$$-strong) topology and ultrastrong-$$\ast$$ (or $$\sigma$$-strong-$$\ast$$) topology. In his book The Theory of Operator Algebras, vol I, in the remark after Corollary 3.11, Takesaki claims that

...the weak topology, the strong topology, and the strong$$\ast$$ topology on a $$W^\ast$$-algebra do not make sense unless we specify on which Hilbert space the algebra acts.

My question is a concrete instance of the more general question Can we recover a Von Neumann algebra from its predual?(perhaps I should add that what Dmitri Pavlov called a von Neumann algebra in that question is what I call a $$W^\ast$$-algebra here, and for me, a von Neumann algebra is an involutive subalgebra of $$\mathcal{B}(H)$$ which coincides with its bicommutant, thus depends on the $$C^\ast$$ algebra $$\mathcal{B}(H)$$, and in particular on the Hilbert space $$H$$). More precisely, I ask to construct explicit examples illuminating the remark of Takesaki quoted above.

Since the weak (resp. strong, strong-$$\ast$$) topology coincides with the ultraweak (resp. ultrastrong, ultrastrong-$$\ast$$) toplogy on bounded parts, the construction of such an example is not trivial.

It would seem that the question is more trivial than I expected. We can just take an infinite dimensional von Neumann algebra $$M$$ acting on a Hilbert space $$H$$, then consider the diagonal map $$\Delta: \mathcal{B}(H) \to \mathcal{B}(H^\infty)$$, where $$H^\infty$$ is the direct sum of an infinite copies of $$H$$. Then obviously, the weak topology on $$(M, H)$$ is different from that of $$(\Delta(M), H^\infty)$$, since the dual space (the space of weakly continuous functionals) for the latter is strictly larger.