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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.

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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.

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