I am studying von Neumann algebras. In the wiki article abelian von Neumann algebras, it mentions that every abelian von Neumann algebras acting on a separable Hilbert space is *-isomorphic to $L^{\infty}[0,1]$, $l^{\infty}(\mathbb{N})$ or their direct sum. I am wondering why $[0,1]$ is special in this case, why not other measure space such as $S^{1}$ or $[0,1]^{2}$? Also, if we consider $L^{\infty}([0,1]^{2})$ acting on $L^{2}([0,1]^{2})$, isn't that a von Neumann algebra acting on a separable Hilbert space? Does that mean that $L^{\infty}[0,1]$ and $L^{\infty}([0,1]^{2})$ are *-isomorphic?
Questions on Maharam’s Classification Theorem – Functional Analysis and Operator Algebras
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Best Answer
The spaces $[0,1]$, $[0,1]^2$, and $S^1$ are all isomorphic as measurable spaces, including their sets of measure 0, as required by the Gelfand-type duality for measurable spaces.
For instance, the isomorphism $[0,1]→S^1$ is given by identifying $S^1=[0,1]/(0∼1)$. Since the points $0$ and $1$ have measure 0, and maps that differ on a set of measure 0 are identified, the above map is indeed an isomorphism.
Likewise, the isomorphism $[0,1]^2→[0,1]$ is given by identifying $[0,1]$ with $\{0,1\}^N$ (e.g., using binary expansions), where $N$ is a countable set, and then taking the isomorphism $[0,1]^2≅\{0,1\}^{N⊔N}≅\{0,1\}^N≅[0,1]$, where $N⊔N$ is isomorphic to $N$ since both are countable sets.
Indeed, the very point of Maharam's theorem is that there are very few measurable spaces up to an isomorphism: any measurable space is isomorphic to the disjoint union of measurable spaces of the form $\{0,1\}^S$, where $S$ is a set of arbitrary cardinality, and this gives a complete classification of measurable spaces up to isomorphism: simply count the (infinite) number of summands for each possible cardinality of $S$, and also count the number of isolated points.