# Abelian von Neumann Algebras on non-separable Hilbert spaces

Is there a classification of Abelian von Neumann algebras on non-separable Hilbert spaces? For a classification of Abelian von Neumann algebras on separable Hilbert spaces, see this link.

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According to this link, for any Hilbert space is isometrically isomorphic to $l^2(I)$ where $I$ is a set having cardinality equal to the dimension of $H$... – Andy Teich Oct 31 '12 at 13:48

## 1 Answer

Every commutative von Neumann algebra $A$ is $^*$-isomorphic to $L_\infty(\Omega,\mu)$ for some compact Hausdorff space and positive measure $\mu$.

Idea of the proof is the following. Decompose $$A\cong\coprod\limits_{\nu\in\Lambda} A_\nu$$ into the coproduct of commutative von Neumann algebras with cyclic vector. For every commutative von Neumann algebra $A_\nu$ with cyclic vector there exist compact Hausdorff space $\Omega_\nu$ and regular bounded Borel measure $\mu_\nu$, such that $A_\nu$ is $^*$-isomorphic to $L_\infty(\Omega_\nu,\mu_\nu)$. Then one can show that $$A\cong L_\infty\left(\coprod\limits_{\nu\in\Lambda} \Omega_\nu,\coprod\limits_{\nu\in\Lambda} \mu_\nu\right)$$

See Theory of operator algebras Part I by M. Takesaki, chapter 3, theorem 1.18.

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Ok, this means that classification of abelian von Neumann algebras reduces to classification of compact Hausdorff spaces, right? But for the latter, there is not really a classification...you can't say that there exist finitely many compact Hausdorff spaces such that any compact Hausdorff space is homeomorphic to one of these...? – Andy Teich Oct 31 '12 at 15:18
You don't need to classify all Hausdorff spaces, just hyperstonean ones. But anyway it is too big class of topological spaces, to be well classified. – Norbert Oct 31 '12 at 15:31