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I read a paper of Alain Connes on "Duality between shapes and spectra" and in page 4, he says

Due to a theorem of von Neumann the algebra of multiplication by all measurable bounded functions acts in Hilbert space in a unique manner, independent of the geometry one starts with.

Question.

  • What is a precise statement and reference for this "mysterious" theorem of von Neumann ?
  • What are nontechnical explanations and justifications of this phenomenon ?
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One possible interpretation of Connes's statement is that up to an isomorphism, there is a unique faithful indecomposable representation of any commutative von Neumann algebra on a Hilbert space.

Indeed, the category of von Neumann algebras is contravariantly equivalent to the category of compact strictly localizable enhanced measurable spaces.

After extracting an enhanced measurable space $(X,M,N)$ from a commutative von Neumann algebra $A$ in this manner, elements of $A$ can be identified with equivalence classes of bounded measurable functions on $X$ modulo equality almost everywhere.

We can now easily describe isomorphism classes of representations of $A$ on a Hilbert space. Such an isomorphism class is specified by partitioning $X$ into almost disjoint (up to a negligible set) nonnegligible measurable subsets $\{X_i\}_{i∈I}$, and assigning a distinct cardinal number $a_i$ to each element of the partition. The corresponding Hilbert space is $$\bigoplus_{i∈I} {\rm L}^2(X_i,M_{X_i},N_{X_i})⊗{\bf C}^{a_i}$$ and $A$ acts on each summand by restricting the corresponding bounded measurable function on $X$ to $X_i$ and then acting via multiplication on the corresponding ${\rm L}^2$-space. Here ${\bf C}^{a_i}$ denotes any complex Hilbert space of dimension $a_i$.

Such a representation is faithful if $a_i≥1$ for all $i$. It is indecomposable if $a_i≤1$ for all $i$. Thus, a faithful indecomposable representation must have $a_i=1$ for all $i$, and there is a unique such a representation, namely ${\rm L}^2(X,M,N)$, also known as the Haagerup standard from of $A$.

The point of all this is that although the (abstract) commutative von Neumann algebra appears to know nothing about an enhanced measurable space or its ${\rm L}^2$-space, all this data can be reconstructed in a unique manner, i.e., it is unique up to a unique isomorphism.

If, furthermore, we know the Dirac operator, we can proceed to refine $X$ to a smooth manifold, as described by Connes.

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  • $\begingroup$ Thanks for the detailed answer. $\endgroup$ – dohmatob Sep 10 at 2:49

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