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Hi, I came across the statement below and I couldn't figure out why it is true. I was hoping someone could explain it or give me a good reference. Thank you in advance. "Let $\pi$ be a non-degenerate representation of a separable C*-algebra $A$ on a separable Hilbert space $H$. Suppose $M$ is an abelian von Neumann algebra contained in the commutant of $\pi(A)$. Then there exists a unique sequence of mutually orthogonal projections $(e_{n})$ in $M$ such that $\sum e_{n}=I$ and $M'e_{n}$ is spatially isomorphic to $L^{\infty}(E_{n})\otimes M_{n}(C)$. Moreover, each $e_{n}$ is invariant under spatial automorphisms of $M$.

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2 Answers 2

Apparently, the data of π and A is irrelevant here. We might as well start with an arbitrary representation of an abelian von Neumann algebra M on a Hilbert space H.

Geometrically, commutative von Neumann algebras are just measurable spaces (the opposite category of the category of commutative von Neumann algebras is equivalent to the category of (localizable) measurable spaces). Likewise, representations of von Neumann algebras on Hilbert spaces can be thought of geometrically as bundles of Hilbert spaces over measurable spaces, i.e., there is a similar equivalence of categories here. Spatial automorphisms of representations of von Neumann algebras are precisely automorphisms f of the underlying measurable space together with an isomorphism f*V→V, where V is the bundle of Hilbert spaces corresponding to the representation.

More generally, a morphism of von Neumann algebras M→N, where M is abelian, can be thought of as a bundle of von Neumann algebras over the spectrum (i.e., the corresponding measurable space) of M. Again, this can be stated as an equivalence of categories. In our case we have a morphism M→M' and M is abelian. The corresponding bundle of von Neumann algebras is precisely the bundle of (fiberwise) endomorphisms of the bundle V of Hilbert spaces corresponding to the representation of M on H.

Denote by E_n the measurable subset of Spec M consisting of points with the dimension of the fiber of V being equal to n. Denote by e_n the corresponding projection in M. Observe that E_n are disjoint and ⋃_n E_n=Spec M, i.e., e_n are mutually orthogonal and ∑_n e_n=1. Moreover, M'e_n is the algebra of endomorphisms of the restriction of V to E_n. The latter bundle is trivial and has dimension n, hence its algebra of endormophisms is just L^∞(E_n)⊗M_n(C) as required. From the geometric interpretation of spatial isomorphisms it follows that every e_n is invariant under any spatial isomorphism.

Finally, the sequence e is unique because the property M'f_n=L^∞(E_n)⊗M_n(C) forces every F_n to be a subset of E_n, and because ⋃_n F_n=Spec M we have F_n=E_n.

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It's a standard result in von Neumann algebras, called the homogeneous decomposition of a type I von Neumann algebra.

To quote results from the literature, I will use Takesaki's book (volume I). Since your $M$ is an abelian von Neumann algebra, it is type I. Then it's commutant $M'$ is also type I (V.1.30). Then Theorem V.1.27 expresses exactly the result you are asking about.

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