# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorDmitri Pavlov
• CommentTimeOct 18th 2021

Created:

## Idea

A measurable field of Hilbert spaces is the exact analogue of a vector bundle over a topological spaces in the setting of bundles of infinite-dimensional Hilbert spaces over measurable spaces.

## Definition

The original definition is due to John von Neumann (Definition 1 in \cite{Neumann}).

We present here a slightly modernized version, which can be found in many modern sources, e.g., Takesaki \cite{Takesaki}.

\begin{definition} Suppose $(X,\Sigma)$ is a measurable space equipped with a σ-finite measure $\mu$, or, less specifically, with a σ-ideal $N$ of negligible subsets so that $(X,\Sigma,N)$ is an enhanced measurable space. A measurable field of Hilbert spaces over $(X,\Sigma,N)$ is a family $H_x$ of Hilbert spaces indexed by points $x\in X$ together with a vector subspace $M$ of the product $P$ of vector spaces $\prod_{x\in X} H_x$. The elements of $M$ are known as measurable sections. The pair $(\{H_x\}_{x\in X},M)$ must satisfy the following conditions. * For any $m\in M$ the function $X\to\mathbf{R}$ ($x\mapsto \|m(x)\|$) is a measurable function on $(X,\Sigma)$. * If for some $p\in P$, the function $X\to\mathbf{C}$ ($x\mapsto\langle p(x),m(x)\rangle$) is a measurable function on $(X,\Sigma)$ for any $m\in M$, then $p\in M$. * There is a countable subset $M'\subset M$ such that for any $x\in X$, the closure of the span of vectors $m(x)$ ($m\in M'$) coincides with $H_x$. \end{definition}

The last condition restrict us to bundles of separable Hilbert spaces. One can also define bundles of nonseparable Hilbert spaces, but this cannot be done simply by dropping the last condition.

## Serre–Swan-type duality

The category of measurable fields of Hilbert spaces on $(X,\Sigma,N)$ is equivalent to the category of W*-modules over the commutative von Neumann algebra $\mathrm{L}^\infty(X,\Sigma,N)$.

(If we work with bundles of separable Hilbert spaces, then W*-modules must be countably generated.)

## References

\bibitem{Neumann} John Von Neumann, On Rings of Operators. Reduction Theory, The Annals of Mathematics 50:2 (1949), 401. doi.

\bibitem{Takesaki} Masamichi Takesaki, Theory of Operator Algebras. I, Springer, 1979.

• CommentRowNumber2.
• CommentAuthorDavidRoberts
• CommentTimeOct 18th 2021

Added whitespace before bullet-point list, to make formatting work

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeOct 18th 2021

How do you reconcile the statements

The last condition restrict us to bundles of separable Hilbert spaces. One can also define bundles of nonseparable Hilbert spaces, but this cannot be done simply by dropping the last condition.

and

(If we work with bundles of separable Hilbert spaces, then W*-modules must be countably generated.)

Isn’t the definition currently given in a way such that the fibres are always separable? I would think the better way to state the Serre-Swan type duality is by default with countably-generated W*-modules, then add a comment about the generalisation to the non-separable case.

• CommentRowNumber4.
• CommentAuthorDmitri Pavlov
• CommentTimeOct 19th 2021

Re #3: Yes, either both fibers and modules are countably generated, or not.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeOct 19th 2021

Reworked the Serre–Swan material to say

The category of measurable fields of Hilbert spaces on $(X,\Sigma,N)$ (as defined above) is equivalent to the category of countably-generated W*-modules over the commutative von Neumann algebra $\mathrm{L}^\infty(X,\Sigma,N)$.

(If we work with bundles of general, possibly nonseparable Hilbert spaces, then the W*-modules do not need to be countably generated.)