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Let $X$ be a measure space, or even a subspace of $\mathbb{R}^n$, and suppose I have a family of finite-dimensional vector spaces $\{V_x\}_{x\in X}$ indexed by $X$. Is there any way to "integrate" this family over $X$ (subject to niceness conditions on it) to obtain some sort of "vector-space-like object" $\int V_x \,dx$ possessing a "dimension" (in general not an integer), such that

$$\mathrm{dim} \left(\int V_x \,dx\right) = \int (\mathrm{dim}\, V_x)\,dx \;?$$

Bonus points if there is some category-theoretic way to view $\int V_x\,dx$ as a "coproduct" of the family $\{V_x\}_{x\in X}$.

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    $\begingroup$ Is there a uniform bound on $\dim(V_x)$? $\endgroup$
    – Yemon Choi
    Commented Sep 25, 2017 at 9:12
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    $\begingroup$ Have you tried the literature on measurable fields of Hilbert spaces (as used in e.g. representation theory of various groups)? I am not sure if it is quite what you are looking for $\endgroup$
    – Yemon Choi
    Commented Sep 25, 2017 at 9:19
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    $\begingroup$ As Yemon says, direct integrals provide one way to do something like this, but they require your $V_x$ to come equipped with an inner product. Is that the case? $\endgroup$
    – Tobias Fritz
    Commented Sep 25, 2017 at 9:51
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    $\begingroup$ Assuming $X$ is endowed with a finite measure and the spaces $V_x$ vary measurably, one can consider the family as a module over the finite von-Neumann algebra $L^\infty(X)$. Then the dimension function you seek is the so called von-Neumann dimension of this module. $\endgroup$
    – Uri Bader
    Commented Sep 25, 2017 at 13:07
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    $\begingroup$ Bob Paré once had a student named Mike Wendt whose thesis contains a wealth of information about such things from a topos theoretic point of view. He is primarily interested in the case of measurable fields of Hilbert spaces, where everything is messed up by the fact that the ell_2 sum does not have a universal property, but there might be something about measurable fields of vector spaces too. (I don't remember.) $\endgroup$
    – Jeff Egger
    Commented Sep 25, 2017 at 16:48

1 Answer 1

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I'll be brief and (happily) add more details on demand (Edit: Some more details were added).

Some Philosophy

Slogan: You can do math fibered over a measured space.

Most of us are already used to the idea of doing algebraic geometry over schemes and topology over topological spaces, but are less familiar with doing math over measured spaces. Yet, this concept has a long history. Maybe its first appearance is in the notion of a bundle of Hilbert spaces over a measured space aka as direct integral of Hilbert spaces. Also in the theory of von-Neumann algebras one decomposes a general algebra into a direct integral of factors (similarly to the way in which an Azumaya algebra is decomposed over its center). I find Furstenberg's pov on Ergodic Theory parallel to Grothendieck's pov on Algebraic Geometry in the way spaces are treated relative to a base space, only that Ergodic Theory is somehow more generous in allowing further constructions, due to the flexibility of measurable functions.

In recent decades Zimmer developed the theory of convex compact spaces, Gaboriau developed the theory of simplicial complexes, Sauer developed the theory of manifolds, all over a base measured space. This pov is quite common nowadays in Ergodic Theory and there are many more examples.

I should probably mention that in all of the above examples, theories were developed for an external sake. Maybe it is about time for approaching these theories as a whole and develop a master theory. I don't know.

Vector spaces over $X$

Given a measured spaces $X$ (that is, a standard Borel space endowed with a measure class), a (complex) vector space over $X$ is a Borel space $V$ endowed with a Borel map $\pi:V\to X$ such that the fibers $V_x$ over (a.e) point is endowed with a vector space structure which varies measurably. A precise axiomatic definition could be given by means of the standard vector space axioms reinterpreted by means of fiber-products constructions. For example you have the addition map $V\times_X V \to V$ and the scalar multiplication $\mathbb{C}\times V \to V$ which commute with the obvious maps to $X$ and satisfy the obvious compatibility relations.

Whatever is ones definition of "a measurably varying $X$-indexed family of vector spaces" it should be equivalent to a vector space over $X$. Unfortunately, I haven't seen this definition published anywhere, so let's say it is a folklore definition.

Note that associated with $X$ we have the algebra of bounded (measurable, defined up to a.e equivalence) $\mathbb{C}$-valued functions $L^\infty(X)$, which is a commutative von-Neumann algebra (aka a W*-algebra), that is a C*-algebra which has a predual ($L^1(X)$).

To a vector space over $X$, $\pi:V\to X$, one associates the vector space of all (classes of) measurable sections of $\pi$, to be denoted $L(V)$ (or $L(\pi)$ if there is a danger of misunderstanding). This is a module over the algebra $L^\infty(X)$.

Dimension

Assume now that $X$ is actually endowed with a finite measure (not merely a measure class). Then integration is a finite trace on the algebra $L^\infty(X)$, and this algebra becomes a finite von-Neumann algebra. For modules over such guys there is a well developed notion of dimension, the von-Neumann dimension. For finitely generated projective modules, this dimension is given by taking the trace of a certain projection in a certain matrix algebra over $L^\infty(X)$ (you can guess which projection: a one associated with a presentation of a the module as a direct summand of a free module, which trace is choice independent). The dimension of a general module is defined as the supremum over the dimensions of its f.g projective submodules. This theory is carried in Lueck's book. For an online survey, see his paper.

Finally, it is an exercise to show that for a vector space over $X$, $\pi:V\to X$, as defined above, we have that the von-Neumann dimension of the $L^\infty(X)$-module $L(V)$ equals exactly $\int_X \dim V_x$.

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  • $\begingroup$ Dixmier (in A.73) only defines direct integrals for measurable fields of Hilbert spaces. The Hilbert space structure here is important in that it defines the relevant square-integrability condition. But as the OP has already mentioned in the comments, his vector spaces don't come equipped with an inner product. I've never seen a notion of direct integral for plain vector spaces without any additional structure. What would be a reference for that? (Of course it may well exist and may even be the simpler concept, but it doesn't seem to be covered by the usual notion of direct integral.) $\endgroup$
    – Tobias Fritz
    Commented Sep 25, 2017 at 15:29
  • $\begingroup$ One can certainly define $\int V_x dx$ by simply omitting any type of integrability condition and this results indeed in an $L^\infty(X)$-module. But unlike in the Hilbert space case, there's no reason to expect it to be finitely generated, which means that von Neumann dimension is generally not applicable. But perhaps one can fix this by considering it as a module over the algebra of all integrable function up to a.e. equivalence (while $L^\infty(X)$ only contains essentially bounded ones). $\endgroup$
    – Tobias Fritz
    Commented Sep 25, 2017 at 15:51
  • $\begingroup$ @TobiasFritz, thanks for the comments. One can define a suitable notion of direct integral of Banach spaces and even more general objects. I am not sure what is the right source for this (maybe Takesaki?), but this is fairly common. $\endgroup$
    – Uri Bader
    Commented Sep 25, 2017 at 16:12
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    $\begingroup$ In any case, regarding finite dimensional vector spaces, measurability of the family also implies that you can impose on it a measurable inner product, so (even if this is not too natural) you could regard it as a family of Hilbert spaces. But let me repeat: this is not necessary, as direct integral of measurable family of f.d vector spaces could be easily defined. I will update my answer with either a reference or an explanation of the construction in the near future. $\endgroup$
    – Uri Bader
    Commented Sep 25, 2017 at 16:12
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    $\begingroup$ This is really interesting, and I look forward to the updated answer too. I'm not sure whether it will do what I want, though, because the category that the object lives in ($L^\infty(X)$-modules) is dependent on $X$, whereas I would hope/expect an "integral over $X$" to eliminate any dependence on $X$. In particular, I want to consider maps between the integrals $\int V_x \;dx$ over different spaces $X$. But maybe I can put these modules overy varying algebras together into one category somehow. $\endgroup$
    – Mike Shulman
    Commented Sep 25, 2017 at 16:35

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