Given an infinite countable group $G$, the category of unitary representations of $G$ admits direct integrals. Namely, given a measure space $(X,\mu)$ and a measurable family of unitary $G$-reps $(H_x)_{x\in X}$, one can form a new representation $$ H=\int^\oplus H_x = \left\{\text{$L^2$ functions $f:X \to \coprod_{x\in X} H_x$}\right\}, $$ called the direct integral of the $H_x$'s.

Conversely, given $H\in \mathrm{Rep}(G)$ and an abelian sub-von Neumann algabra $A\subset \mathrm{End}_G(H)$ of the von Neumann algebra of endomorphisms of $H$, one can find a measure space $X$ s.t. $A=L^\infty X$, and a measurable family of unitary representations $(H_x)_{x\in X}$ such that $H=\int^\oplus H_x$. The category of unitary representation of $G$ is said to admit direct integral decompositions.

What does it mean for a $W^*$-category to admit direct integrals?

What does it mean for a $W^*$-category to admit direct integral decompositions?

Presumably, "admitting direct integrals" is not a property of a category, but is extra structure instead. What is this extra structure? What axioms does this extra structure satisfy?

Is "admitting direct integral decompositions" a property of a category that admits direct integrals?

Here are some examples of categories that admit direct integral decompositions.

Example1: The category of representations of a separable C* algebra.

Example2: Let $A$ be a separable C* algebra, let $(p_n)_{n\in\mathbb N}$ be an increasing sequence of projections in $A$, and let $p\in A$ be yet another projection which is bigger than all the $p_n$. Then the category of all representations $(H,\pi)$ of $A$ such that $\pi(p_n)\to \pi(p)$ in the strong topology is another example of a category that admits direct integral decompositions.

Example3 (very slight generalisation of Example2): Let $A$ be a separable C* algebra. Let $S$ be a countable set. For every $\alpha\in S$, let $(p^\alpha_n)_{n\in\mathbb N}$ be an increasing sequence of projections, and let $p^\alpha\in A$ be another projection which is bigger than all the $p^\alpha_n$. Then we can consider the category of all representations $(H,\pi)$ of $A$ such that $\forall \alpha\in S,\pi(p^\alpha_n)\to \pi^\alpha(p)$ in the strong topology.

  • $\begingroup$ Pardon my ignorance, but what is a $W^\ast$-category? Is it just a category enriched in the category of $W^\ast$-algebras and $*$-homomorphisms with the tensor product inherited from Hilbert spaces? If so I'm a bit confused because at least if the $W^\ast$-algebras are commutative, this tensor product should be the coproduct and so categories enriched in it should be kind of trivial. Also in this case I'm not sure I see how unitary $G$-representations are an example. $\endgroup$ – Tim Campion Oct 18 '18 at 18:21
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    $\begingroup$ @Tim Campion. A $W^*$-category is a "von Neumann algebra with many objects". It's a category such that for every finite collection of objects $(a_i)$, the algebra $\bigoplus_{i,j} Hom(a_i,a_j)$ is a von Neumann algebra. $\endgroup$ – André Henriques Oct 18 '18 at 20:37
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    $\begingroup$ @TimCampion Apparently we can equally think of it as a complex dagger-category satisfying a few axioms that horizontally categorify the axioms of a $W^*$-algebra: projecteuclid.org/download/pdf_1/euclid.pjm/1102703884 $\endgroup$ – Kevin Arlin Oct 19 '18 at 3:21
  • $\begingroup$ @AndréHenriques Let $\oplus$ be the $L^2$-"direct sum". If $(H_x)_{x \in X}$ is constant at $H$, then $\int_x^\oplus H_x = L^2(X) \otimes H = H^{\oplus \kappa}$ where $\kappa = dim(L^2(X))$. More generally, if the family is locally trivial, i.e. $X = \amalg_{n \in\mathbb N} X_n$ with $H_x$ being constant on each $X_n$, then $\int^\oplus_{x \in X} H_x = \oplus_n \int_{x \in X_n} H_x$. So locally trivial direct integrals reduce to $\oplus$. And at least according to Wikipedia, direct integrals reduce "in some sense" to locally trivial ones. $\endgroup$ – Tim Campion Apr 9 '20 at 17:59
  • $\begingroup$ @Tim Campion. There exists non-locally-trivial families too. For example, take the category of unitary representations of $\mathbb R$. Then, there are examples of non-locally-trivial families of 1-dimensional representations. $\endgroup$ – André Henriques Apr 10 '20 at 23:17

This is a question that I have been working on recently together with Robert Furber and Bas Westerbaan. Let me sketch what we know so far, starting with the case of (infinite) direct sums, treated in this paper. These are a special case of direct integrals which already display some of the essential features.

In the following, it may help to keep the example of the category of Hilbert spaces in mind, or more generally the category of normal representations of any W*-algebra. These categories contain Hilbert spaces of all dimensions, regardless of separability.

Infinite direct sums

Let $I$ be an indexing set for a family of objects $(A_i)_{i \in I}$ of which we want to take the direct sum $\oplus_{i\in I} A_i$. For another object $B$, consider the set $$\oplus_i \mathsf{C}(A_i,B) := \left\{(f_i : A_i \to B)_{i\in I} \mid \sum_i f_i f_i^* < \infty \right\}$$ as a normed space under the $\ell^2$-norm $$\|(f_i)\| := \|\sum_i f_i f_i^* \|^{1/2}.$$ It is not hard to show that this is actually a Banach space. Moreover, this construction is clearly functorial in $B$.

In line with the reindexing which has already been discussed, we use:

Definition: The direct sum $\oplus_i A_i$ is any object which represents the $\mathsf{Ban}$-enriched functor $\oplus_i \mathsf{C}(A_i,-) : \mathsf{C} \to \mathsf{Ban}$.

While the representability of a $\mathsf{Ban}$-enriched functor a priori characterizes the representing object up to unique isomorphism, we have shown that this unique isomorphism must actually be unitary. So although our universal property does not assume any compatibility with the involution/dagger, this turns out to come out naturally. This will be obvious to anyone who knows that if $u$ is an invertible element in a C*-algebra with $\|u\| = \|u^{-1}\| = 1$, then $u$ is unitary.

Although our definition may look like the definition of a (weighted) limit at first, I do not think that this is actually the case, even for finite $I$.

We have then shown the equivalence with the earlier definition of direct sums in W*-categories given by Ghez, Lima and Roberts:

Proposition: An object $A$ is a direct sum $\oplus_i A_i$ if and only if there is a family of morphisms $(\kappa_i : A_i \to A)_{i\in I}$ such that $\kappa_i^* \kappa_j = \delta_{ij}$ and $$\sum_i \kappa_i \kappa_i^* = 1,$$ with convergence in the ultraweak topology on $\mathsf{C}(A,A)$.

As one would expect, this is an equivalence of structures on $A$ rather than properties of $A$. The "if" part is the more difficult direction.

In the category of Hilbert spaces, this clearly recovers the usual direct sums, and more generally so in any category of normal representations of a W*-algebra.

As was already known to Ghez, Lima and Roberts, we thus conclude that the existence of direct sums is still a property of a W*-category rather than extra structure.

Direct integrals

Having treated direct sums as a warm-up, we can now try to tackle direct integrals. This seems to be more of the same, but with many tricky pitfalls that we need to steer clear of, and I'm not sure whether we've managed to do that yet. The following items describe where we're currently at. I will update this answer as soon as we have an actually working definition. Further contributions by MOers will be welcome, just drop me an email.

  • The role of the indexing set $I$ is now played by a complete strictly localizable compact measure space $(X,\Sigma,\mu)$. This class is large enough for every commutative W*-algebra to be of the form $L^\infty(X,\Sigma,\mu)$, regardless of separability.

  • For a given W*-category $\mathsf{C}$, we then want to construct a W*-category $L^\infty(X,\mathsf{C})$, whose objects are suitably defined measurable families of objects $(A_x)_{x\in X}$ indexed by our measure space. To this end, we can try to work with a generalization of measurable fields of Hilbert spaces, as follows. For $(A_x)_{x\in X}$ an arbitrary family of objects and $B$ any other object, let us say that a family of morphisms $(f_x : B \to A_x)_{x\in X}$ is measurable if the map $x\mapsto f_x^* f_x$ is measurable. The measurable families of morphisms form a subfunctor of the functor $\prod_{x\in X} \mathsf{C}(-,A_x) : \mathsf{C}^{\mathrm{op}} \to \mathsf{Set}$. A measurable field of objects could now be a further subfunctor of this measurable families functor, satisfying properties for every $B$ analogous to those of measurable fields of Hilbert spaces (see e.g.~Dixmier A69).

  • One could then hope that the definition of direct integral will be similar to the direct sum case: any object which represents the $\mathsf{Ban}$-enriched functor $$B \longmapsto \int_X^\oplus \mathsf{C}(A_x,B)\, d\mu(x) := \left\{ (f_x : A_x \to B)_{x\in X} \textrm{ in the subfunctor} \: \bigg| \: \int_X f_x f_x^* \, d\mu(x) < \infty \right\},$$ again carrying the $L^2$-norm. As per the discussion below, starting with Simon Henry's comment, it doesn't quite work like this yet.

  • Since working with this definition of measurable fields of objects will be cumbersome in practice, one can try to add an additional layer of abstraction by assuming that $\mathsf{C}$ itself is a category internal to (large and suitably nice) measurable spaces, and then simply require the map $x\mapsto A_x$ to be measurable. Then a family $(f_x : B \to A_x)$ is part of the further subfunctor if and only if it is measurable as a map $X \to \mathrm{Mor}(\mathsf{C})$.

  • The resulting category of measurable families of objects, $L^\infty(X,\mathsf{C})$, should again be a W*-category. For $\mathsf{C} = \mathsf{Hilb}$, we should recover Yetter's categories of measurable fields of Hilbert spaces at least insofar as we should have good arguments for any potential differences. Then forming direct integrals should correspond to the analogous universal property as in the case of direct sums.

  • An important consistency check will be the following: if $\mathsf{C}$ has direct integrals $\int_X^\oplus A_x \, d\mu(x)$, then such a direct integral should come equipped with a normal representation of $L^\infty(X)$. Moreover, we then expect to get an equivalence of W*-categories between $L^\infty(X,\mathsf{C})$ and normal representations of $L^\infty(X)$ in $\mathsf{C}$, generalizing the spectral theorem.

Summary: We don't have a complete treatment of direct integrals yet. Concerning the necessary extra structure on the W*-category, it seems that we either need no extra structure at all, resulting in a cumbersome definition of direct integrals; but probably we want to require the collections of objects and morphisms to come equipped with suitable $\sigma$-algebras, in a way that makes the structure maps measurable, in order to get a potentially more usable definition.

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    $\begingroup$ Concerning the second bullet: In the motivating example of $Rep(G)$ where $G$ is an infinite discrete group, you certainly want to include $X$-parametrised families of representations which are all pairwise non-isomorphic. This is already relevant in the very simplest case of $G=\mathbb Z$. $\endgroup$ – André Henriques Oct 21 '18 at 16:46
  • $\begingroup$ @AndréHenriques: yes, you're right, I've just been thinking about that as well. So then we seem to be led to require some additional structure on the W*-category. I'm currently trying to match this up with Simon's answer, which claims that one does not need any additional structure. I will update my answer accordingly once this is clear. $\endgroup$ – Tobias Fritz Oct 21 '18 at 16:48
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    $\begingroup$ Yes, I think that "category internal to measurable spaces" is going in the right direction. But "stack over the site of measurable spaces" might be slightly better. Lots to be worked out still... $\endgroup$ – André Henriques Oct 24 '18 at 22:46
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    $\begingroup$ @TobiasFritz : Unless I'm missing something in your definition there seem to be a problem: Assume I have a direct integral $A = \int A_x$ in your sense. Then the identity of $A$ should corresponds to an element of your Ban-valued functors i.e. to "measurable familly of functions" $A_x \rightarrow A$. But when the measure is diffuse, there is in general absolutely no non-zero such functions. For example the representation of $C([0,1])$ on $L^2([0,1])$ has no non zero morphism to the pointwise evaluation representation despite being their integrals. $\endgroup$ – Simon Henry Oct 25 '18 at 16:20
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    $\begingroup$ @TobiasFritz : Maybe. I have added an answer which some obstruction for defining direct integral by a universal property. It is not totally clear that it will completely prevent this type of approach but it does impose some important restrictions. $\endgroup$ – Simon Henry Oct 25 '18 at 21:00

The following is an argument for showing that "having direct integral" is definitely not a property nor a "property-like-structure" of $W^*$-categegories, but a real, non-trivial additional structure. I.e. that in examples where they exists they are not intrinsic to the $W^*$ structure. That does not answer the question, but I felt like it was worth mentioning as it shows that lots of approach to this question are bound to fail, and hopefully give some hints for what kind of additional structure a solution should have.

Let $C=C([0,1])$, and I'm considering the category of representations of $C$, it admit direct integral, but the $W^*$-category have a lot of automorphisms that do not preserves these representations.

Indeed the category of representation of $C$ is equivalent to the category of normal representation of its enveloping Von Neuman algebra $C''$. As any commutative von Neuman algebra $C''$ can be split into an atomic part and a diffuse part.

One has one atom for each element of $[0,1]$, and they corresponds to the irreducible representations $\chi_x$ of $C$.

So for any bijection of $[0,1]$ there is an automorphism of $C''$ that justs permute the atom and leave the diffuse part unchanged. On the $W^*$-category it permutes the irreducible representation (and their orthogonal sums) while leaving the diffuse part of each representation unchanged.

As any representation is a direct integral of irreducible, these direct integral are clearly not going to be preserved by the automorphisms:

for example if I simply chose the automorphism $1-x$ of $[0,1]$ then the representation $L^2([0,1/2]) = \int_0^{1/2} \chi_x$ should become (if integral were preserved) under the automorphism $L^2[1/2,1] = \int_{1/2}^{1} \chi_x$ and these representation are not even isomorphic.

Even worse (but I have not checked this last part in details), in the case (as above) where the automorphism of $[0,1]$ is chosen measurable, it seems to me that the isomorphisms of the $W^*$-category preserve the natural notions of measurable fields of representations and morphisms. Indeed unless I'm mistaken the splitting of representations in "atomic+diffuse" can be applied to the measurable fields and if the permutation of the atomic part is measurable it can be applied to fields as well. So this type of structure will be insuficient. (this seem to for example completely prevent what is suggested at the end of Tobias Fritz answer).


The following is basically a guess, which may or may not work out.

It seems to me that a $W^\ast$-category $C$ is a special type of category enriched in the monoidal category $\mathcal V$ of Hilbert spaces. I suspect that the notion of an "$X$-indexed family of objects of $\mathcal C$" allows one to extend $C$ to a category $\underline{C}$ enriched in the bicategory $\mathcal W$ of commutative $W^\ast$-algebras, bimodules, and bilinear maps.

The difference between a category $C$ enriched in a monoidal category $\mathcal V$ and a category $\underline C$ enriched in a bicategory $\mathcal W$ is that an object of a category enriched in a bicategory has an "extent" -- an object of $\mathcal W$ that it "lies over". One thinks of $\underline C$ as being like a sheaf over $\mathcal W$, with the "extent" of an object being its support-- and indeed, Walters showed that Cauchy-complete dagger categories enriched in the bicategory of relations in a site are the same thing as sheaves on the site.

More specifically, an object of $\underline C$ with extent $\mathcal O(X)$ is an $X$-indexed family of objects of $C$. A morphism $(H_x)_x \to (K_x)_x$ of these lying over $X$ is a family of morphisms of $C$ $(H_x \to K_x)_x$, where we identify two families if they agree for almost every $x \in X$.

Anyway, the direct integral construction is then right adjoint to reindexing between different extents, so it amounts to a completeness condition on the $\mathcal W$-enriched category $\underline C$.

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    $\begingroup$ A $W^*$-category is not enriched over Hilbert spaces. A one-objet $W^*$-category is just a $W^*$-algebra (a.k.a. a von Neumann algebra). And the underlying vector space of a $W^*$-algebra is not a Hilbert space. $\endgroup$ – André Henriques Oct 21 '18 at 12:08
  • $\begingroup$ I agree with you that the direct integral wants to be then right adjoint to what you call "reindexing between different extents". But it's not a right adjoint: think about the identity map $\int^\oplus H_x\to \int^\oplus H_x$, and where this map would go to under the adjunction. $\endgroup$ – André Henriques Oct 21 '18 at 12:14
  • $\begingroup$ @AndréHenriques Sorry, my thinking was that a $W^\ast$-algebra is in particular a $C^\ast$-algebra, and any $C^\ast$ algebra is in particular a Hilbert space. Is that not correct? Alternatively, I thought I'd heard talk of a bicategory of $W^\ast$-algebras, $W^\ast$-bimodules, and maps. I was under the impression that a $W^\ast$-bimodule was in particular a Hilbert space, and that the unit bimodule was the $W^\ast$-algebra itself, which would mean that a $W^\ast$ algebra was in particular a Hilbert space. Is that not correct either? $\endgroup$ – Tim Campion Oct 21 '18 at 12:16
  • $\begingroup$ The unit bimodule is not the $W^*$-algebra itself. Have a look at §6 (and also the previous sections) of my paper arxiv.org/pdf/1110.5671.pdf. Also, a C^*-algebra is not a Hilbert space (the space of continuous functions on a compact space is not a Hilbert space). $\endgroup$ – André Henriques Oct 21 '18 at 12:23
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    $\begingroup$ If an analyst may be permitted to butt in, I know Andre has already pointed out that a $C^*$-algebra is not a Hilbert space (unless one-dimensional), but it bears repeating in 10-foot high letters of fire that a $C^*$-algebra is not a Hilbert space $\endgroup$ – Yemon Choi Oct 21 '18 at 12:32

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