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Is there a well-defined notion of connection on a measurable bundle of Hilbert spaces?

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  • $\begingroup$ If it did, then one could do differential calculus in an only measurable context. While some tormented notions of differentiability can exist in non-smooth settings (the metric derivative, for instance), only measurability is probably too little to get anything useful. $\endgroup$
    – Alex M.
    Jul 4, 2021 at 13:51
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    $\begingroup$ Does it get any better if the base space can be topologized? In my case, the base space is the unitary dual of a locally compact topological group and is a topological space, say, equipped with Fell topology. It is a measure space equipped with the Plancherel measure. $\endgroup$
    – Hasib
    Jul 4, 2021 at 15:25
  • $\begingroup$ @user78032: Is your topological group a real or complex Lie group, or do you also want to allow more general objects, such as p-adic groups? $\endgroup$ Jul 4, 2021 at 16:05
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    $\begingroup$ @DmitriPavlov My topological group is a 7-dimensional nilpotent real Lie group which is topologically R^7. It's triple central extension of the abelian group of R^4 given in the paper:arxiv.org/pdf/1309.7086.pdf $\endgroup$
    – Hasib
    Jul 4, 2021 at 16:18
  • $\begingroup$ @user78032: Is the bundle of Hilbert spaces finite-dimensional or infinite-dimensional? $\endgroup$ Jul 4, 2021 at 16:27

1 Answer 1

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This is an answer to the refined question formulated in the comments: the base space is the unitary dual of a Lie group $G$.

The definition can be carried out in the setting of stacks in groupoids (or simplicial sets) on the site of cartesian spaces ($\def\R{{\bf R}} \R^n$ with smooth maps, for all $n≥0$).

Specifically, we define a stack $R_G$ that gives the unitary dual of a Lie group $G$ and a stack $B_∇$ that gives bundles of Hilbert spaces with connectiom. Then a morphism $R_G→B_∇$ is precisely a bundle of Hilbert spaces with connection over $R_G$.

To define the stack $R_G$ that gives the unitary dual of a Lie group, assign to a cartesian space $T$ the following groupoid $R_G(T)$.

Objects of $R_G(T)$ are given by a Hilbert space $H$ together with a smooth $T$-indexed family of irreducible unitary representations of $G$ on $H$. This is simply a smooth map $\def\Hom{\mathop{\rm Hom}} f\colon T→\Hom(G,U(H))$ (landing in irreps), where $U(H)$ is equipped with the ultraweak topology and $\Hom(G,U(H))$ is the space of continuous group homomorphisms equipped with the compact-open topology. Smooth means that the adjoint map $T⨯G→U(H)$ composed with the inclusion $U(H)→B(H)$ is smooth as a map from a finite-dimensional smooth manifold to a topological vector space $B(H)$.

Morphisms of $R_G(T)$ from $(H,f)$ to $(H',f')$ are given by a smooth map $h\colon T→U(H,H')$ that intertwines the action of $G$. Here $U(H,H')$ is the topological space of unitary isomorphisms $H→H'$ equipped with the ultraweak topology.

Next, we define stacks $B$ and $B_∇$ of bundles of Hilbert spaces, (equipped with connection in the case of $B_∇$) as follows. Given a cartesian space $T$, we define a groupoid $B_∇(T)$ as follows.

Objects of $B_∇(T)$ are pairs $(H,∇)$, where $H$ is a Hilbert space and $∇\colon T→Hom(T,I(H))$ is a smooth map, where $I(H)$ denotes the space of unbounded skew-adjoint operators on $H$ (equivalently, one-parameter unitary groups on $H$) and $\Hom(T,I(H))$ is the space of linear maps $T→I(H)$.

Morphisms of $B_∇(T)$ from $(H,∇)$ to $(H',∇')$ are smooth maps $p\colon T→U(H,H')$ such that $∇'=p^*\theta+\mathop{\rm Ad}_{p^{-1}}∇$, where $\theta$ is the Maurer–Cartan form and Ad denotes the adjoint action.

The stack $B$ is defined analogously, but dropping $∇$ and the condition on $p$. We have a canonical forgetful map $B_∇→B$.

The obvious forgetful map $W\colon R_G→B$ defines the canonical bundle $W$ over $R_G$. In particular, the fiber of $W$ of a point of $R_G$ given by an irreducible representation $ρ$ of $G$ is simply $ρ$ itself.

Now, a connection on $W$ is a lift of the map $W\colon R_G→B$ through the forgetful map $B_∇→B$.

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  • $\begingroup$ A naive and basic question: in the setting of the OP's question, what is the space T? How is its smooth structure related to the Fell topology on the usual unitary dual of G? $\endgroup$
    – Yemon Choi
    Jul 4, 2021 at 19:59
  • $\begingroup$ On further thought, I see that I have a more basic misunderstanding: it is $R_G$ which is the unitary dual, not any particular $R_G(T)$. But could you include in your answer some indication of how one comes down from the stack $R_G$ to what harmonic analysts usually understand by the unitary dual of a locally compact group? $\endgroup$
    – Yemon Choi
    Jul 4, 2021 at 20:17
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    $\begingroup$ @YemonChoi: And points in $R_G$, i.e., isomorphism classes of objects in $R_G({\bf R}^0})$ are precisely points in the unitary dual. $\endgroup$ Jul 4, 2021 at 20:23
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    $\begingroup$ @DavidRoberts: Ultraweak topology is the default topology for von Neumann algebras, it is the weak-* topology induced by the predual. For instance, morphisms of von Neumann algebras are defined as ultraweakly continuous *-homomorphisms, or, equivalently, *-homomorphisms that admit a predual. $\endgroup$ Jul 5, 2021 at 3:20
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    $\begingroup$ @DavidRoberts: The Maurer–Cartan form on U(H) can be defined directly: to a smooth plot P: T→U(H) we must assign a differential 1-form on T, defined as follows: given a point p∈T with a tangent vector q, consider some curve c: R→T such that c(0)=p, c'(0)=q, and take the derivative of P∘c at 0. $\endgroup$ Jul 5, 2021 at 3:30

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