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If $A$ is an involutive algebra over the complex numbers $\mathbb{C}$, then a Fredholm module over $A$ consists of an involutive representation of $A$ on a Hilbert space $H$, together with a self-adjoint operator $F$, of square $1$ and such that the commutator $[F,a]$ is a compact operator for all $a \in A$.

Now the slogan is the "Fredholm modules" are abstractions of differential operators, or more explicitly, a bounded version of a differential operator constructed by some functional calculus argument.

In this context, what is the geometric motivation behind the compact commutator condition? In other words, what property of a differential operator is it abstracting?

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    $\begingroup$ A quick comment for lack of time: the compact commutator condition encodes the notion that $F$ is the bounded transform $F := D \lvert D \rvert^{-1}$ of a first-order elliptic differential operator $D$, e.g., a Dirac-type operator. The corresponding properties of the unbounded operator $D$ are that it has bounded commutators with $A$ (first-order) and compact resolvent (elliptic). $\endgroup$ Jun 8 '19 at 22:12
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    $\begingroup$ If you're comfortable enough with global analysis, you might want to give the following survey by Carey, Phillips, and Rennie a try: ro.uow.edu.au/cgi/… $\endgroup$ Jun 8 '19 at 22:14

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