# Geometric motivation behind the Fredholm module definition

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?

• 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). – Branimir Ćaćić Jun 8 '19 at 22:12
• 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/… – Branimir Ćaćić Jun 8 '19 at 22:14