A spectral triple $({\cal A},{\cal H},D)$ consists of a unital $*$-algebra ${\cal A}$ represented as bounded operators on a Hilbert space ${\cal H}$, together with an unbounded operator $D$ having compact reslovent and such that the operator $[D,a]$ is bounded, for all $a \in {\cal A}$.

Roughly speaking, when ${\cal A}$ is commutative, then such objects correspond to compact Riemannian spin manifolds, with ${\cal H}$ reducing to the square integrable spinors, ${\cal A}$ the smooth functions, and $D$ the Dirac operator.

Motivated by the commutative picture, people seem to usually work with spectral triples such that ${\cal H}$ is separable. But the definition seems to make perfect sense without this assumption. Is there anything that goes seriously wrong, or becomes much more difficult, when one passes from the separable to the non-separable situation.

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