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.

  • Not being an expert in spectral triples, but a guess: Separable Hilbert spaces are sufficiently big to represent all separable algebras (in question). Spectral triples are often accompanied with $K$-theory and $KK$-theory, and there it's often easier to have separable algebras. Also, underlying Hilbert modules in KK-theory need to be countably generated. In other words, in case of a Hilbert space it would need to be sparable. – hänsel Sep 29 at 11:14

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.