Spectral triple

Remarks

Example: let $A = C^\infty(S^1)$ act on $H = L^2(S^1)$ by multiplication and let $D = -i \, d/d\theta$ on the Sobolev space of functions with square-integrable derivative. The functions $e_n(\theta) = e^{in\theta}$, $n \in \Z$, form an orthonormal basis of eigenvectors with $D e_n = n e_n$, so $(D - i)^{-1}$ has eigenvalues $1/(n - i) \to 0$ and is compact. For $f \in A$ the product rule gives $[D, f] = -i f'$, which is multiplication by a bounded function. More generally, smooth functions on a compact Riemannian spin manifold, acting on square-integrable spinors, with the Dirac operator, form a spectral triple.

Non-example: if $H$ is infinite-dimensional and $D = 0$, every commutator is $0$, so clause 1 holds. But $(D - i I)^{-1} = i I$ is not compact, so clause 2 fails.

Often $A$ is a dense $*$-subalgebra of a C*-algebra, and $A$ acts on $H$ through a faithful $*$-representation. A spectral triple is called even if there is a self-adjoint unitary $\gamma$ on $H$ that commutes with $A$ and anticommutes with $D$. For two states $\varphi$ and $\psi$ of the norm closure of $A$, Connes' distance formula is $$d(\varphi, \psi) = \sup \set{ \abs{\varphi(a) - \psi(a)} : a \in A, \ \|[D, a]\| \le 1 }.$$ It recovers the geodesic distance when $\varphi$ and $\psi$ are evaluations at two points of a manifold.