Definition of spectral triples

We start with the definition of the main concept in noncommutative geometry.

For now, and until further notice, all spectral triples will be defined over unital algebras. The compact-resolvent condition must be modified if $\sA$ is nonunital: as well as enlarging $\sA$ to a unital algebra, we require only that the products $a(D - \la)^{-1}$, for $a \in \sA$ and $\la \notin \spec(D)$, be compact operators.

Note that, if $\la,\mu \notin \spec(D)$, then $(D - \la)^{-1} - (D - \mu)^{-1} = (\la - \mu)(D - \la)^{-1}(D - \mu)^{-1}$ --this is the famous ``resolvent equation''-- since

$$(D - \la) \bigl((D - \la)^{-1} - (D - \mu)^{-1}\bigr) (D - \mu) = (D - \mu) - (D - \la) = \la - \mu.$$

Thus $(D - \la)^{-1}$ is compact if and only if $(D - \mu)^{-1}$ is compact, so we need only to check this condition for one value of $\la$. In the same way, we get the following useful result.

By the spectral theorem $(D^2 + 1)^{1/2}-\vert D\vert = f(D)$, where $f\: \bR \to \bR$ is the continuous function $f(\la) := \sqrt{\la^2+1} - \vert\la\vert = \frac{1}{\sqrt{\la^2+1} + \vert\la\vert}$; and $0 < f(\la) \leq 1$ for all $\la \in \bR$, so that $\Vert f(D)\Vert \leq 1$. Or more precisely: the operator $f(D) := (D^2 + 1)^{1/2} - \vert D\vert$, defined initially on $\Dom D$, extends to a bounded operator on $\sH$, of norm at most 1.

In many arguments to come, we shall employ $\vert D\vert$ and $\vert D\vert^{-1}$ as if we knew that $\ker D = \{0\}$. However, even if $\ker D \neq \{0\}$, we can always replace $\vert D\vert$ by $(D^2 + 1)^{1/2}$ and $\vert D\vert^{-1}$ by $(D^2 + 1)^{-1/2}$, at the cost of some extra calculation.