TITLE: Spectral triples which are not $\theta$-summable
QUESTION [9 upvotes]: I would like to see an example of a spectral triple $(A,H,D)$ such that the underlying algebra $A$ is commutative but this spectral triple is not $\theta$-summable in the sense that $e^{-tD^2}$ is never traceclass operator. 
EDIT: Spectral triple is defined as a unital $*$-algebra $A$ represented faithfully on some Hilbert space $H$ together with the (usual unbounded) self adjoint operator $D$ with the property that $D$ has compact resolvent and such that for all $a$ in $A$ the commutator $[D,a]$ is (extends to) a bounded operator

REPLY [2 votes]: One can produce an operator in a less artificial way.  Consider the Sturm-Liouville operator $Du=-u''+V(x)u$, where $V(x)$ grows at infinity as $\log\log|x|$.
The eigenvalues grow very slowly, also as $\log\log$, and therefore the exponent of the operator is not trace class (the details on the asymptotics of eigenvalues can be found, eg, in the book by S.Levendorskii, 'Asymptotic distribution of eigenvalues of differential operators. Kluwer Academic Publishers Group, Dordrecht, 1990.)