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 



Spectral triples which are not $\theta$-summable
So, under this generality, it seems to me that we can choose $A$ in a clever way: if we make $A$ smaller, then it gets "easier" to satisfy the condition. So why not take $A$ to be the scalar multiples of the identity-- then $D$ commutes with all of $A$ on the nose.


Now let $H=\ell^2$ and let $D$ be multiplication by a sequence of real numbers $(d\_n)$. If $z\in\mathbb C$ not an accumulation point of the $(d\_n)$ then $(zI-D)^{-1}$ exists and is the multiplication operator by the sequence $(z-d\_n)^{-1}$. If $d\_n\rightarrow\infty$ then $(z-d\_n)^{-1}\rightarrow 0$ and so $(zI-D)^{-1}$ is compact and so $D$ has compact resolvant.


Similarly, $e^{-tD^2}$ is the multiplication operator by the sequence $(\exp(-td\_n^2))$. This will be trace class if and only if
$$ \sum\_n \exp(-td\_n^2) < \infty $$
So you just need to let $(d\_n)$ grow very slowly. For example, set
$$ d\_n = \big( \log(1/e\_n) \big)^{1/2} \implies \exp(-td\_n^2)
= e\_n^{t} $$
where we now just need that $e\_n\rightarrow 0$. Let $e\_n = 1/2$ for the first $N\_2$ terms, then $e\_n=1/3$ for next $N\_3$ terms, and so on. Then
$$ \sum\_n \exp(-td\_n^2) = \sum\_{k\geq 2} \frac{N\_k}{k^t}. $$
Pick $N\_k \geq k^k$ so that for any $t>0$ if $K>t$ then
$$ \sum\_n \exp(-td\_n^2) \geq \sum\_{k\geq K} \frac{N\_k}{k^t}
\geq \sum\_{k\geq K} 1 = \infty. $$


In this example, you could also take $A=c\_0$ for a less trivial algebra.


