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

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    $\begingroup$ I deleted a non-answer after deciding that surely the definition of "spectral triple" includes the condition that the unbounded operator $D$ appearing should have compact resolvant. It might be worth clarifying exactly what you mean by "spectral triple" in this context. $\endgroup$ – Matthew Daws Mar 2 '17 at 22:03

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.

  • $\begingroup$ It's worth noting that every self-adjoint unbounded operator with compact resolvant is, up to unitary equivalence, of the form I give here. See en.wikipedia.org/wiki/Resolvent_formalism#Compact_resolvent $\endgroup$ – Matthew Daws Mar 3 '17 at 19:47

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.)


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