Summability of spectral triples

For instance, if $A \geq 0$ with $s_k(A) := \dfrac{1}{(k + 1)^{1/p}}$, then $A \in \sL^{p+}$ by the integral test:

$$\sg_N(A) \sim \int_1^N t^{-1/p} dt \sim \frac{p}{p-1} N^{(p-1)/p}, \as N \to \infty.$$

Indeed, since $p > 1$, $T \in \sL^{p+}$ implies $s_k(T) = O((k+1)^{-1/p})$. To see that, recall that $s_0(T) +\cdots+ s_k(t) = \sg_{k+1}(T)$; since $\{s_k(T)\}$ is decreasing, this implies $(k + 1)s_k(T) \leq \sg_{k+1}(T)$, and thus $s_k(T) \leq \frac{1}{k+1} \sg_{k+1}(T) \leq C(k + 1)^{-1/p}$ for some constant $C$.

Therefore, $T \in \sL^{p+}$ implies $s_k(T^p) = O(\inv{k + 1})$ and then $\sg_N(T^p) = O(\log N)$, so that $T^p \in \sL^{1+}$, which serves to justify the notation $\sL^{p+}$. It turns out, however, that there are, for any $p > 1$, positive operators $B \in \sL^{1+}$ such that $B^{1/p} \notin \sL^{p+}$, so the implication `` $T \in \sL^{p+} \implies T^p \in \sL^{1+}$'' is a one-way street. For an example, see [GVF, Sec. 7.C].

For positivity of all Dixmier traces, it suffices that $\liminf_{N\to\infty} \inv{\log N} \sg_N((D^2 + 1)^{-p/2}) > 0$. Note that, in view of Corollary , this can happen for at most one value of $p$.

The assumption that $D$ is invertible in the statement of Proposition  is not essential (though the proof does depend on it, of course). With some extra work, we can modify the proof to show that $(D^2 + 1)^{-1/2} \in \sL^{p+}$ implies that all $[F,a] \in \sL^{p+}$, where $F$ is redefined to mean $F := D (D^2 + 1)^{-1/2}$, in contrast to (). This is proved in [CPRS], in full generality.