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## When is a Pseudo-differential operator trace class or in Dixmier ideal?

Let's denote the set of all Pseudo-differential operators with symbol of “order” $d$ by $\Psi_d(M)$ and Sobolev space on $M$ by $H_s(M)$. It is known that

If $P\in\Psi_d(M)$ Then $P$ extends to a continuous map $P:H_{s}(M)\to H_{s-d}(M)$ for all $s$. Moreover, since the natural inclusion $H_s\to H_t$, for $s>t$ is compact, $P:H_{s}(M)\to H_{t}(M)$ is compact operator if $t< s-d$.

See for example Lemma 1.3.4, Gilkey's book Invariance Theory: The Heat Equation and the Atiyah-Singer Index Theorem.

In special case, when $s=0$, $L^2(M)=H_0(M)$, $P:L^2(M)\to L^2(M)$ is continuous if $d\leq 0$. and it is compact if $d<0$.

Now my question is

when is $P:L^2(M)\to L^2(M)$ trace class? and when is it in Dixmier ideal $\mathcal{L}^{1,\infty}(L^2(M))$ or in general $\mathcal{L}^{(p,q)}(L^2(M))$?

Thanks

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It is trace class when it is an operator of order $-k$, $k>\dim M$. For a proof see Section 4.3 of these notes www3.nd.edu/~lnicolae/Pseudo.pdf – Liviu Nicolaescu Jan 26 at 14:43
Moreover, by the Connes trace formula (alainconnes.org/docs/action88.pdf), if your operator is of order $-k$ for $k = \dim M$, then it is in the Dixmier ideal; indeed, it is measurable (in the sense of the theory of Dixmier traces), and the (unique value of the) Dixmier trace is given by the Wodzicki residue of your operator. – Branimir Ćaćić Jan 28 at 10:30
Thanks for the useful references. However, I still wondering if there is a bound like $l$ such that pseudo differential operator of the order $d$ is in the Dixmier ideal (not necessary measurable) when $d<l$. of course $l$ should be in $[-k,0)$ where $k=dim M$. – Asghar Ghorbanpour Jan 28 at 17:24

An operator is trace class whenever it is the product of Hilbert-Schmidt operators. There is a simple characterization of Hilbert-Schmidt operators pseudodifferential operators: a pseudodifferential operator with symbol $a$ is HS if and only if $a$ belongs to $L^2$ of the cotangent bundle (this is equivalent also to the fact that the kernel of the operator is $L^2$).
For a comprehensive account of what you are looking for, see the book by Simon Scott Traces and determinants of pseudodifferential operators''