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Let $M$ be a compact manifold. If we have a pseudodifferential operator $P$ of order $0$ on $M$, then $P$ is pseudolocal, i.e., every commutator $[f,P]$ with a continuous function $f \in C(M)$ is a compact operator.

Let $M$ be a compact manifold. Denoting the pseudodifferential operators of order $0$ on $M$ by $\Psi^0(M)$, and the pseudolocal operators by $D^\ast(M)$, do we have that the closure of $\Psi^0(M) \subset \mathfrak{B}(L^2(M))$ coincides with $D^\ast(M)$?

There was already the question What is the smallest $C^*$-algebra containing the "standard" pseudodifferential operators? asking about the closure of $\Psi^0(\mathbb{R})$. But this question seems to be harder to answer since in this case we certainly have $\overline{\Psi^0(\mathbb{R})} \subsetneqq D^\ast({\mathbb{R})}$.

Note that it is important here to work with pseudodifferential operators from Hörmanders class $S_{1,0}^0(M)$ and not with pseudodifferential operators with a homogeneous symbol (i.e., not with polyhomogeneous pseudodifferential operators), since it was shown by Melo that $\overline{\Psi^0_{phg}(M)} \subsetneqq \overline{\Psi^0(M)}$.

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  • $\begingroup$ I think the answer to my own question here should be no. Note that we have for the commutator $[\Psi^0, \Psi^0] \subset \Psi^{-1}$ and this inclusion passes of course to the completions. Now $\overline{\Psi^{-1}}$ are exactly all the compact operators $\mathfrak{K}$ since $M$ is compact. So the quotient $\overline{\Psi^0} / \mathfrak{K}$ is abelian, whereas $D^\ast(M) / \mathfrak{K}$ should not be abelian. $\endgroup$
    – AlexE
    Commented Sep 5, 2014 at 7:39

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