We denote by $Cl^\mu$ the class of classical pseudo-differential operators of order $\mu$. Consider the notation
$$[Cl^{a},Cl^{b}]\hookrightarrow [Cl^{a'},Cl^{b'}]$$
which means that a commutator on the left hand side can be written as a sum of commutators of the right hand side, i.e.
$$\forall A\in Cl^a\ \text{and}\ B\in Cl^b\ \exists\ P_i\in Cl^{a'}\ \text{and}\ Q_i\in Cl^{b'}: [A,B]=\sum_i [P_i,Q_i]$$
where we have $a+b=a'+b'$.
The question is for which $a,b,a',b'$ we can construct these inclusions.
As an example, let $a\leq 0$ and consider a differential operator $D$ of order $1$, such that $D^a\in Cl^a$ and $D^{-a}\in Cl^{-a}$. Then we can show
$$[Cl^{0},Cl^{2a}]\hookrightarrow [Cl^{a},Cl^{a}]$$
since for $A\in Cl^0$ and $B\in Cl^{2a}$ the operators $AD^a, D^aA, D^a, BD^{-a}, D^{-a}B, ABD^{-a}$ and $D^{-a}BA$ are in $Cl^{a}$ and
$$2[A,B] = [AD^a,D^{-a}B] + [D^aA,BD^{-a}]+[ABD^{-a},D^a]+[D^{-a}BA,D^a].$$
My problem is to show whether or not there holds
$$[Cl^{1},Cl^{m}]\hookrightarrow [Cl^{0},Cl^{m+1}].$$
My intuition says no but I could not find a counterexample.
For $m<0$ the classes on the RHS form an algebra and I can still show that given $A\in Cl^1$ and $B\in Cl^m$ using the same $D$ as above
$AB$ differs from
$[AD^{-1},DB]$ by an operator $-DBAD^{-1}$,
$[ABD^{-m-2},D^{m+2}]$ by an operator $-D^{m+2}ABD^{-m-2}$ and
$-[D^{-m-1}B,AD^{m+1}]$ by an operator $-D^{-m-1}BAD^{m+1}$.
Similarly, $-BA$ differs from
$-[BD^{-m-1},D^{m+1}A]$ by an operator $D^{m+1}ABD^{-m-1}$,
$[D^{-1}A,BC]$ by an operator $D^{-1}ABD$ and
$-[BAD^{-m-2},D^{m+2}]$ by an operator $D^{m+2}BAD^{-m-2}$.
Other operators in the commutators do not allow to write $AB$ or $BA$.
So by repeating this argument ($DBAD^{-1}$ differs from $[DAD^{-2},D^2B D^{-1}]$ by an operator $-D^2BAD^{-2}$) we have that $[A,B]$ can be written as an infinite sum of commutators on the RHS.
Does this make sense? Can this sum be really infinite? Probably, one can give a hit in the right direction or knows a counterexample?