Let $M$ be a closed $n$ dimensional manifold. I would like to know why is it true that there are functions $A_1,...,A_N$ and vector fields $B_1,...,B_N$ such that $\sum_i[B_i,A_i]=nI$ and if $D$ is any differential operator of order $q$ then $$(n+q)D=\sum_i[B_iD,A_i]+R$$ where $R$ is of order less than $q$. I was given a hint that in the flat case one should take $A_i=x_i$ and $B_i=\frac{\partial}{\partial x_i}$. I also managed to figure out that using Leibniz identity for Lie bracket the problem boils down to show that $\sum_i B_i[D,A_i]$ is equal to $qD+R'$ where $R'$ is of order less that $q$.
EDIT: Just to give some motivation behind this question: it is technical step in the proof that there is no nonzero traces on the algebra of differential operators.

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    $\begingroup$ Try doing this locally, then maybe patching things together using partitions of unity. $\endgroup$ – Avi Steiner Oct 10 '16 at 4:55
  • $\begingroup$ Indeed, one has to prove this locally first: however I don't see why it is true. $\endgroup$ – truebaran Oct 10 '16 at 6:04
  • $\begingroup$ Prove it first when $q = 1$. Then when $q = 2$. At that point, you should have a sense of how to do the general case. If not, try $q = 3$, too. $\endgroup$ – Deane Yang Oct 11 '16 at 3:18
  • $\begingroup$ Locally it suffices to show that $[x_{l_1},[\ldots,[x_{l_q},[\sum \partial_i[D,x_i]-qD]\cdots]=0$ for any choice of coordinate functions $x_{l_j}$. This basically reduces to the Leibniz rule. $\endgroup$ – Michael Bächtold Oct 11 '16 at 18:20

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