This puzzles me from some time and is in parts connected to the questions Symmetrized derivatives version and Symmetrized derivatives version II.
For me the linear DO between vector bundles $E$ and $F$ over compact base $M$ is $$P: \Gamma(E) \rightarrow \Gamma(F)$$ linear and support, weakly, decreasing. From the Peetre theorem we know that the
First equivalent definition is that there is a linear bundle map $q: J^k(E) \rightarrow F$, where $k$ is the degree of $P$, such that $$P = q \circ j^k.$$
As was beautify explained in 1 this definition is equivalent to the following because there is an explicit isomorphism, induced by any covariant derivative on $M$ and $E$, from $J^k(E)$ to $\oplus_{l=0}^k \left(\vee^lT^*M \otimes E \right)$. On sections it is given by assigning symmetrized covariant derivatives.
Second equivalent definition (applied for example in Aubin's book) is that there is a linear bundle map $Q: \oplus_{l=0}^k \left(\vee^lT^*M \otimes E \right) \rightarrow F$ such that $$P = Q \circ D.$$
I have encountered in some places, for example Tian's or Joyce's or Besse's books, a
Third (equivalent) definition, it says that there is a linear $\tilde{Q}: \oplus_{l=0}^k \left(\oplus^lT^*M \otimes E \right) \rightarrow F$ such that $$P=\tilde{Q} \circ \nabla .$$
Of course having $P$ as in second definition it is easy to find $\tilde{Q}$ by extending $Q$ as $0$ on non totally symmetric part of any tensor, then $$\tilde{Q} \circ \nabla = \tilde{Q} \circ Sym \circ \nabla = Q \circ Sym \circ \nabla = Q \circ D.$$ On the other hand having an operator like in the third definition it is clearly support decreasing so it is, by the equivalence of the first and second definition, of the form $P = Q \circ D$.
My question (Edited) is having $\tilde{Q}$ such that $P= \tilde{Q} \circ \nabla$ is there a way, not necessarily canonical, to automatically produce $Q$ or $q$ such that $P=Q \circ D$ or $P=q \circ j^k$?
