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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 $P=q \circ j^k$?

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  • $\begingroup$ Why not simply map the symmetric algebra of T*M into the tensor algebra of T*M via symmetric tensors? This will produce Q from Q̃. $\endgroup$ Dec 25, 2019 at 19:36
  • $\begingroup$ because that does not work. If for all symmetric tensors $t$ you put $Q(t):=\tilde{Q}(t)$ then $\tilde{Q} \circ D \not = Q \circ \nabla$ unless $\tilde{Q}$ was already $0$ on non symmetric tensors. $\endgroup$
    – J.E.M.S
    Dec 28, 2019 at 15:03
  • $\begingroup$ Maybe you should explain first what do you mean by "automatically produce Q or q". The dimension of the tensors used in the third definition is much bigger than those in the 1st and 2nd definition, so a bijective correspondence is simply impossible. $\endgroup$ Dec 28, 2019 at 19:51
  • $\begingroup$ Well of course there is no bijection between linear bundle maps from symmetric tensors to $F$ and from all tensors to $F$ but the question was different and I thought it was clear, just in case I edited it. There is no point in looking for "some $Q$" not related to the operator $P$. Of course we are looking for $Q$ which factorizes $P$ through symmetrized covariant derivative knowing $\tilde{Q}$ factorizes it through covariant derivative. $\endgroup$
    – J.E.M.S
    Dec 30, 2019 at 10:51

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