Suppose on a manifold $M$ we have a differential operator $D$ of order 1 from the smooth sections $C^\infty(M, E)$ to smooth sections $C^\infty(M, F)$, where $E$ and $F$ are vector bundles of rank $n$ and $m$ respectively. If $u \in C^\infty(M, E)$ and $f \in C^\infty(M)$ is a smooth function, how would $D(fu)$ look like? I am trying to understand what the product/Leibniz rule could be in this case.
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$\begingroup$ By definition you know that $u \mapsto fD(u) - D(fu)$ is function linear and hence a section of the endomorphism bundle $\mathsf{End}(E, F)$. But without additional information you can not say much more than that. $\endgroup$– Stefan WaldmannCommented Jul 9, 2014 at 6:06
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The "Generalized Leibniz rule" can be written in terms of commutators: see http://en.wikipedia.org/wiki/Differential_calculus_over_commutative_algebras and references therein.
In this perspective, the linear function in Stefan Waldmann's comment can be rewritten "without mentioning the $u$" as follows: $$ \delta_f(D):=[f,D]\in\textrm{Hom}_{C^\infty(M)}(C^\infty(M,E),C^\infty(M,F)) $$ So, it is true that at this point you cannot say anything else.
Nevertheless, by "commuting again" you get zero: $$ \delta_g(\delta_f(D))=0 $$ as "linearity" means "commuting with the multiplication". Now you can bring in the $u$ and get something resembling a Leibniz rule: $$ D(fgu)= gD(fu)+fD(gu)-gfD(u). $$ When $E$ is the trivial bundle, i.e., $C^\infty(M,E)=C^\infty(M)$, you can set $u=1$: then, $D$ is a derivation if and only if $D(1)=0$, and above formula become the usual Leibniz rule.
