Let $M$ be a smooth manifold and $\tau M$ its second-order tangent bundle. A second-order vector field $A\in \Gamma(\tau M)$ can locally be expressed as a finite sum of operators $C^\infty(M)\rightarrow C^\infty(M)$ of the form


$$ L\_X+L\_YL\_Z,$$ 


where $L\_V:C^\infty(M)\rightarrow C^\infty(M)$ denotes the Lie derivative along the vector field $V$.


The smooth functions can be regarded as a subspace of the exterior algebra over $M$. Moreover, the Lie derivative defined on this subspace can be extended to the entire exterior algebra by requiring $L\_V$ to be a wedge-product derivation that commutes with the exterior derivative. Thus, an operator $C^\infty(M)\rightarrow C^\infty(M)$ of the form $L\_X+L\_YL\_Z$ extends to the entire exterior algebra as well.


Do second-order vector fields determine differential operators on the exterior algebra in a natural way? Given a decomposition of $A$ as a sum of operators of the form given above, the answer is yes. I'm having trouble understanding if the particular decomposition really matters.



Are sections of $\tau M$ differential operators on the exterior algebra?
If I understand correctly, you ask if any second order linear differential operator $A:C^\infty(M)\to C^\infty(M)$ of the form $A=L\_X+L\_Y L\_Z$ can be made to act on differential forms by choosing any such decomposition and applying it by Lie derivatives to differential forms. This does not work, as it depends on the choice of decomposition. Here is an explicit counterexample:


Note that $x\partial\_x+xy\partial\_y\partial\_x=(x\partial\_y)(y\partial\_x)$ hence as differential operators on functions we have $L\_{x\partial\_x}+L\_{xy\partial\_y}L\_{\partial\_x}=L\_{x\partial\_y}L\_{y\partial\_x}$, but if you apply the left and right side to $dx\wedge dy$ you find $dx\wedge dy$ on the left and $0$ on the right.


I believe, but am not sure, that there is also no other way to make such differential operators act naturally on differential forms of arbitrary degree. Maybe @petermichor or @robertbryant know a definite answer. On the other hand, there is a natural right action of scalar differential operators on top degree forms.


