TITLE: How to look at the Lie derivative as a partial differential operator?
QUESTION [12 upvotes]: In Gromov's book Partial Differential Relations, a partial differential operator (PDO) is defined as follows. Given a fibration $X\to V$ and a vector bundle $G\to V$ over a manifold $V$, a PDO of order $k$ is a map $D:\Gamma(X)\to \Gamma(G)$, satisfying $D(x)=\Delta\circ j^k_x$ for every $x\in \Gamma(X)$, where $\Delta:X^{(k)}\to G$ is a bundle map. Here $X^{(k)}$ is the $k^{\text{th}}$ jet bundle of sections of $X\to V$ and $j^k_x$ is the $k^{\text{th}}$ jet of $x$.
Now, say $G$ is some tensor bundle on $V$ and $X=TV$. Then for a fixed section $\omega\in\Gamma(G)$, we have an operator,
\begin{align*}
D_\omega:\Gamma(TV)&\to \Gamma(G)\\
x&\mapsto \mathcal{L}_x(\omega)
\end{align*}
where $\mathcal{L}_x(\omega)$ is the Lie derivative of the tensor $\omega$ along the vector field $x$.
How do I see that this $D_\omega$ is indeed a partial differential operator of order 1, as defined by Gromov? In particular, I would like to find out the bundle map $\Delta=\Delta_w$ such that $D_\omega(x)=\Delta\circ j^1_x$.
For the case $G=\Lambda^k T^*V$, I can use the Cartan's formula and get, $$D_\omega(x)=\mathcal{L}_x(\omega)=d\iota_x(\omega)+\iota_xd\omega$$ I do know that the exterior derivative map $d$ can be seen as a PDO. But I still can't see how to get to the map $\Delta$. And I'm in complete loss when $G$ is an arbitrary tensor bundle, since in that case the Lie derivative is defined using flows.
Any help regarding this appreciated.

REPLY [2 votes]: The statement can be proven by induction.
Let $\omega_1$ and $\omega_2$ be tensors which satisfy the property. That is, the operators $X\mapsto\mathcal{L}_X\omega_1$ and $X\mapsto\mathcal{L}_X\omega_2$ are differential operators of order $1$. By the Leibniz rule,$$\mathcal{L}_X(\omega_1\otimes\omega_2)=(\mathcal{L}_X\omega_1)\otimes\omega_2+\omega_1\otimes(\mathcal{L}_X\omega_2),$$and it follows immediately that $X\mapsto\mathcal{L}_X(\omega_1\otimes\omega_2)$ is also a differential operator of order $1$.
You already know the statement is true for $1$-forms. It is also true for vector fields, as $\mathcal{L}_XY=[X,Y]$. Since the tensor algebra is generated by vector fields and $1$-forms, we are done.