Coordinate-free description of the kernel of the jet projection?

Question

Given a vector bundle $E \to M$ we can take the $r$-jet prolongation $J^r E \to M$. This is equipped with a vector bundle morphism called the jet-projection $\pi^r_{r-1}: J^r E \to J^{r-1} E$. It is often stated that $\ker(\pi^r_{r-1}) \cong S^r T^* M \otimes E$. It is easy to see why this is true if we pick coordinates, but I have not been able to find a nice coordinate-free proof of this. Does anyone know of a nice proof?

Answer 1

Note that $J^r(E)=J^r(\bf R) \otimes E$ : two sections of $E$ have the same $r$ jets if there coordinates in some (any) local trivialisation have. So you just have to check it when $E$ is the trivial rank one bundle over $M$. The obvious map from $\ker \pi _{r-1}^r$ to $S^rT^*M$ is injective : a $r$-jet is $0$ iff its $r-1$ art and it last part are. It is also surjective, but for this you need coordinates, I am afraid.