We know this fact that the first jet Bundle $J^1M$ is diffeomorphic with $T^\*M×\mathbb{R}$.i.e,
($J^1M=T^\*M×\mathbb{R}$)


Is there something like this identity for higher jet bundle $J^kM$?


I editted my question after comment of Michael Murray



looking for an identity for higher jet bundle $J^kM$?
First, note that there is a natural exact sequence of bundle maps
$$
Sym^kT^\*M \rightarrow J^kM \rightarrow J^{k-1}M
$$
but there is no natural splitting of this sequence. You can, however, do it by choosing a Riemannian metric and defining a map $J^{k-1}M \rightarrow J^kM$ by extending a $(k-1)$-jet to the $k$-jet whose symmetrized $k$-th order covariant derivatives all vanish at the point. From this you can conclude that there exists a bundle isomorphism
$$
J^kM = \mathbb{R} \oplus T^\*M \oplus Sym^2T^\*M \oplus\cdots \oplus Sym^kT^\*M.
$$
This is all explained in the algebraic geometric context in section 2.4 of [these notes](http://math.stanford.edu/~vakil/files/jets.pdf). My guess is that it is all explained in the book "The Geometry of Jet Bundles" by Saunders.


