Given a vector bundle $E \to M$ with a corresponding $k$-th jet bundle $J^kE \to M$, denote by $j^k : \Gamma(E) \to \Gamma(J^kE)$ the $k$-th jet prolongation $(k \in \mathbb{N} \cup \{0\})$ and recall that a section $\sigma \in \Gamma(J^kE)$ is holonomic if it is in the image of $j^k$. Is there an easy way to see why $\Gamma(J^kE)$ is generated as a $C^\infty(M)$-module by its holonomic sections?

It's fairly clear that pointwise, this is true, because if given a collection of jet-data in $J^k_xE$ at a point $x \in M$, I can always locally integrate it back to a single function using, say, polynomials. But I think the generation as a $C^\infty(M)$-module is somewhat stronger.

I remember having encountered a similar problem like this, asking whether Hamiltonian vector fields of a symplectic manifold generate all tangent sections, and apparently this could be solved using tubular neighbourhoods, which I am not fluent with. Might something like that yield results here as well, and is a kind, jet-fluent soul available to explain it to me? Thank you!

(Disclaimer: I first posted this on math.stackexchange, and due to low interest, I thought it might be more appropriate here. I deleted the math.stackexchange post.)

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    $\begingroup$ Using a partition of unity, you can reduce to the trivial vector bundle over a subset over $\mathbb R^n$, where polynomials do the job: There is a basis of the dual bundle of the $k$-th jet bundle indexed by multiindices $(i_1,\dots,i_m)$ with $1\le i_1\le \cdots\le i_m\le n$ and $0\le m\le k$ given by taking the derivative $\partial_{i_1}\cdots\partial_{i_m}$ of a jet, and evaluating these on the holonomic sections $j^k(x^{i_1}\cdots x^{i_m})$ gives an upper-triangular matrix, so these sections form a basis of the $k$-th jet bundle. $\endgroup$ – Bertram Arnold Feb 19 at 13:01
  • $\begingroup$ I see now, it is actually pretty simple. Thanks! I only now realized with your comment that jet sections can locally still be viewed as Taylor series with varying coefficients, and then it indeed is not that difficult to see. $\endgroup$ – Lukas Miristwhisky Feb 20 at 10:27

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