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I work with a smooth $f: M \to \Bbb C$ and I would like to have an object mimicking the concept of "$k$-th order differential" from multivariate calculus. For various reasons that are not important here I had to rule out covariant derivatives and I am now contemplating jets. If $p \mapsto J^k _p (f)$ is the jet of $f$, what is the natural object to pair it with in order to obtain a smooth function? Warden ("Foundations of Differential Manifolds and Lie Groups") talks about $k$-th order tangent vectors, but they are quite abstract and I don't get the feel of them without passing to coordinates, which I would like to avoid in order to get a better understanding.

To be even clearer, if I have tangent fields $X_1, \dots, X_k$, is there any natural object constructed with them that pairs naturally with $J^k (f)$?

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  • $\begingroup$ Two vector fields $X$ and $Y$ have the same $k$-jet at a point $p$ if their difference vanishes to order $k$ as an operator on smooth functions, i.e. $Xf-Yf$ has vanishing $k$-jet at $p$ as a function. Once you understand function $k$-jets, you understand $k$-jets of sections of any vector bundle. $\endgroup$
    – Ben McKay
    Commented Mar 14, 2016 at 13:14
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    $\begingroup$ I'm not sure whether it satisfies the criterion of avoiding "too abstract", but see $\S$12 of the standard reference by Kolar, Michor, & Slovak: emis.de/monographs/KSM $\endgroup$
    – Travis Willse
    Commented Mar 14, 2016 at 13:21
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    $\begingroup$ What do you mean by "pair"? You mean in the linear sense? Then you have no hope. The Jet bundle is affine, but not linear; you can't hope to have a naturally well-defined linear mapping from Jets to functions. On the other hand, you can do it modulo lower order terms, in which case you are basically thinking about the principal part of a partial differential operator. $\endgroup$
    – Willie Wong
    Commented Mar 14, 2016 at 14:00
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    $\begingroup$ @WillieWong A Jet bundle of a vector bundle is naturally a vector bundle with the jet of the zero section as the zero section of the jet bundle. So I would say for any vector bundle the $k$-th jet pairs naturally with $k$ differential operators (almost by definition). $\endgroup$
    – Saal Hardali
    Commented Mar 14, 2016 at 14:13
  • $\begingroup$ @WillieWong: I think that what you say is precisely what I am after: working modulo lower order terms, the highest-order "homogeneous" term would be precisely the analogue of the $k$-th order differential from calculus - but how to formalize this? $\endgroup$
    – Alex M.
    Commented Mar 14, 2016 at 14:31

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