Timeline for How to visualize the dual objects of jets of functions?
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2016 at 17:24 | comment | added | Alex M. | @HeleneSigloch: No, unfortunately it doesn't. I've been around for a while now to know that SE communities prefer questions to be closed by accepting an answer. Unfortunately, this cannot be done here. Thanks for asking, though. | |
| Apr 12, 2016 at 15:08 | comment | added | Helene Sigloch | @AlexM: Is your question answered now? In this case it would be nice if Willie Wong posted the answer as an answer and if you accepted it. | |
| Mar 14, 2016 at 17:03 | comment | added | Willie Wong | Note however the above definition requires knowing up to the $k-1$ jets of the vector fields $X_*$. If you want the definition to be tensorial in the $X$'s, however, you start running into problems. (By the inverse function theorem, if $f$ is a scalar function on $M$ such that $\mathrm{d}f_p\neq 0$ then there exists a coordinate system $\{x^1, \ldots, x^m\}$ near $p$ such that $\partial_1 f \equiv 1$ in the neighborhood. A modification of this argument tells you that any purely top order, with $k \geq 2$, object can only act trivially. Note that this is already a problem for $f:R\to R$.) | |
| Mar 14, 2016 at 16:45 | comment | added | Willie Wong | @AlexM.: given tangent fields $X_1, \ldots, X_k$, you can obviously just consider the mapping $f \mapsto X_1(X_2(\cdots(X_k(f))\cdots))$ which lifts to a $\mathbb{R}$-linear mapping from $J^k$ to scalars. If $f$ is a section of a fibre bundle you can treat the action as that of the covariant Lie derivative (assuming you have a connection). If you don't want to worry about the ordering, you can sum/average over all permutations of indices. | |
| Mar 14, 2016 at 16:40 | comment | added | Saal Hardali | @WillieWong Ah I see. I completely agree. The only thing you can define coordinate independently is "differential operator of degree at most $k$" where $k$ is some natural number. Under a suitable coordinate transformation you can turn any "pure" $k$ differential operator to a "non-pure" one. | |
| Mar 14, 2016 at 16:37 | comment | added | Willie Wong | @SaalHardali: I didn't express myself clearly. The problem is that $J^k$ is not naturally a vector bundle over $J^{k-1}$; it is only an affine bundle. (See Prop 12.11 in the book that Travis linked to in his comment.) Were $J^k$ a vector bundle over $J^{k-1}$ you can easily define "pure" $k$-th order derivative operators, which seems to be what the OP is after. The whole point of the development of the theory of jet bundles is precisely that such a naive generalisation from the coordinate-centric development on Euclidean spaces is impossible. | |
| Mar 14, 2016 at 15:49 | comment | added | Alex M. | @SaalHardali: You mean, if I have a $k$-contravariant tensor $X_1 \otimes \dots \otimes X_k$, then its natural pairing with $J^k (f)$ is given by $(X_1 \dots X_k) (f)$ where the fields derive successively from right to left? It is that easy? | |
| Mar 14, 2016 at 14:31 | comment | added | Alex M. | @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? | |
| Mar 14, 2016 at 14:13 | comment | added | Saal Hardali | @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). | |
| Mar 14, 2016 at 14:00 | comment | added | Willie Wong | 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. | |
| Mar 14, 2016 at 13:21 | comment | added | Travis Willse | 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 | |
| Mar 14, 2016 at 13:14 | comment | added | Ben McKay | 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. | |
| Mar 14, 2016 at 12:59 | history | asked | Alex M. | CC BY-SA 3.0 |