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Feb 16, 2014 at 22:06 history edited Andrés E. Caicedo
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Jun 18, 2013 at 0:23 comment added S. Carnahan @Ricardo: Yes, you're correct that if you have a global trivialization of your vector bundle (i.e., taking jets with values in a vector space), you get an additive structure. However, without a fixed trivialization, you don't get linear maps on jets of order greater than one. For example, the transition between the north-pole and south-pole stereographic projections on $S^1$ produces a quadratic term on tangent 2-jet coordinates: if $z = 1/w$, then $d^2z = -\frac{2}{w^3} (dw)^2 - \frac{1}{w^2}d^2w$.
Jun 17, 2013 at 1:17 comment added Ricardo Andrade @Scott: Please correct me if I am wrong, but I think that the bundle of jets on a manifold with values in a vector space does have a canonical vector bundle structure (given by adding jets). In fact, the transition functions are linear, as pre-composing with a smooth map gives a linear map on jets. It seems that more generally the jets of sections of a vector bundle themselves form a vector bundle.
Jun 16, 2013 at 6:44 answer added S. Carnahan timeline score: 4
Jun 16, 2013 at 3:54 comment added S. Carnahan The $k$-jet bundle isn't a vector bundle for $k \geq 2$. The fibers are vector spaces, but the transitions are not linear maps. In particular, it cannot be dual to a co-jet vector bundle.
Jun 14, 2013 at 18:45 comment added Steven Gubkin Any update on this?
Jun 14, 2013 at 9:45 history edited Sebastien Palcoux
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Sep 19, 2012 at 9:12 answer added Peter Michor timeline score: 6
Apr 6, 2011 at 15:36 comment added Deane Yang I don't think Michael is mistaken. His comments seem on target. You might also want to take a look at Griffiths' work on doing calculus of variations with exterior differential systems.
Apr 5, 2011 at 6:23 comment added Michael Bächtold Maybe I'm mistaken but if $\Gamma(N)$ is the module of sections of a vector bundle over $M$ then $\Gamma(J^K(N)^*$ is "the same" as the module of linear differential operators from $\Gamma(N)$ to $C^\infty(R)$. I.e. differential operators associating a function to a section. There is a differential d on these objects (if you allow the order k to change) called the Spencer complex (see for example the articles of Spencer on overdetermined system or the book cited by Dean or books by Vinogradov and Lychagin). But i don't know if this d is the one you are looking for.
Apr 5, 2011 at 0:31 comment added Toby Bartels I'm pretty much restricting to the case where $M$ has $1$ dimension and $N$ is $\mathbb{R}$ (the trivial line bundle), but I would like to understand at least higher-dimensional $M$ (which is the usual context for exterior differential forms). So far, I'm working things out for myself, but I'd hate to think that this is new.
Apr 5, 2011 at 0:28 comment added Toby Bartels For jets, you can see Wikipedia en.wikipedia.org/wiki/Jet_%28mathematics%29 but the basic idea is that a $k$-jet from $M$ to $N$ at a point $p$ is an equivalence class of smooth functions to $N$ from a neighbourhood of $p$, where two such functions are equivalent if their derivatives up to and including order $k$ are the same. If $N$ is a vector bundle and we restrict to jets that map $p$ to $0$, then these form a vector space. The $k$-cojet space at $p$ is the dual space, and these join to form a vector bundle whose sections I call the degree-$k$ cojet differential forms.
Apr 4, 2011 at 16:07 comment added Michael Bächtold Could you add the definition for cojet? Or give a reference?
Apr 4, 2011 at 1:43 comment added Deane Yang As far as I know, using differentials do not provide any advantage when computing higher derivatives by hand. However, it does work beautifully in designing very simple recursive algorithms for computing higher derivatives of functions in software. Just search for descriptions of "automatic differentiation". It is a lot of fun to implement this using, say, C++ templates.
Apr 4, 2011 at 0:18 comment added Toby Bartels But I'm not sure what exactly I calculated there. At the moment, to be sure that all of my steps are rigorous, I must divide by $\mathrm{d}{x}$ to get $\mathrm{d}{y}/\mathrm{d}{x} = 3x^2 - 3$, then differentiate that to get $\mathrm{d}(\mathrm{d}{y}/\mathrm{d}{x}) = 6x \,\mathrm{d}{x}$. And the test isn't quite whether this is positive or negative; I must divide by $\mathrm{d}{x}$ again first. So this application has lost much of the beauty of the differential-based approach.
Apr 4, 2011 at 0:13 comment added Toby Bartels So here's a problem from freshman calculus: Given that $y = x^3 - 3x$, for which values of $x$ does $y$ reach a local maximum or minimum? We compute $\mathrm{d}{y} = (3x^2 - 3) \,\mathrm{d}{x}$, set this to $0$ and solve for $x$ to get two critical points, which we test using the second derivative. Starting from $\mathrm{d}y$ above, we compute $\mathrm{d}^2{y} = 6x \,(\mathrm{d}{x})^2 + (3x^2 - 3) \,\mathrm{d}^2{x}$. Plugging in $x = \pm{1}$, we get $\mathrm{d}^2{y} = \pm{6x^2} \,(\mathrm{d}x)^2$. So $y$ has a local minimum when $x = 1$ and a local maximum when $x = -1$.
Apr 3, 2011 at 23:29 comment added Toby Bartels I didn't want to get into the context, in case people started discussing that instead. But there is a place to discuss that, in this old thread. I want to understand the theory behind a differentials-based approach to teaching freshman calculus, as advocated by Dray and Manogue (pdf). For first derivatives, I know how to make everything that they write formally correct. But what about higher derivatives?
Apr 3, 2011 at 22:54 comment added Deane Yang If you want to see a modern approach to the formal theory of PDE's (i.e., a cohomological approach to the Cartan-Kahler theorem, which was developed originally using exterior differential systems), look at the work of Hubert Goldschmidt, which builds on work by Spencer, Quillen, Guillemin, and Guillemin-Sternberg. See, for example, Chapters IX and X of the book "Exterior differential systems" by Bryant, Chern, Goldschmidt, and Griffiths.
Apr 3, 2011 at 19:17 comment added Deane Yang Could you provide an example of a calculation that you would like to be able to do or do more easily using such a calculus?
Apr 3, 2011 at 18:55 history asked Toby Bartels CC BY-SA 2.5