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  Jets and vertical differential

+ 1 like - 0 dislike
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For a vector bundle $(E,\pi, M)$ let $\phi :M\mapsto E$ be a section of $\pi $, $x\in M$ and $u=\phi (x)$. The vertical differential of the section $\phi$ at point $u\in E$ is the map: \begin{equation} d^v_u\phi :T_uE\mapsto \mathcal V_u\pi \end{equation} In coordinates on $E$ $(x^i,u^\alpha)$ we write; \begin{equation} d^v_u\phi =\bigg(du^\alpha -\frac{\partial \phi ^\alpha }{\partial x^i}dx^i\bigg)\otimes \frac{\partial }{\partial u^\alpha} \end{equation} Apparently it is obvious from this that $d^v_u\phi$ depends only on the first order jet space $j^1_x\phi$.

What is $\mathcal V_u\pi$ in this case? It is clearly related to the jet manifold $J^1\pi$ whose total space is the product $T^*M\otimes _E\mathcal V\pi$ . But I don't really understand what an associated vector bundle is!

References:

  1. C.M. Campos, Geometric Methods in Classical Field Theory and Continuous Media, pages 24-25.


This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user Janet the Physicist

asked Apr 14, 2015 in Mathematics by Janet the Physicist (15 points) [ revision history ]
edited Aug 12, 2015 by Dilaton
Janet the Physicist. Looks like your bundle is endowed with the connection, i.e., family of "horizontal" subspaces, while the vertical differential is the projection of $T_uE$ to vertical fibers $V_\pi$ of the bundle. How else you can define projection $d^V$?

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user user2612
i.e., $V_u \pi$ is the vector space tangent to the fiber of the bundle $\pi: E\to M$. In the book local coordinates $(x,u)$ provide $T_u E$ with the splitting $V_u\pi + H_uE$, where (horizontal) subspace is identified with $T_uM$. so that taking vertical differential of a section equals the projection of the ordinary differential $d\phi$ to $V_uE$ along this $H_u$. Jet manifold $J^1_\pi$ has projections on $E$ and $M$ which makes it bundle, it is called associated since its structure group is defined by the structure group of the initial bundle $\pi$, see (3.5).

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user user2612
@user2612 So the tangent space to a point $u\in E$, $T_uE$ is the sum of vertical and horizontal spaces $\mathcal V_u\pi+H_uE$. The horizontal space in this case are $H_uE$ which is also (?) identified with $T_uM$. Taking the vertical differential of a section of an element in $T_uE$ maps to the vertical space of the tangent space? Thank you so much for your help :)

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user Janet the Physicist
Yes, you are welcome! I will put this in more details below as an answer.

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user valeri

1 Answer

+ 1 like - 0 dislike

$V_u\pi $ is the vector space tangent to the fiber $\pi^{-1}(x)$ of the bundle $\pi : E\to M$ going through $u$. In the book local coordinates $(x',u')$ provide $T_u E$ with the splitting $V_u \pi + H_u E$, where (horizontal) subspace is identified with $T_u M$. We may think of the horizontal space as the tangent to locally constant sections $\phi: M\to E$ going through the point $u$, i.e., $\phi(x')\equiv u$. Now, taking vertical differential of a section $\phi$ equals the projection of the ordinary differential $d\phi$ to $V_u E$ along this $H_u$. Jet manifold $J^1_π$ has projections on both $E$ and $M$ which makes it bundle, it is called associated since its structure group is defined by the structure group of the initial bundle $\pi$, see (3.5).

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user valeri
answered Apr 16, 2015 by valeri (10 points) [ no revision ]
thank you for your answer, is $V_u\pi$ the same as $V_uE$ in this case?

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user Janet the Physicist
yes, probably, I should write the splitting above as $T_u E = V_u E + H_u E$.

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user valeri
or. stack to authors notations - i.e., use $V_u\pi$ everywhere, sorry for confusion!

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user valeri
I can't begin to thank you enough, I have been here three days without a clue! :D

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user Janet the Physicist
Janet the Physicist welcome!

This post imported from StackExchange Mathematics at 2015-08-12 17:47 (UTC), posted by SE-user valeri

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