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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.
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  • $\begingroup$ Please don't use latex commands in place of italics, the asterisk is meant for that (e.g., *this* would give this). $\endgroup$
    – Kyle Kanos
    Apr 14, 2015 at 15:03
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    $\begingroup$ This seeems to be a pure math question. $\endgroup$ Apr 14, 2015 at 15:06
  • $\begingroup$ Sorry about the latex Kyle, I'm not very proficient with computers. I don't think it belongs on the maths forum personally. I am after a physics based answer primarily! $\endgroup$ Apr 14, 2015 at 15:34
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    $\begingroup$ 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 $d^V$? $\endgroup$
    – user2612
    Apr 14, 2015 at 20:17
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    $\begingroup$ 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). $\endgroup$
    – user2612
    Apr 14, 2015 at 21:28

1 Answer 1

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$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).

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  • $\begingroup$ thank you for your answer, is $V_u\pi$ the same as $V_uE$ in this case? $\endgroup$ Apr 16, 2015 at 9:06
  • $\begingroup$ yes, probably, I should write the splitting above as $T_u E = V_u E + H_u E$. $\endgroup$
    – valeri
    Apr 16, 2015 at 9:33
  • $\begingroup$ or. stack to authors notations - i.e., use $V_u\pi$ everywhere, sorry for confusion! $\endgroup$
    – valeri
    Apr 16, 2015 at 9:38
  • $\begingroup$ I can't begin to thank you enough, I have been here three days without a clue! :D $\endgroup$ Apr 16, 2015 at 9:44
  • $\begingroup$ Janet the Physicist welcome! $\endgroup$
    – valeri
    Apr 16, 2015 at 10:05

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