I need some help for establishing a connection between two definitions of $k$-jets:

Algebraic Definition: Let $E\rightarrow M$ be a smooth vector bundle and define the ideal of $\Gamma(E)$: $$I_p(M):=\{f\in \Gamma(E): f(p)=0\}.$$ Here $\Gamma(E)$ is the $C^\infty(M)$-module of smooth section over $E$. Then $$I^{k+1}_p(M):=\{\sum_{finite} f_1\cdots f_{k+1}: f_i\in I_p(M)\},$$ is again an ideal of $\Gamma(E)$. Since $\Gamma(E)$ is a $C^\infty(M)$-module we might consider the submodule $$I^{k+1}_p(M)\cdot \Gamma(E):=\{\sum_{finite} f_iu_i: f_i\in I_p^{k+1}(M), u_i\in \Gamma(E)\},$$ and consequently the quotient, $$J^k(E)_p:=\Gamma(E)/Z^k_p(E),$$ where $Z^{k+1}_p(E):=I^{k+1}_p(M)\cdot \Gamma(E)$. The class of $f\in \Gamma(E)$ is denoted by $j^kf(p)$ and is called a $k$-jet of $f$ in $p$.

Geometrical Definition: I also have a geometrical definition of $k$-jets of a section $f\in \Gamma(E)$ in $p$: $$j^kf(p)=\{g\in \Gamma(E): \partial^\alpha f=\partial^\alpha g, \forall |\alpha|\leq k\},$$ where I'm using multi-index notation above.. In other words this geometrical definition says all derivatives of $f$ and $g$ coincide up to order $k$.

Can anyone explain me how the above definitions are related?

  • $\begingroup$ I only know half of the expressions used above (in particular I do not know what the range of $f\in\Gamma(E)$ is), but this looks like an application of product rules to me. Except for the fact that I would prefer $Z_{p}^{k}(E):=I_{p}^{k+1}\cdot\Gamma(E)$ but I might be wrong there. $\endgroup$ – M. Luethi Nov 22 '13 at 12:20
  • $\begingroup$ (Aside: while you can use up to five tags, you don't have to use five. I removed some of the less relevant ones.) As for the relation: the two definitions agree. $\endgroup$ – Willie Wong Nov 22 '13 at 12:26
  • $\begingroup$ Before I write an answer: is it clear to you that the algebraic definition implies the geometric definition? (Think local coordinates). $\endgroup$ – Willie Wong Nov 22 '13 at 12:30
  • $\begingroup$ Also, are you familiar with the similar result if $E$ is the trivial $\mathbb{R}$ bundle? (So that sections of $E$ are just functions from $M \to \mathbb{R}$?) $\endgroup$ – Willie Wong Nov 22 '13 at 12:39
  • $\begingroup$ @M.Luethi your right as to the notation.. $\endgroup$ – PtF Nov 22 '13 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.