Global analysis, analysis on manifolds Definition

Jet

Builds on Smooth manifold Smooth function Partial derivative Taylor polynomial Germ of a smooth function

Definition

Let $M$ and $N$ be smooth manifolds, let $p \in M$ and let $k \ge 0$ be an integer. Let $f$ and $g$ be smooth maps into $N$ , each defined on an open neighbourhood of $p$ in $M$ . Then $f$ and $g$ have the same $k$ -jet at $p$ if both conditions hold.

$f(p) = g(p)$ .
Take a chart $x$ of $M$ around $p$ and a chart $y$ of $N$ around $f(p)$ . Then $y \circ f \circ x^{-1}$ and $y \circ g \circ x^{-1}$ have equal partial derivatives of every order $1, \dots, k$ at $x(p)$ .

This is an equivalence relation. The $k$ -jet of $f$ at $p$ , written $j^k_p f$ , is the equivalence class of $f$ . The set of all $k$ -jets at $p$ of maps from $M$ to $N$ is written $J^k_p(M, N)$ .

Remarks

Condition 2 does not depend on the charts, by the chain rule. Indeed, the derivatives up to order $k$ of a composite depend only on those of its factors. The jet $j^k_p f$ depends only on the germ of $f$ at $p$ . For $M = \R^n$ and $N = \R$ , the $k$ -jet of $f$ at $p$ is the same data as its Taylor polynomial of degree $k$ at $p$ .

Example: on $\R$ , $f(x) = \sin x$ and $g(x) = x$ have the same $2$ -jet at $0$ . Indeed $f(0) = g(0) = 0$ , $f'(0) = g'(0) = 1$ and $f''(0) = g''(0) = 0$ .

Non-example: $f(x) = e^x$ and $g(x) = 1 + x$ have the same $1$ -jet at $0$ but not the same $2$ -jet. Condition 2 fails for order $2$ , since $f''(0) = 1$ and $g''(0) = 0$ .

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Tag	`TRKEGEZGGZ`
URI	`m2://TRKEGEZGGZ`
Kind	definition
Writing	W0 stub
MSC	58A20
Defines	Jet
Uses	Smooth manifold, Smooth function, Partial derivative, Taylor polynomial, Germ of a smooth function

Jet

Definition

Remarks

Reading path

Further reading

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