# On the definition of jets

I have some problems with the definition of jets and it would be great if someone could help me here:

In many books it is written, that the $r-th$ order jet $j^r_xf$ of a smooth function $f:M \rightarrow N$ between smooth manifolds 'depends only on the germ of $f$ at $x'$.

What does it mean?

First I thought this means that all functions in the equivalence class $j^r_xf$ have the same germ at $x$, but this is wrong as the following counterexample shows:

Let

$$\begin{eqnarray} f: \mathbb{R} &\rightarrow& \mathbb{R}^2 \\ x &\mapsto& (x,x) \end{eqnarray}$$

and

$$\begin{eqnarray} g: \mathbb{R} &\rightarrow& \mathbb{R}^2 \\ x &\mapsto& (x,\sin(x)) \end{eqnarray} .$$

Then $f$ and $g$ have the same first order jet at zero, that is $f,g \in j^1_0f$, but they don't define the same germ at $x$ since there is no neighbourhood of $(0,0)$ in $\mathbb{R}^2$ such that $f$ equals $g$ on that neighbourhood.

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## migrated from meta.math.stackexchange.comJan 16 '12 at 10:04

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The phrase that "the $r$-jet depends only on the germ" means that

If two functions $f,g$ have the same germ at $x$, the two functions will have the same $r$-jet.

It does not mean what you interpreted, which is the converse of the statement, and can be written as "the germ depends only on the $r$-jet", or "if two functions have the same $r$-jet, they have the same germ." This statement, as your example shows, is false.

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Ok. Thanks for the fast help. Problem solved – Mark Neuhaus Jan 16 '12 at 10:30

Jets are essentially the same thing as coefficients of Taylor expansions, but done in a coordinate free way. To say that something depends only on the germ of $f$ at $x$ means that, if $f$ and $g$ agree on some neighborhood of $x$, then that something will also agree. A less formal way of saying that a property depends only on the germ of $f$ at $x$ is to say "the property is local."

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Ok, clearly if $f$ and $g$ ave the same germ, then they have the same $r$-jet. But as my counterexample shows this is just an implication, not an equivalence. – Mark Neuhaus Jan 16 '12 at 10:23
@Mark Neuhaus: Of course not. Just like the value of $f(x)=x^2$ depend only on $x$, there is not an equivalence between values of $x$ and values of $f(x)$. A closer analogy would be that the germ of a function is NOT determined by the value of it's first derivative at a point. "A only depends on B" means (1) changing A will change B, and (2) knowing A is enough to determine B. This is not like "if and only if", which is where I assume your confusion comes in? – Aaron Jan 16 '12 at 10:33