I am studing Ehresmann's jet bundles on manifolds and I came up with a (maybe silly) question.
In order to make it easy I skip the details of the definition and go directly to the part that I don't quite understand.
Given an $n$-dimensional manifold $M$ and two local diffeomorphisms from a nbh of $0\in\mathbb{R}^n$ to a nbh of a point $x\in M$
$$
f_1: V_1\subset \mathbb{R}^n \rightarrow U_1 \subset M\\
f_2: V_2\subset \mathbb{R}^n \rightarrow U_2 \subset M\
$$
it is said that $f_1$ and $f_2$ define the same $r$-jet at $x\in M$ if for any chart $\varphi$ around $x$ all the components of the maps $\varphi \circ f_1$ and $\varphi \circ f_2$ (which are functions from $\mathbb{R}^n$ to $\mathbb{R}$), have the same $k$-th order derivatives at $0$ for any $0 \leq k \leq r$. (In particular $f_1(0) = f_2(0) = x$).
My question is: can we change "for any chart" for "for some chart" in this definition? In other words, does the property above depend on the choice of the chart?
If $r=1$ the answer is yes, since we can interpret the condition above in terms of the differentials of the maps. Since the charts are local diffeomorphisms we can multiply and divide by their differentials and the condition for one chart implies the condition for another chart (since what we really have is that the differentials of $f_1$ and $f_2$ at $0$ coincide).
Can we do something similar for general $r$?
Best Answer
You seem to have the right idea. Just as the differential can be computed in any system of local coordinates, the collection of all partial derivatives of order $\le k$ contains the same information regardless of the charts used to compute it. Showing this requires a bit of bookkeeping, though.
For a smooth function $f:\mathbb{R}^n\to\mathbb{R}^m$, denote by $D^kf$ the collection of $k$-th partial derivatives of $f$, regarded as a map $\mathbb{R}^n\to\mathbb{R}^{A(n,m,k)}$ for a suitably defined combinatorial function $A$. We can likewise define $D^{\le k}f:=(f,D^1f,\cdots,D^kf)$ as the collection of all partial derivatives of order $\le k$.
Let $f:\mathbb{R}^n\to\mathbb{R}^m$ and $g:\mathbb{R}^m\to\mathbb{R}^l$ be smooth maps. The operator $D^{\le k}$ has an analog of the chain rule, where $D^{\le k}(g\circ f)(p)$ can be expressed explicitly in terms of $D^{\le k}f(p)$ and $D^{\le k}g(f(p))$. To avoid some of the messy combinatorics, we can use a more abstract version of this result:
Lemma: $D^{\le k}(g\circ f)$ can be expressed as $D^{\le k}(g\circ f)(p)=\mathcal{P}\left(D^{\le k}f(p), D^{\le k}g(f(p))\right)$, where $\mathcal{P}$ is a polynomial whose coefficients depend only on $(n,m,l,k)$.
This can be proven by induction on $k$: each component of $D^{k+1}(f\circ g)$ can be written as $\frac{\partial h}{\partial x^i}$ where $h$ is a component of $D^k(f\circ g)$. Using the product and chain rules, one can show that if $h$ can be written as a polynomial in $(D^{\le k}f, (D^{\le k}g)\circ f)$ , then $\frac{\partial h}{\partial x^i}$ can be written as a polynomial in $(D^{\le k+1}f, (D^{\le k+1}g)\circ f)$.
This result implies a useful corollary:
Corollary: Let $f_1,f_2:\mathbb{R}^n\to\mathbb{R}^m$ and $g:\mathbb{R}^m\to\mathbb{R}^l$ be smooth functions.If $D^{\le k}f_1(p)=D^{\le k}f_2(p)$, then $D^{\le k}(g\circ f_1)(p)=D^{\le k}(g\circ f_2)(p)$.
It is straightforward to apply this result to show that if $f_1$ and $f_2$ have the same $k$-jet at $p$ in one chart, they have the same $k$-jet in any other chart around $p$.