What does the "$k$-th tangent mapping" refer to?

47 Views Asked by At

I'm reading the section on jet bundles in the book Manifolds of Differentiable Mappings. The author defines a k-jet from $X$ to $Y$ as an equivalence class of pairs $$(f : X \to Y, x \in X),$$ where two pairs $(f, x)$ and $(f', x')$ are equivalent if their Taylor series agrees up to the $k$-th order.

The author later remarks:

For formalists there is another definition of the equivalence relation $\sim_k$: $[f, x]_k = [f', x']_k$ iff $x = x'$ and $T_x^k f = T_x^k f'$ ($T^k$ denotes the $k$-the tangent mapping).

What is the "$k$-th tangent mapping" that the author refers to? And how is it equivalent to the Taylor series definition?

1

There are 1 best solutions below

0
On BEST ANSWER

Given $f:X\to Y$, the tangent map is $Tf:TX\to TY$ (also called the push-forward or differential), the second tangent map is $T^2f=T(Tf):T^2X\to T^2Y$, and so on, for each $k\geq 1$, we have $T^kf:T^kX\to T^kY$. Each $T^kX$ is a fiber bundle over the base $X$, and over this base, we have $T^kf$ restricts to a fiber-preserving map over $f$, i.e $T^kf$ maps the fiber $T^k_xX$ over $x\in X$ to the fiber $T^k_{f(x)}Y$ over $f(x)\in Y$. The information contained in this map is essentially all partials up to $k^{th}$ order.