I don't quite get how to use the word "jet" in differential topology.

Question. Are jets more like tangent vectors, or are they more like vector fields?

In other words, are $k$-jets elements of the order-$k$ jet bundle, or are they sections of the order-$k$ jet bundle?

Also, what do we call the other thing? For example, if "$k$-jet" means a section of the order-$k$ jet bundle, then what do we call the elements of the order-$k$ jet bundle? And vice versa.


1 Answer 1


Given smooth manifolds $M$ and $N$ (without boundary), consider pairs $(f, m)$ consisting of a smooth mapping $f:M \to N$ and a point $m \in M$. Fix an integer $k \geq 0$. Two such pairs $(f, m)$ and $(\bar{f}, \bar{m})$ are said to be equivalent if $\bar{m} = m$ and there agree the Taylor polynomials of order $k$ of $f$ and $\bar{f}$, in any local charts centered at $m$ and $f(m)$. (One has to check that this defines an equivalence relation.) A $k$-jet is defined to be an equivalence class $[f, x]$. It is often written with some notation such as $j^{k}f(x)$. The space of $k$-jets of mappings from $M$ to $N$ is written $J^{k}(M, N)$.

The usual usage of $k$-jets therefore refers to an element rather than to a section. However, one can also refer to the $k$-jet of a smooth mapping $f$, meaning $j^{k}f(x)$ for all $x$ in the domain of definition of $f$. However, one has to keep in mind that not every section arises in this way: There are well-defined source and target maps from $J^{k}(M, N)$ to $M$ and $N$, respectively, associating with $[f, x] \in J^{k}(M, N)$ the source, $x \in M$, and the target, $f(x) \in N$. When $N$ has linear structure (for example $N = \mathbb{R}^{n}$), the space $J^{k}(M, N)$ can be given a structure as a vector bundle over $M$. Note, however, that a section $s$ of a such a jet bundle need not have the form $s(x) = j^{k}f(x)$ for some smooth mapping $f$ (in general $s(x) = [f, x]$, but the $f$ depends on the $x$). Such sections are called holonomic. They are characterized by being tangent to a certain canonically defined distribution on $J^{k}(M, N)$.

The basic example is the case of $1$-jets. A $1$-jet of a smooth mapping $f:M \to \mathbb{R}$ carries the information of the values, $f(x)$, of $f$, and the values of its differential, $df(x)$. So a section of the bundle $J^{1}(M, \mathbb{R}) \to M$ is more like a one-form than it is like a vector field. In this sense, jets are more like forms than they are like vector fields (they transform covariantly rather than contravariantly). A choice of local coordinates, $x^{1}, \dots, x^{m}$, on $M$ determines induced local coordinates $(x^{1}, \dots, x^{m}, z, y_{1}, \dots, y_{m})$ on $J^{1}(M, \mathbb{R})$ defined by $x^{i}(j^1f(p)) = x^{i}(p)$, $z(j^{1}f(p)) = f(p)$, and $y_{i}(j^{1}f(p)) = \tfrac{\partial f}{\partial x^{i}}(p)$. The kernel of the one-form $dz - y_{i}dx^{i}$ does not depend on the choice of local coordinates; it is a contact distribution on $J^{1}(M, \mathbb{R})$. A section of $J^{1}(M, \mathbb{R}) \to M$ is holonomic if and only if it is tangent to this contact distribution.

  • $\begingroup$ Thanks for the answer; your enthusiasm for the material definitely shines through! $\endgroup$ Jun 20, 2017 at 8:57

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.