Take the 2-minute tour ×

Consider a smooth vector bundle $\pi: E\rightarrow M$, the associated infinite jet bundle $J^\infty(\pi)$, and evolutionary vector fields $\partial_\varphi = \sum_{i,\sigma}(D_\sigma\varphi^i)\frac{\partial}{\partial u^i_\sigma}$. Here $D_\sigma$ is the composition of total derivatives corresponding to the multi-index $\sigma$. As is well known, these are the vector fields which leave the Cartan distribution invariant and are vertical.

The question is this: what does the (local) flow of such a vector field look like? (Recall that the flow of a vector field $V$ is a map $\theta$ from a subset of $\mathbb{R} \times M$ to $M$ such that $\left.\frac{d}{dt}\right|_{t=0}\theta(\cdot,x) = V_x$ for all $x$.) I have read at various places (e.g. Symmetries and Conservation Laws for Differential Equations of Mathematical Physics by Krasil'shschik and Vinogradov, and here) that in the case of evolutionary vector fields, these flows are sections of the bundle $\pi$, i.e. smooth maps $s: M \rightarrow E$ such that $\pi\circ s = \text{id}$. (Thus these evolutionary vector fields gain the interpretation as specifying the evolution of sections of the bundle.)

Can anyone explain how this works? Why are these flows in fact sections of the bundle?

share|improve this question

2 Answers 2

Suppose you have a vertical vector field $\varphi^i \frac{\partial}{\partial u^i}$ just on the total space $E$ of the bundle. Suppose that, for simplicity, $\varphi^i$ are independent of $u$. Then its flow is just $\theta:\mathbb{R}\times E \to E$, given by $(t,x,u)\mapsto (x,u+t\varphi(x))$. The evolutionary vectorfield $\partial_\varphi$ is the prolongation of the one I defined to $J^\infty(E)$. The prolongation will commute with the integration of the vector field to a flow, hence the flow of $\partial_\varphi$ will be the prolongation of the flow of $\varphi^i \frac{\partial}{\partial u^i}$. In the case that $\varphi^i$ depend on $u$, the only thing that changes is the formula for $\theta$, it will not be as simple, since one has to integrate an ODE for each $x\in M$.

One way to get a section from this kind of flow is to consider the zero section $M\to E$ and compose it with $\theta(1,\cdot)$. This will give a section $M\to E$, given by $x\mapsto (x,\phi(x))$, at least for the example I gave above. I don't know for sure that this is the way you were alluding to, but it sounds plausible.

share|improve this answer
up vote 0 down vote accepted

I believe I have found the answer. I think it works like this, but I still have to verify it. In case of any other people that might have the same question, I will outline it here. It relies on the following proposition (which can be found in, for example, the book I cited in my question):

Proposition. Let $\mathcal{C}$ be the Cartan connection on $J^\infty(\pi)$ and denote with $j^\infty(s)$ the infinite jet of some section $s$. A submanifold of $J^\infty(\pi)$ is a maximal integral manifold of $\mathcal{C}$ if and only if it is the graph of $j^\infty(s)$ for some section $s$.

Now take some evolutionary vector field $\partial_\varphi$. Then $\partial_\varphi$ is an infinitesimal automorphism of the Cartan distribution, i.e. $[\partial_\varphi, X] \in \mathcal{CX}(\pi)$ whenever $X \in \mathcal{CX}(\pi)$ (where $\mathcal{CX}(\pi)$ is the space of vector fields whose values lie in $\mathcal{C}$). This implies (I think) that its flow $F$ leaves $\mathcal{C}$ invariant, i.e. if $\theta \in J^\infty(\pi)$ then $(F_t)_*\mathcal{C}\_\theta = \mathcal{C}_{F_t(\theta)}$. Therefore it maps maximal integral manifolds to maximal integral manifolds.

Take a section $s$. Then the argument above implies that the image of the map $x \mapsto F_t(j_x^\infty(s))$ is another maximal integral manifold of $\mathcal{C}$. Therefore, it comes from some other section $s_t$, i.e. there is a section $s_t$ such that $F_t(j_x^\infty(s)) = j^\infty_x(s_t)$ for all $x$. In this way, the flow $F$, when restricted to the jet of some section $s$ becomes (the jet of) another section $s_t$. Moreover, denote with $F_{t,\sigma}^i$ the $\sigma,i$-component of $F_t$, where $\sigma$ is a multi-index. Then the equation that determines that $F$ is a flow of $\partial_\varphi$ is $\left.\frac{d}{dt}\right|\_{t=0}F^i_{t,\sigma}(\theta) = (\partial_\varphi)^i_\sigma(\theta) = (D_\sigma\varphi^i)(\theta)$ for $\theta \in J^\infty(\pi)$. When restricted to $j^\infty(s)$, this finally becomes

$$\left.\frac{d}{dt}\right|\_{t=0}\frac{\partial^{|\sigma|}s_t^j}{\partial x^\sigma}(x) = (D_\sigma \varphi^i)(j^\infty(s)),$$

which is the equation that inspired my question.

share|improve this answer
This is "virtually" correct reasoning as Vinogradov once pointed out to me. In reality the flow $F_t$ does not exist for most generating functions $\varphi$. And it exists neither on the level of the infinite dimensiona mainofold $J^\infty$ nor on the space of of sections $s$. The first statement is a result due to Chetverikov, who proved that vector fields on $J^\infty$ posses a local flow iff they are of bounded shift, and the second statement follows from the non uniqueness of solutions to evolutionary equations. See for example mathoverflow.net/questions/82408 –  Michael Bächtold Oct 15 '12 at 14:37

Your Answer


