Flow of evolutionary vector fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:57:51Z http://mathoverflow.net/feeds/question/86198 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86198/flow-of-evolutionary-vector-fields Flow of evolutionary vector fields Sietse Ringers 2012-01-20T11:14:51Z 2012-01-26T15:46:08Z <p>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.</p> <p>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. <em>Symmetries and Conservation Laws for Differential Equations of Mathematical Physics</em> by Krasil'shschik and Vinogradov, and <a href="http://staff.www.ltu.se/~norbert/home_journal/electronic/1_3art1.pdf" rel="nofollow">here</a>) 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.)</p> <p>Can anyone explain how this works? Why are these flows in fact sections of the bundle?</p> http://mathoverflow.net/questions/86198/flow-of-evolutionary-vector-fields/86235#86235 Answer by Igor Khavkine for Flow of evolutionary vector fields Igor Khavkine 2012-01-20T17:40:42Z 2012-01-20T17:40:42Z <p>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 $x\in M$.</p> <p>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.</p> http://mathoverflow.net/questions/86198/flow-of-evolutionary-vector-fields/86727#86727 Answer by Sietse Ringers for Flow of evolutionary vector fields Sietse Ringers 2012-01-26T15:09:24Z 2012-01-26T15:46:08Z <p>I believe I have found the answer. I <em>think</em> 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):</p> <blockquote> <p><strong>Proposition.</strong> 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 $s$.</p> </blockquote> <p>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.</p> <p>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</p> <p>$$\left.\frac{d}{dt}\right|_{t=0}\frac{\partial^{|\sigma|}s_t^j}{\partial x^\sigma}(x) = (D_\sigma \varphi^i)(j^\infty(s)),$$</p> <p>which is the equation that inspired my question.</p>