Prolongation formula vector fields

A vector field

$$ Y=\xi_{i} \partial x^{i}+\eta^{J}_{a} \partial u^{\alpha}_{J} $$

in the jet bundle $J^k(E)$ (think of $E$ as a generalization of the trivial bundle of dependent and independent variables $X\times U$) may be a symmetry of a distribution of the Cartan distribution. From the characterization of symmetries from the point of view of Lie derivative of 1-forms (see here), it can be shown that this vector field satisfies the formula:

$$ \eta^{J, i}_{a}=D_{i} \eta^{J}_{a}-u_{J, m}^{a}\left(D_{i} \xi_{m}\right) $$

where $D_i$ is the total derivative operator associated to $x_i$ (it is said in Gaeta_2005 page 2).

Reciprocally, any vector field satisfying the prolongation formula is such that

$$ L_Y(\theta)\in \mathcal{E}^* $$

being $\mathcal{E}$ the Cartan distribution and $\mathcal{E}^*$ its dual description.

If the projection to the independent and dependent variable space $E$ is a vector field $V\in \mathfrak{X}(E)$ which does not depend on derivatives, we obtain a proper vector field on $E$, which is candidate to be a Lie symmetry of a system of DEs. We say that $Y$ is the prolongation of $V$. It can be shown that every vector field in $E$ has an unique prolongation (stated in Anderson_1992 Definition 2.2, but for $J^{\infty}(E)$).

Meaning

The idea behind this is that given the bundle $E\rightarrow M$, a vector field $X\in \mathfrak{X}(M)$ is the infinitesimal generator of a one-parameter local group of transformations. The action of this group can be prolonged to act on $J^kE.$ The infinitesimal generator os this action is the prolonged vector field.

So if a vector field in the jet space is a prolonged one what we know is that it carries Cartan distribution to the Cartan distribution. Since vector fields corresponding to ODEs "live inside" Cartan distribution, the prolonged vector fields are candidates to be Lie point symmetrys of the ODE.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es


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