Symmetry of an exterior differential system

Given a exterior differential system $\mathcal{I}$ on a manifold $M$, a vector field $X\in \mathfrak{X}(M)$ is called a symmetry of $\mathcal{I}$ if

for every $\omega \in \mathcal{I}$.

Remarks

(see @ivey2016cartan exercises 6.1.2)

• Symmetries of an EDS form a Lie algebra.

• To show that $X$ is a symmetry we only need to check the condition above in a set of form which generate $\mathcal{I}$ differentially (see formulas for Lie derivative, exterior derivatives, bracket, interior product).

A particular case of symmetries is given by the Cauchy characteristic vector fields.

Theorem (Th 2.3.3 Barco thesis). Given an ideal $\mathcal{I}$ and suppose that the Cauchy characteristic space $A(\mathcal{I})$ is not the 0 modulus. If a vector field $X$ is a symmetry of $\mathcal{I}$ then it is also a symmetry of the distribution $A(\mathcal{I})$.$◼$

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es