Total vector field

(Anderson_1992 Definition 2.3)

Definition

Let $E\to M$ be a vector bundle and $X\in \mathfrak{X}(M)$. Then there is a unique vector field in the jet bundle $J^{\infty}(E)$, called the total vector field of $X$ and denoted by $\mbox{tot}X$, such that:

1. $X$ and $\mbox{tot}X$ agree on functions on $M$.

2. $\mbox{tot}X$ annihilates all contact 1-forms, that is, if $\omega$ is a contact form then

$$ \mbox{tot}X \,\lrcorner\,\omega=0. $$

$\blacksquare$

If $X$ is given by $X=X^i \partial x_i$ then

$$ \mbox{tot}X=X^i D_{x_i} $$

where $D_{x_i}$ are the total derivative operators. Observe that this definition gives us an intrinsic definition of this total derivative operators.

Idea: A vector in a manifold $M$ is a kind of little displacement. In the jet space $J^{\infty}(E)$ we have little displacement, too. For example, you can go from $(x,u,u_1)\in J^{\infty}(E)$ to:

$$ (x,u,u_1,\ldots) \mapsto (x+0.1,u+0.3,u_1-0.04,\ldots). $$

But if you think of $(x,u,u_1,\ldots)$ as the class of functions defined on $M$ (sections of $E$, indeed) such that $f(x)=u$, $f'(x)=u_1$ and so on, then the little displacement $x\mapsto x+0.1$ induce a natural displacement in $(x,u,u_1)$ since:

$$ u\mapsto u+0.1 u_1 $$ $$ u_1\mapsto u_1+0.1 u_2 $$

because of the definition of derivative:

$f(x+0.1)\approx f(x)+0.1 f'(x)$,

$f'(x+0.1)=f'(x)+0.1 f''(x)$

...

So the induced natural displacement is

$$ (x,u,u_1,\ldots) \mapsto (x+0.1,u+0.1 u_1,u_1+0.1 u_2,\ldots). $$

This induced displacement would be a total vector field.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es


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