How to Interpret "Lie Derivative of a Form Equals 0"(Contact-Reeb Vector Field )

Question

Everyone:

I'm trying to understand better the meaning of a diff. form being constant along a flow; more specifically:

One of t properties of a Reeb vector field X associated to a contact form w (meaning that the contact distribution is globally-generated by Kerw ) is that :

$L_X w=0$

(This is because : $L_X w= i_X dw+ d(i_X w) =dw(X,.)+d(W(X))=0$ , since X is in $Ker(w)$, and w (X)==1 )

i.e., the Lie derivative of the form w about the Reeb vector field X is $0$, meaning that the form w is constant along the Reeb flow, or maybe that the tensor field (assignment of a 1-form at each point of the flow, or at each $TM^* _p$ ) is constant along the flow of $X$

Does anyone have any insights into the meaning of a form being constant along a flow, or the meaning of a tensor field being constant ?

Thanks.

Answer 1

The vanishing of the Lie derivative $\mathcal L_X s$ means that the flow $\psi_t$ of $X$ is a symmetry of the tensor field $s$ - if you "change your coordinates" by flowing along $X$, then $s$ looks exactly the same. When $s$ defines some kind of structure on the manifold, this means the flow of $X$ is a symmetry of that structure; e.g. when $s=g$ is a metric the $\psi_t$ are isometries, and when $s=w$ is the contact structure the $\psi_t$ are isomorphisms of the contact structure.

For differential forms in particular you can interpret the Lie derivative in terms of integrals, though I'm not sure how natural this is in contact geometry. Using the definition of the Lie derivative in terms of the pullback of the flow, we see

$$\int_{\gamma} \mathcal L_Xw = \frac d{dt} \int_{\gamma} {\psi_t}^*w= \frac d{dt} \int_{\psi_t \gamma} w.$$

Thus the vanishing of the Lie derivative means that if you take any loop $\gamma$ and let it move with the flow, the integral $\int_\gamma w$ is constant - alternatively you can fix the loop and flow the form using the pullback.