I am trying to get a better feel for both the exterior derivative of a form and the contraction of a form by a vector field $X$.
Basically, when are these inverses? If I have a one-form $\omega$ and I compute $dw$, getting a 2-form. When can I find a vector field $X$ so that $w = i_Xdw$? Is there a general result describing the relation between forms $w, v$ such that $w=i_Xdv$?
I am aware of the issue of closed and exact forms, but I am having trouble framing this question in this format.
EDIT: Also, is there a "nice" geometric way of seeing the interior product? We can see the exterior derivative in terms of geometric algebra, by creating "parallelepipeds" or their $n$-dimensional equivalent, as volume elements. Is something similarly geometric going on when we contract by a vector field?
Thanks.
Best Answer
The exterior derivative and $i_X$ for a vector field $X$ are never inverses. Let $\omega$ be a non-zero closed form, then $i_Xd\omega = 0 \neq \omega$. One can still try to find out when a given one-form $\omega$ satisfies $i_Xd\omega = \omega$ for some vector field $X$. What follows doesn't completely answer your question, but it does give a necessary condition.
If $i_X d\omega = \omega$, then $i_X\omega = i_X(i_Xd\omega) = 0$ as $i_X\circ i_X = 0$. Now by Cartan's magic formula, we have $\mathcal{L}_X\omega = i_Xd\omega + d(i_X\omega)$, but $i_X\omega = 0$, so $\mathcal{L}_X\omega = i_Xd\omega = \omega$.
So, if there a vector field $X$ such that $i_Xd\omega = \omega$, it must satisfy $\mathcal{L}_X\omega = \omega$.