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?



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 $\mathcal{L}_X\omega = \omega$.

  • $\begingroup$ Thank you. Is there a unique solution (when it exists) to v=dw? $\endgroup$ – RikOsuave Dec 25 '14 at 0:44

A good example I think is the case of the unit sphere ($S^{2}\to\mathbb{R}^{3}$) area element defined as $$\omega=\iota_{n}\gamma\tag{1}$$ where $\gamma$ is the volume form on $\mathbb{R}^{3}$, $n$ is a smooth unit normal vector field to $S^{2}$.

Using $n=(x,y,z)$, and the linearity of $\iota$, $$ \omega=x\ \iota_{\partial_{x}}\gamma+y\ \iota_{\partial_{y}}\gamma+z\ \iota_{\partial_{z}}\gamma. $$ Remembering that $\gamma=dx\wedge dy\wedge dz=dy\wedge dz\wedge dx=dz\wedge dx\wedge dy$; $$ \omega=x\ dy\wedge dz+y\ dz\wedge dx+z\ dx\wedge dy $$ Applying the exterior derivative we recover the volume form, albeit with a factor, $$ d\omega=3dx\wedge dy\wedge dz= 3\gamma $$ and the eq. (1) can now be written as $$ \omega=\tfrac{1}{3}\iota_{n}d\omega. $$


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