Is the curl of a vectorfield expressable in terms of a Lie derivative?

341 Views Asked by At

I am loving differential forms and the general version of Stokes theorem. However, I am also fond of the intuitive pictures of my older physics days of $grad$, $div$ and $curl$.

I recently stumbled upon the following simple identity: Write $\Omega := d x^1 \cdots dx^n$ for the standard volume form on $\mathbb R^n$. Given a vectorfield $\xi$ on $\mathbb R^n$ we have,

$L_\xi \Omega = div(\xi) \Omega$

I derived it by accident while doing some differential geometry exercise. Now, this is certainly well known to every differential geometer; however I was surprised no one told me about it yet. The physical interpretation of the divergence was always along the lines of "infinitesimal change of volume". Well, the above formula makes it precise: The divergence of $\xi$ is the factor by which the volume changes under the flow induced by $\xi$.

I am wondering if there is a similar equation for the $curl$ of a vector field in $\mathbb R^3$? Namely, as a factor that appears when applying the Lie derivative to a $2$-form or bivector?

EDIT: A more general version of this question for arbitrary oriented Riemannian 3-folds would be:

How are $(\ast \text{d} \xi^\flat)^\sharp$ and $L_\xi$ related?

However, I expect that an answer for $\mathbb R^3$ will give away the general case with it.