Let $(M,g)$ be some smooth, Riemannian manifold. Let $d$ be the exterior derivative and $\delta$ the codifferential on forms. For a smooth vector field $X$, let $L_X$ be the Lie derivative associated to $X$. We know from Cartan formula that $L_X = d \iota_X + \iota_X d$ where $\iota_X$ is the interior derivative associated to the vector field $X$. So it is well-known that $L_X$ and $d$ commute: for any arbitrary form $\omega$, we have that $L_Xd\omega = dL_X\omega$.
This is, of course, not true for codifferentials. In general $[L_X,\delta]\neq 0$. For certain cases the answer is well known: if $X$ is a Killing vector ($L_Xg = 0$) then since it leaves the metric structure in variant, it commutes with the Hodge star operator, and so $L_X$ commutes with $\delta$. Another useful case is when $X$ is conformally Killing with constant conformal factor ($L_X g = k g$ with $dk = 0$). In this case conformality implies that the commutator $[L_X,*] = k^\alpha *$ where $\alpha$ is some numerical power depending on the rank of the form it is acting on (I think… correct me if I am wrong), so we have that $[L_X,\delta] \propto \delta$.
So my question is: "Is there a general nice formula for the commutator $[L_X,\delta]$?" If it is written down somewhere, a reference will be helpful. (In the Riemannian case, by working with suitable symmetrisations of metric connection one can get a fairly ugly answer by doing something like $\delta \omega \propto \mathop{tr}_{g^{-1}} \nabla\omega$ and use that the commutators $[L_X, g^{-1}]$ and $[L_X,\nabla]$ are fairly well known [the latter giving a second-order deformation tensor measuring affine-Killingness]. But this formula is the same for the divergence of arbitrary covariant tensors. I am wondering if there is a better formula for forms in particular.)
A simple explanation of why what I am asking is idiotic would also be welcome.
Best Answer
Doing along what Deane and Jose suggest one can make the following observations. First is that the Hodge star operator can be expressed in terms of the inverse metric $g^{-1}$ and volume form $\epsilon$. That is to say, formally for a form $\omega$ we have $$ *\omega = \epsilon \cdot (g^{-1}) \cdot \omega $$
Take $L_X$, when it falls on $\epsilon$ we have by definition of divergence $$ L_X \epsilon = (\mathrm{div} X) \epsilon.$$ Where $\mathrm{div} X$ is also half the metric trace of the deformation tensor $L_X g$. We also have the formula $$ L_X g^{-1} = g^{-1}(L_X g) g^{-1} .$$ Plugging this in we have that schematically $$ L_X *\omega = (\mathrm{div} X) *\omega + * [(L_X g)\cdot \omega] + * L_X \omega$$ from which the computation can be completed.