Commutator of Lie derivative and codifferential? 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. 
 A: 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. 
A: You can find a formula for the commutator of the codifferential and the Lie derivative in http://arxiv.org/pdf/gr-qc/0306102. The important point to get the commutator is to notice that the codifferential is a derivation on the Schouten-Leibniz algebra. You can see the details in the paper above. I hope this is what you are looking for.
A: Closely related to your question is what is the commutator of Lie derivative and Hodge dual *. I recently cam across a nice answer to that question, in a 1984 article by Trautman, which is referenced here: http://inspirehep.net/record/206126?ln=en
The answer is $[L_X,*]\alpha = [i(h) - \frac12 Tr(h)]*\alpha$, where $\alpha$ is a form, $h$ is the 1-1 tensor defined by the Lie derivative of the metric, contracted on one index with the inverse metric (i.e., $\nabla_a X^b$, where $\nabla_a$ is the derivative operator determined by the metric) and $i(h)$ is the derivation generated by $h$ sending 1-forms to 1-forms.
A: There is a perfectly nice formula that can be found by a straightforward exercise. It is not hard even in local co-ordinates. If you do it first for a 1-form and a 2-form, the general pattern becomes apparent. In fact, you'll know what to do for an arbitrary tensor. It is helpful to keep in mind that the codifferential is just a divergence (a trace of the covariant derivative of the tensor).
