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.

  • $\begingroup$ Naively I'd say that for general vector fields one would not expect a nice formula, simply because a general vector field does not respect the additional structure which allows you to define the codifferential. Basically it comes down to a formula for the Lie derivative of the Hodge operator. I'm not aware of any memorable formula. I think one has to get one's hands dirty and the 'ugly' formula you have gotten is the best you can hope for. $\endgroup$ – José Figueroa-O'Farrill Feb 11 '10 at 12:50
  • $\begingroup$ Right. Also, I forgot to mention that a bit of computations show that actually there is a "not too bad" commutator for interior derivative and the codifferential. Which pushes the work to the commutator of the codifferential with the exterior derivative. Now, the anti-commutator is of course the Hodge laplacian. Is there any use for the commutator? $\endgroup$ – Willie Wong Feb 11 '10 at 13:27
  • $\begingroup$ I am voting to close this question, because I am now convinced that the "nice" formula I was looking for doesn't exist, and I should just suck it up and do the computations. $\endgroup$ – Willie Wong Jul 11 '10 at 16:56

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.


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.

  • 2
    $\begingroup$ Not quite. I've seen that paper before, and it is not all that useful for the case I am considering. There are some problems with their definitions that make it hard to apply, one of the most obvious ones being that their algebraic definition does not use the metric structure! They define the Hodge dual not as a map from p-forms to (n-p)-forms as the standard way; rather, as a map from p-forms to (n-p)-vectors through a (dual) volume form. So the formula that they derive only applies to contravariant, and not covariant objects. To get it to work for forms I'd need to commute with the metric... $\endgroup$ – Willie Wong Apr 11 '10 at 22:09
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    $\begingroup$ ... to lower the index again. Just to illustrate, the formula they give on page 8, if you apply to Lie bracket of vectors, is a triviality. But if you try to act the operator "on forms" the stupid way, you'd see that it is quite obviously wrong. In other words, the formula does not commute with the isomorphism between covariant and contravariant tensors induced by the metric tensor. Thanks for trying to help! $\endgroup$ – Willie Wong Apr 11 '10 at 22:22
  • $\begingroup$ The linked article: Coll and Ferrando - On the Leibniz bracket, the Schouten bracket and the Laplacian. $\endgroup$ – LSpice Nov 19 '20 at 19:17

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.


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).

  • $\begingroup$ How is this different from what I wrote in the third paragraph of my question, pray tell? $\endgroup$ – Willie Wong Feb 12 '10 at 14:11
  • $\begingroup$ Sorry, it's exactly the same. I did consider just deleting my answer. $\endgroup$ – Deane Yang Feb 18 '10 at 23:53

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