# What is the meaning and importance of the Hodge codiferential?

In differential geometry given a smooth manifold $M$ we can define the exterior derivative $d$ acting on $k$ forms giving back $k+1$ forms. It is a map $d : \Omega^k(M)\to \Omega^{k+1}(M)$ which is in principle not so hard to define. It is indeed the natural way to define a derivative of differential forms and it leads directly to Stokes' theorem:

$$\int_M d\omega=\int_{\partial M}\omega.$$

That is all fine, but now comes the thing. If one has a metric and can define the Hodge dual $\star$ it is possible to define the Hodge codifferential on $k$ forms by

$$\delta = (-1)^{k} \star^{-1}d\star.$$

When defining the product $(\cdot,\cdot)$

$$(\eta,\zeta)=\int_{M} \eta \wedge \star \zeta$$

it turns out $\delta$ is a sort of adjoint of $d$, in the sense that

$$(d\eta,\zeta)=(\eta,\delta \zeta)$$

and finaly from it one defines the operator $\Delta = \delta d+ d\delta$ which would be a generalization of the laplacian on general differential forms.

It is all nice, but there is a thing here. I don't understand the meaning of any of this in the sense of how to interpret these things geometricaly and recognize their real importance.

Actually I'm interested on this because of theoretical Physics. In some texts about Clifford bundles this operator appears quite a lot, the reason why I'm interested on it.

So my question is: what is the meaning of this operator? How do we understand it geometricaly? Why would one introduce it and what is its importance? And if possible, why this operator would be so considered in Physics anyway?

• This is only a partial answer, but for a $1$-form $\alpha$, $\delta \alpha$ is (maybe up to sign) just the divergence of the vector field $\alpha^{\sharp}$, so one can view $\delta$ as a generalization of the usual divergence operator to forms of general degree. – Travis Apr 29 '17 at 15:40

If you like physics, there is an important example. In the space time endowed with the Lorentz (or Einstein) metric, the Maxwell equations reduces to $dF=0, \delta F= \mu_0 J$, where $F$ is the 2-form on the space time (a 4-dimensional space) whose 6 components are the 3 components of the electric field, and the 3 other the magnetic field (after the choice of a coordinate systems!). From this formalism you can easily prove that a good choice of coordinate enable you to "kill" the electric or the magnetic part of the field. See https://en.wikipedia.org/wiki/Maxwell%27s_equations, and the fantastic book by Penrose.