# Thread: differential forms and Hodge star

1. ## differential forms and Hodge star

I am trying to figure out these relations (Hodge operator acts on the following r-forms)
$\displaystyle *(dx\wedge dy)=dz$
this one I do not get
$\displaystyle *(dy\wedge dz)=dx$
$\displaystyle *(dx\wedge dx)=-dy$ how do they work?

They only way I can explain it is that the wedge product work like the cross product for example
$\displaystyle x\times y=z$
$\displaystyle x\times z=-y$ and so on
is it a correct way of thinking?

2. You need to be careful beucase the wedge product can be used in on any forms in the exterior algebra but the cross product only works in three dimientions.

The more general idea is

The Hodge dual is a map from

$\displaystyle \bigwedge^2(\mathbb{R}^3) \to\bigwedge^{3-2}(\mathbb{R}^3)$

Loosely speaking If we are given an r form w we are looking for a 3-r form v
such that ()

$\displaystyle w \wedge v=|J| dx \wedge dy \wedge dz$

Where J is the Jacobian matrix (the volume element of the curvilinear coordinate system.)

In rectangular coordinates |J|=1

So in your first case you have

$\displaystyle w=dx \wedge dy \wedge v= dx \wedge dy \wedge dz \implies v=dz$

2nd

$\displaystyle dy \wedge dz \wedge v=dy \wedge dz \wedge dx$

Notice the the order is wrong we need to move dx to the front and since the wedge product is antisymmetric we get a factor of (-1) each time we move through a 1-form this gives

$\displaystyle dy \wedge dz \wedge dx=(-1)[dy \wedge dx \wedge dz]=(-1)^2[dx \wedge dy \wedge dz]$

So the Hodge dual is

$\displaystyle dx$

and for the last one I assume you have a typo and that you mean

$\displaystyle dx \wedge dz \wedge v=dx \wedge dz \wedge dy =(-1)[dx \wedge dy \wedge dz=dx \wedge (-dy) \wedge dz$

So the Hodge dual is

$\displaystyle -dy$