Quote by mathwonk
d of a one form IS the "curl" of that one form.
i.e. no matter WHAT coordinates u,v are,
the curl of fdu + gdv is ALWAYS df^du + dg^dv, in THOSE coordinates.
e.g. the curl of the angle form dtheta, in polar coords, is ddtheta = 0.
d of a one form, i.e. the curl, is always computed the same way, in all coordinates. there is no need to transform anything.
all i was doing was proving this.
there is nothing else to memorize. by introducing hodge duals you are just making life needlessly more complicated.
the only time a metric is needed is when one wants to study harmonic functions, i.e.laplacians.
somehow i feel we are still not communicating. you are trying to express something which is natural in all coordinate systems, i.e. exterior derivatives d, in terms of something less natural, i.e. components gij using a metric.
perhaps you have a need for this, but i do not see it.

I think that it's because we are talking about different curls.
My goal (and it may be that the goal in itself seems useless but that's a different discusssion) is to recover, from a differential form approach, the following expression that we learned in elementary calculus:
[tex] \nabla \times \vec{A} = \frac{1}{r sin \theta} \bigl( \frac{\partial ( sin \theta A_{\phi})}{\partial \theta}  \frac{\partial A_{\theta}}{\partial \phi} \bigr) \hat{r} + \ldots [/tex]
This is what I am trying to obtain starting from differential forms.
regards