Differential forms on R^n vs. on manifold

[Kris-L]

First time looking at differential forms. What is the difference of the forms over R^n and on manifolds? Does the exterior product and derivative have different properties? (Is it possible to exaplain this difference without using the tangent space?)

 

[fresh_42]

First time looking at differential forms. What is the difference of the forms over R^n and on manifolds?

Can you define "over $\mathbb{R}^n$" versus "on a manifold"?

Does the exterior product and derivative have different properties?

Depends on what you consider a property.
$$\begin{equation*} d(\omega_1 \wedge \omega_2)= d\omega_1 \wedge \omega_2 + (-1)^k \omega_1 \wedge d\omega_2 \\ \textrm{ for all }\omega_1 \in \wedge^k(U)\, , \,\omega_2 \in \wedge^l(U) \end{equation*}$$
is the rule for multiplications in the exterior (Graßmann) algebra, whereas derivatives obey the product (Leibniz) rule
$$
d(f\cdot g)=df \cdot g + f \cdot dg
$$
This can be considered as a difference, although the comparison doesn't match exactly.

(Is it possible to explain this difference without using the tangent space?)

Not really. You can use a different language: sections, pullbacks, and vector bundles. But in the end, we are talking about tangent spaces.

You might want to read:
https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/#toggle-id-1

 

[stevendaryl]

The exterior derivative business is a way to systematize the sorts of derivatives that are already used in Euclidean 3-D vector calculus: the gradient, the divergence, the curl.

Then some of the amazing facts about vector calculus can then be seen as special cases of much more general facts. For example, we know that the curl of the gradient of a scalar is always zero. Similarly, the divergence of the curl of a vector is always zero. In exterior calculus, these are two instances of the more general fact that $d^2 = 0$. Applying the exterior derivative twice always produces zero.

Then there are two facts relating integrals of different numbers of dimensions:

$\int \vec{A} \cdot \vec{dl} = \int (\nabla \times \vec{A}) \cdot \vec{dS}$
(integrating a vector field around a closed loop produces the same result as integrating the curl of the vector over the surface enclosed by the loop)

$\int \vec{A} \cdot \vec{dS} = \int (\nabla \cdot \vec{A}) dV$
(integrating a vector field over a closed surface produces the same result as integrating the divergence of the vector over the volume enclosed by the surface)

In the exterior calculus, these are two different instances of the more general "Stokes theorem", which applies to objects of any degree.

 

[wrobel]

Only global properties of the forms can be different in R^n and in other manifold. Locally defined operations ( like different differentiations, exterior product, i_v etc) have the same properties

 

[jbergman]

First time looking at differential forms. What is the difference of the forms over R^n and on manifolds? Does the exterior product and derivative have different properties? (Is it possible to exaplain this difference without using the tangent space?)

I am not sure what the question is. I would say that the answer is no, they are the same as R^n is a manifold. The only thing that changes are the charts that define our manifold. On R^n there happens to be one global chart that is the identity map so that makes things easier.

For a general manifold only patches of it are diffeomorphic to R^n so you have charts for these different patches. When doing computations with forms one often needs to work with local coordinates which involves mapping patches of manifold to R^n and doing the computations in the exact same fashion a in R^n.

Different patches have different chart maps so that the form can end up being a different function in different charts. So the main differences are this. On a general manifold, you have to do computations in multiple local charts and the local coordinate form can vary form chart to chart. This isn't an issue in R^n.