Talk:Differential form

How exactly is a differential 1-form dual to a vector space?

The duality page doesn't explicitly talk about how differential forms are dual to anything, or how differential relate to duals. In fact, outside this line: "are naturally dual to vector fields on a differentiable manifold", I cannot find any other source communicating the same detail. GeraldMeyers (talk) 14:45, 12 April 2022 (UTC)

A differential form is a section of the cotangent bundle, and a vector field is a section of the tangent bundle. These two vector bundles are dual to each other, in the sense that each is the dual bundle of the other. In particular, this means that every fiber of the cotangent bundle is the dual vector space of the corresponding fiber of the tangent bundle. Ozob (talk) 15:08, 12 April 2022 (UTC)

I understand that the dual is more general than the transpose, but in simple cases the dual is the transpose, correct? E.g. the tangent space (bundle?) of a unit sphere is the collection of all tangent planes. The dual of a tangent basis is the cotangent basis, which are just transposes of each other? GeraldMeyers (talk) 15:43, 12 April 2022 (UTC)

Closed form redirect

I think that closed form should redirect here, rather than to de Rham cohomology as at present; and also should be disambiguated with respect to the 'closed form solution' meaning.

Charles Matthews 14:03, 11 Nov 2003 (UTC)

Disagree with merging closed and exact differential forms into here

See Talk:Closed and exact differential forms (unsigned comment by Oleg Alexandrov (talk))

A question about a calculating technics of differential forms (using abstract index notation)

In appendice B Differential Forms, Integration, and Frobiniu's Theorem of the book General Relativity written by Robert M. Wald, the book discussed the way to apply Stokes' Theorem to the embedded sub-manifold D of an n-dim (pseudo) Riemannian manifold M,

Wald gave the formula (B.2.24):

${\epsilon }_{ba_{1}a_{2}...a_{n-1}}=n_{b}\wedge {\tilde {\epsilon }}_{a_{1}a_{2}...a_{n-1}}(=-n_{a_{1}}\wedge {\tilde {\epsilon }}_{ba_{2}...a_{n-1}})$

then he derived (B.2.25):

$v^{b}{\epsilon }_{ba_{1}a_{2}...a_{n-1}}=(v^{b}n_{b}){\tilde {\epsilon }}_{a_{1}a_{2}...a_{n-1}}$

where:

• $v^{a}$ is an arbitrary tangent vector field on M;

• $n_{a}$ is the normal covector of the hypersurface $\partial D$ (which is the boundary of the embedded sub-manifold D), and $n_{a}$ is normalized $g^{ab}n_{a}n_{b}=\epsilon =\pm 1$;

• ${\epsilon }_{ba_{1}a_{2}...a_{n-1}}$ is the adapted volumn element of the (pseudo) Riemannian manifold M;

• ${\tilde {\epsilon }}_{a_{1}a_{2}...a_{n-1}}$ is the adapted volumn element on $\partial D$ ($\partial D$ has an induced mtric $h_{ab}=g_{ab}-\epsilon n_{a}n_{b}$);

So the question is: How to derive (B.2.25) from (B.2.24)?

Aphysicsstudent (talk) 08:38, 28 April 2022 (UTC)

Infix notation is fundamentally evil

As a person who just randomly cruised in with essentially zero preexisting knowledge beyond rusty elementary calculus, I'm presented with f(x, y, z) dx ∧ dy + g(x, y, z) dz ∧ dx + h(x, y, z) dy ∧ dz . After staring at the article, I think that's meant to be read as (f(x, y, z) (dx ∧ dy)) + (g(x, y, z) (dz ∧ dx)) + (h(x, y, z) (dy ∧ dz)) ... but honestly, I would most naturally read it as (f(x, y, z) dx) ∧ (dy + g(x, y, z) dz) ∧ (dx + h(x, y, z) dy) ∧ dz. That ∧ thingy looks like something that should bind very loosely. Maybe it's just me. But, hey, I asked somebody else and he agreed, so that's two of us...

I know that the notation is not going to change, and obviously I'm not actually suggesting using something even more confusing... but are parentheses allowed? If not, perhaps an explanation? I don't dare add either myself because I could easily be dead wrong.

I wish I had a more satisfying answer, but: Wikipedia goes by the usage of reliable sources, and in all reliable sources I'm aware of, ∧ binds more tightly than +. Ozob (talk) 21:02, 11 December 2022 (UTC)

It doesn't actually matter, because $f(dg\wedge dh)=(fdg)\wedge dh$ by definition, for any scalar functions $f,g,h$. pony in a strange land (talk) 18:47, 2 July 2023 (UTC)