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))

wedge product

The current part considering wedge product is only slightly helpfull. Better would be the full definition (s. e.g. ). ChristianTS (talk) 17:24, 7 November 2016 (UTC)

I've expanded the article. Does this help? Ozob (talk) 03:56, 8 November 2016 (UTC)

References

A mistake in the embedding map of Wedge power into tensor space?

Under the Intrinsic Definision section, I think the embedding map is written in the wrong direction.

I.e. it should be:

$\operatorname {Alt} \colon \bigotimes ^{k}T^{*}M\to \bigwedge ^{k}T^{*}M$

The mapping in the currently written direction is trivial. The given map takes an arbitrary Tensor and extracts the Totally-Antisymmetric part of it.

Did I miss anything? תום ה (talk) 05:26, 5 April 2018 (UTC)

The direction in the article is correct. The mapping from the tensor power into the exterior power is the quotient mapping, not the alternation mapping. The alternation map takes a tensor $\omega$ to another tensor $\operatorname {Alt} (\omega )$ and is constant on the cosets of the ideal I in the tensor algebra, so factors through a mapping from the exterior algebra to the tensor algebra. Sławomir Biały (talk) 11:00, 5 April 2018 (UTC)