Talk:De Rham cohomology

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Ok, what we have here in technical terms is called a mess. The Hodge decomposition ought to be exported to Hodge theory, although I'm not quite sure how to do that -- especially since de Rham is already discussed in detail there. Furthermore, Hodge's theorem on elliptic operators is not actually necessary to prove the de Rham theorem (despite the fact that it is a very elegant application of this theorem). There are many simpler techniques which do not rely on Sobolev spaces and other kludges (such as the Eilenberg-Steenrod axioms). So the options are, really clarify how the Hodge decomposition applies to this theorem (at the end of the article), or find a way to incorporate this material into some other article (preferably without making anyone angry). Silly rabbit


Typography: de Rham or De Rham[edit]

I think the rules in French are to capitalize De Rham if it is used as a stand-alone name, and write "Georges de Rham" or "The professor de Rham said that..." if it follows a name or title designating this person. In brief, I think most of the occurences of "de Rham" in this article should be replaces by "De Rham" (so that De Rham's theorem has been proved by Georges de Rham). I am unsure whether the same capitalization conventions apply in English, so I didn't do it myself. — Preceding unsigned comment added by 2001:700:300:1470:5EF9:DDFF:FE73:5D0F (talk) 10:26, 26 April 2013 (UTC)


de Rham's theorem[edit]

since the statement of de Rhan's theorem requires a compact manifold, perhaps an example of a non-compact manifold where the theorem fails would enrich this entry.

Actually it does not require that (or orientability for that matter). The proof is simple: The Poincaré lemma shows that the de Rham sheaf complex is a resolution of the constant sheaf R, and due to the existence of smooth partitions of unity the de Rham sheaves \Omega^k are fine, hence soft and thus acyclic. Therefore they calculate the cohomology of the constant sheaf R. (The link to the singular cohomology definition of the same is provided by the fact that singular sheaf complex is also a soft (hence acyclic) resolution of the constant sheaf R and teherefore also computes the cohomology with constant coefficients R. As the article already makes use of sheaf cohomology, I'll modify it according to these remarks. I'll also incorporate sheaf cohomology methods into the general cohomology articles at a later stage to provide more structure to the comparison of different theories. Stca74 06:21, 15 May 2007 (UTC)

removed reference to Lie coalgebra[edit]

It was in the definition section of the article and seemed out of place. Perhaps it could be added back with a fuller explanation whenever this article is rewritten. Akriasas (talk) 21:57, 25 December 2008 (UTC)

Laplacian[edit]

In Euclidean space, does the definition of Laplacian given here differ from the ordinary notion by a sign? TotientDragooned (talk) 19:01, 4 March 2010 (UTC)

Some phrasing that needs fixing[edit]

In the section on Harmonic forms, this sentence appears:

"If M is a compact Riemannian manifold, then each equivalence class in contains exactly one harmonic form. That is, every member ω of a given equivalence class of closed forms can be written as ω = dα + γ where α is some form, and γ is harmonic: Δγ=0."

But the "that is" clause is *not* a restatement of the previous one -- since it says nothing about the harmonic form's uniqueness. This sentence needs to be rewritten, preferably by someone who is an expert in the subject (which excludes me).

ALSO: The section about de Rham cohomology of the 2-torus -- correct me if I'm mistaken -- seems to assume that there are only two cohomology classes of the torus:

"For example, on a 2-torus, one may envision a constant 1-form as one where all of the "hair" is combed neatly in the same direction (and all of the "hair" having the same length). In this case, there are two cohomologically distinct combings; all of the others are linear combinations."

But there is no reason to say there are "two cohomologically distinct combings" when in fact there are infinitely many integer cohomology classes, each containing a unique harmonic 1-form. What is wanted here is that a basis for integer, or de Rham, cohomology is provided by (any) two harmonic 1-forms that are linearly independent over R. So this, too, needs rewriting.

Likewise, the follow-up sentence on the n-torus uses equally inexact wording:

"More generally, on an n-dimensional torus T^n, one can consider the various combings of k-forms on the torus. There are n choose k such combings that can be used to form the basis vectors for (H^k)_dR(T^n); the k-th Betti number for the de Rham cohomology group for the n-torus is thus n choose k."

There are certainly not merely n choose k such combings that "can be used" for form the basis vectors for (H^k)_dR(T^n). Rather, that is the size of any basis. And so that, too, needs rewriting.Daqu (talk) 20:50, 22 October 2010 (UTC)

Simpler introduction?[edit]

According to T. Tao: "The integration on forms concept is of fundamental importance in differential topology, geometry, and physics, and also yields one of the most important examples of cohomology, namely de Rham cohomology, which (roughly speaking) measures precisely the extent to which the fundamental theorem of calculus fails in higher dimensions and on general manifolds."

From here: http://www.math.ucla.edu/~tao/preprints/forms.pdf

This may be a better introduction -- at least laypeople that are just interested in mathematics can get a gist of what is happening, and I do not believe that this will sacrifice the overall rigour/details of the article. 128.54.66.73 (talk) 20:46, 5 June 2013 (UTC)

Section Ordering[edit]

I believe the section "Harmonic forms" should follow the section "Hodge decomposition". One could then easily see that if you believe in the Hodge decomposition then you get the Hodge theorem easily for elements whose classes belong in the k-th dimensional cohomology. — Preceding unsigned comment added by 12.31.71.58 (talk) 01:16, 7 December 2013 (UTC)