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
[edit] de Rham's theorem
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 Ωk 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)