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 23:24, 18 November 2005 (UTC)[reply]

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)[reply]

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 ${\displaystyle \Omega ^{k}}$ 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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

I have added this quote in a "Quote frame" to the introduction. I agree that it provides a very accessibly introduction for a passing layman. Beyond that, the quote itself is an anecdotally interesting tidbit which I feel many students of mathematics would appreciate. I left the original introduction in place as I feel this quote alone would not suffice as the full introduction for the article. PhysicsSean (talk) 14:35, 24 August 2017 (UTC)[reply]

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)[reply]

This page should discuss compactly supported de-Rham cohomology and it's relation with Borel-Moore homology using integration. — Preceding unsigned comment added by 71.212.185.82 (talk) 22:59, 17 August 2017 (UTC)[reply]

I 100% agree, the relation of Homology and Cohomology is a topic that is unfortunately lacking from Wiki's discussions of algebraic topology. If someone has the knowledge to write this up then that would be great.PhysicsSean (talk) 22:46, 24 August 2017 (UTC)[reply]

I'll start with a disclaimer: I am extremely shaky on this topic. With this in mind, this edit adds the qualification "(depending on the metric)", which seems to me to be completely superfluous: the codifferential is defined in terms of exterior derivative and the Hodge star. Aside from orientation, the Hodge star has strictly less structure than the symmetric bilinear form (the metric tensor), or otherwise said, the metric tensor determines the Hodge star up to a sign, but the Hodge star determines very little about the metric tensor (it reduces the real degrees of freedom in the metric by only 1). Given that the codifferential is assumed in the statement, by this logic the qualification should be removed as potentially confusing.

What I find less obvious is that the Hodge decomposition is really unique. I find the text of various textbooks opaque, which probably only speaks to my own maths illiteracy. But I see this, and Hodge_theory#De_Rham_cohomology, which is from where I copied the uniqueness claim (guilty as charged: I copied from a WP article without verification using a source). I do not find it obvious that the decomposition into three parts is unique from the Frank Warner text but this probably implies it:

6.8 The Hodge Decomposition Theorem For each integer p with 0 ≤ p ≤ n, H^p is finite dimensional, and we have the following orthogonal direct sum decompositions of the space E^p(M) of smooth p-forms on M:

E^p(M) = Δ(E^p) ⊕ H^p

= dδ(E^p) ⊕ δd(E^p) ⊕ H^p

= d(E^p−1) ⊕ δ(E^p+1) ⊕ H^p.

If my interpretation is correct, this might be nicer to capture in words as "Every p-form decomposes uniquely as the sum of an exact, a co-exact and a harmonic p-form." —Quondum 21:15, 21 July 2018 (UTC)[reply]

Well, the article text mentions L^2 and L^2 implies there is an inner product (on functions) and that inner product has to come from somewhere; Riemannian metric, no? As for the Hodge star operator, from what I can see from Warner, the definition involves a volume form and so that depends on the metric in that way (but I admit I'm not an expert on this). So I thought it is necessary to mention that the decomposition involves a choice of metric; in fact, in Ch 6 of Warner, he only considers compact oriented Riemannian manifolds. (I suppose oritentation can be handled by either using density or orientation sheaf.) Ok. Maybe you use metric sheaf to suppress choices of metrics??? -- Taku (talk) 00:22, 22 July 2018 (UTC)[reply]

As for the uniqueness, yes, the direct sum decomposition implies uniqueness (since each component is a projection). So I think it's a good idea to make it explicit. (My problem with uniqueness was something different.) -- Taku (talk) 00:28, 22 July 2018 (UTC)[reply]

>> "that inner product has to come from somewhere; Riemannian metric, no?"

No, it does not need any kind of metric tensor, and my reasoning is as follows (OR alert). The volume form is all that is needed: it implies and determines an orientation on the manifold, it determines the Hodge star, and determines nothing about the metric tensor aside from an overall scale (i.e., for every nondegenerate metric tensor, only a scalar field multiplier is needed to make it compatible with an arbitrary nonzero volume form, as I tried to say in my opening paragraph. It determines an L² metric: ... an L² metric on the vector spaces Ω^k(M) [is] defined by integrating over M with respect to the volume form .... (I forget where I saw the explicit definition; as I remember it, it is ${\displaystyle \textstyle \|\alpha \|_{2}=\int _{M}\alpha \wedge {\star \alpha }}$) Even if there is a metric tensor, whether it is Riemannian or only pseudo-Riemannian makes no difference whatsoever. Which comes to my thesis: we get everything needed for the Hodge decomposition from a differential manifold with a volume form, which is distinctly less structure than an oriented Riemannian manifold has. —Quondum 11:55, 22 July 2018 (UTC)[reply]

I'm more familiar with the complex case. So if you allow me to quote a complex (but very simple!) textbook, the Hodge *-operator is defined as

${\displaystyle \alpha \wedge *\beta =\langle \alpha ,\beta \rangle {\text{vol form}}}$.

(Daniel Huybrechts, Complex geometry: an introduction, pg 32). So the definition explicitly involves an inner product (metric). I understand up to scaling but that's up to scalars point-wise; thus varying the point seems to result in a metric; unless we have different things in mind. -- Taku (talk) 23:09, 22 July 2018 (UTC)[reply]

Also, the Hodge decomposition is stated explicitly with the Riemannian metric g (Theorem A.0.16. of loc. cit.) -- Taku (talk) 23:14, 22 July 2018 (UTC)[reply]

Ah – my bad. I have been remembering properties of the metric dual of the Hodge star (which is in some sense more primitive than the Hodge star), and attributing them to the Hodge star: bad memory. (Defined in Vaz, Jayme; da Rocha, Roldão (2016), An Introduction to Clifford Algebras and Spinors, ISBN 978-0-19-878292-6.) Yes, the Hodge star (and hence presumably the codifferential) does depend on the pseudo-Riemannian metric tensor in a non-trivial way.

My remarks should thus be restricted to: In the context of Hodge theory, pretty much everything remains valid on a pseudo-Riemannian manifold. Despite the restriction of the interest of many authors to only Riemannian manifolds, we should not echo this restriction without good reason, especially since this is relevant in general relativity. This suggest reviewing the previous section (Harmonic forms). It is possible any additional premise may be a bit subtler than a local condition such as a pseudo-Riemmannian metric tensor field due to the global character of the L² metric used for the Hodge decomposition. I would also prefer to avoid the potentially ambiguous term "metric". It seems to me that a careful literature review is needed to get the requisite conditions. E.g. must the manifold be compact? Must it be oriented? (I think there is an obvious generalization of the L² metric to unorientable manifolds that involves a double cover.) —Quondum 14:10, 23 July 2018 (UTC)[reply]

As I alluded a bit above, "oriented" can be dropped by use of density (see orientation sheaf). The idea is very simple: we use the L^2 space of (equivalence classes) of smooth densities as opposed to differential forms. I don't know of a similar trick for metric (and that's why I added "metric" but omitted orientation). "compact" can also be dropped: again I'm only familiar with the complex literature but see for example [1] Ch VIII, § 3. Though the extensions like these lie outside the scope of the article, in my opinion. As for dropping "positive-def", I doubt that works: since you need a positive-definite sesqui-linear form (a.k.a. an inner product) for a Hilbert space structure. -- Taku (talk) 02:33, 24 July 2018 (UTC)[reply]

Orientation is not important, and I agree with that part. On ambiguity, in this context the reader could understand "metric" to refer to L² metric or Riemannian metric, which should be made clear, I thought. I presume you mean the latter.

The Hodge star, differential, codifferential and Laplace–de Rham operator are all defined for pseudo-Riemannian manifolds. I am out of my depth on the L² inner product/metric, though. I presume that the Hilbert space structure you refer to is the space of L² functions with the L² inner product. It is unclear to me how an indefinite metric tensor (bilinear form) impacts this. —Quondum 03:51, 24 July 2018 (UTC)[reply]

Hum. I myself never use the term "L^2 metric" but only "L^2 norm"; I'm pretty sure the latter is more standard. So I didn't see the ambiguity. (It's hard to decide how much details on Hodge theory; this article is about de Rham cohomology after all.) By "Hilbert space structure", yes, I meant that on the L^2 space of square-integrable differential forms; so by definition the inner product must be positive definite (in fact, for me, an inner product is a positive-def sesqui-linear form). The definiteness is needed so that for example a Hilbert space is a metric space not just pseudo-metric space. -- Taku (talk) 23:12, 24 July 2018 (UTC)[reply]

Are you suggesting that the Hodge theory part be trimmed down? I does duplicate a lot that is already in that article.
Are you comfortable with how I've changed it (though the Hodge decomposition could be removed here, if you think it should only be in Hodge theory).
I can see that there may be issues with L² in the pseudo-Riemannian case if a Hilbert space is needed for uniqueness to work. —Quondum 02:25, 25 July 2018 (UTC)[reply]

No I'm not suggesting anything concrete (since I'm not sure about the appropriate amount of details on Hodge theory here). But I think it makes sense to mention Hodge decomposition here since it gives a concrete tangible way to understand de Rham cohomology. -- Taku (talk) 23:16, 25 July 2018 (UTC)[reply]

I'm ok with the changes; I think we should just mention orientation to be precise. -- Taku (talk) 23:26, 25 July 2018 (UTC)[reply]

I wish I wasn't taxing my brain cells quite so much. I can see that I've gone against some of your observations (constraint of compactness, orientedness, etc.). Before this discussion, I hadn't even been aware that a metric on functions over the whole manifold, as opposed to properties local to a neighbourhood, played a role (yep, that's how new I am to this). Feel free to undo unnecessary restrictions that I've obtained from less general treatments/sources.

The de Rham cohomology seems to precede and be the basis of Hodge theory. Importing results from a derived topic to help with understanding is fine, but I am mildly concerned that there might be too much imported (and hence repeated) where referencing the derived topic through links might be sufficient for substantiation and explanation. There are two aspects to this: undue duplication, and creating the impression that Hodge theory is considered part of or preceded de Rham cohomology. (I didn't, and probably still don't, clearly understand the distinction between them.) —Quondum 00:55, 26 July 2018 (UTC)[reply]

– Ah, I've just figured out why the L² norm was confusing me, especially with regard to locality. The definition of δ that I'm familiar with is purely local: δ = ∗ ∘ d ∘ ∗ (up to a sign). One then derives that δ and d are mutually adjoint with respect to the L² norm. This would immediately seem to sidestep many of the constraints on the manifold (such as smooth, closed, complete, or compact). But I may be making sloppy inferences – the existence of ∗ implies global properties such as orientability (though I think it is easy to work around this one, as mentioned beofre). —Quondum 23:14, 26 July 2018 (UTC)[reply]

The last thing first: you can recover the volume form from the Hodge *-operator easily: for any zero-form f, *f = f \text{vol form}; so *1 = \text{vol form}.

As for the relationship between Hodge theory and de Rham cohomology; perhaps this is my personal view but I don't view de Rham cohomology as a basis for Hodge theory. But, rather, Hodge theory (depending on a metric) determines a cohomology-type theory and it so happened (theoretically not historically) that it coincides with de Rham cohomology. This is to be expected since a cohomology theory is characterized by univer properties. For editorial considerations, I personally don't believe having duplicated materials is problematic. Yes, we can provide links. But many readers may prefer not to follow the links and find the recalling of basic facts and definitions convenient. -- Taku (talk) 04:22, 27 July 2018 (UTC)[reply]

Yes, the Hodge star provides a lot of information that can be recovered (volume form, hence orientation, and even the metric tensor). My realization related to the way Hodge decomposition uniqueness is stated: if the co-differential is defined locally, the L² norm is absent (one does not need the statement about orthogonality). Hence, it would seem to me that the decomposition should hold on any Riemannian manifold for forms that are say twice-differentiable, even separately on any neighbourhood in it.

I'm not suggesting that all the material should be removed. But the presentation gives so much detail, it took me a long time to realize that Hodge theory (or this part of it) is not contained in de Rham cohomology. The section intro does imply this, but somehow this was not sufficient at first. It is this balance that feels wrong to me. And yes, your characterization seems good: it seems like it would be better to Hodge theory as historically inspired by rather than based on de Rham cohomology, though I'm no clearly authority. —Quondum 11:01, 27 July 2018 (UTC)[reply]

Given a real vector space ${\displaystyle V}$ we can compute the cohomology of ${\displaystyle GL(V)}$ using the endomorphism-valued form from the injection ${\displaystyle g:GL(V)\to End(V)}$. The details are included in https://web.archive.org/web/20130124054444/http://www.indiana.edu/~jfdavis/teaching/m721/mano.pdf but this could easily be worked out in the case of dimension 2 (or even 3). The key is that ${\displaystyle \Theta =g^{-1}dg}$ can be easily represented. If we take

${\displaystyle g={\begin{bmatrix}x&y\\z&w\end{bmatrix}}}$

then

${\displaystyle \Theta ={\frac {1}{xw-yz}}{\begin{bmatrix}w&-y\\-z&x\end{bmatrix}}\cdot {\begin{bmatrix}dx&dy\\dz&dw\end{bmatrix}}}$

Using this we can find the ${\displaystyle \gamma }$ classes, ${\displaystyle \gamma _{1},\gamma _{3}}$ which generate the cohomology ring. — Preceding unsigned comment added by 50.246.213.170 (talk) 23:57, 26 July 2018 (UTC)[reply]

In this section, F and underlined blackboard F denote the same sheaf, only one should be used, right? 80.254.78.161 (talk) 18:34, 1 July 2022 (UTC)[reply]

In the edit https://en.wikipedia.org/w/index.php?title=De_Rham_cohomology&diff=next&oldid=1086029520, someone edited the text "One prominent example when all closed forms are exact is when the underlying space is contractible to a point, i.e., it is a star domain (no-holes condition)" and replaced "star domain" with "simply connected space". Isn't this wrong? It would say that simply connected sets have trivial de Rham cohomology but there are counterexamples right below, e.g. S^2, or punctured R^3. 76.131.96.107 (talk) 05:34, 18 May 2023 (UTC)[reply]