# Talk:Differential form

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Field: Analysis

## Serious Revisions Needed!!!

This page has too many problems. It should begin with a disclaimer. —Preceding unsigned comment added by MephJones (talkcontribs) 20:06, 12 March 2010 (UTC)

Maybe you should be more specific... I for one think the concept section (at least) is terrific! —Preceding unsigned comment added by 66.97.107.188 (talk) 01:04, 3 April 2010 (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))

## symplectic form

http://en.wikipedia.org/skins-1.5/common/images/button_math.png Mathematical formula (LaTeX) Hi, sorry if this is a mixture or request/comment/question, but i was just wondering if symplectic differential form has any relation with this, and what exactly the correct definition would be. The preceding unsigned comment was added by Evilbu (talk • contribs) .

Either a 2-form, or a closed 2-form - probably the latter, in contemporary literature. Charles Matthews 22:25, 6 February 2006 (UTC)
In addition to being closed, a symplectic form must also be nondegenerate. See symplectic manifold for details. There are words like almost-symplectic, which means not necessarily closed, etc. Orthografer 00:57, 3 November 2006 (UTC)

## multilinear map from $\wedge^n\ TM$?

The article claims "At any point p on a manifold, a k-form gives a multilinear map from the k-th exterior power of the tangent space at p to R." Wouldn't this be just a linear map from the exterior power (which itself, however, might be thought of as a alternating multilinear map on $TM\times\cdots\times TM$)? Tesseran 05:59, 24 July 2006 (UTC)

We cannot consider a multilinear map on $TM\times\cdots\times TM$ because this is not a linear space. Commentor (talk) 02:00, 27 March 2008 (UTC)

## Why

you people don't explain why you use indexes below and above without any care? Why don't you explain that there is an advantage by indexing above for coordinated function? --kiddo 01:56, 2 November 2006 (UTC)

## Page complexity

I think this page is too complex for general wikipedia users. Any one expert on this topic please make the page more comphrensive. —Preceding unsigned comment added by 76.184.2.133 (talk) 11:19, 18 October 2007 (UTC)

## notation

Why is a basic 1-form written in two different ways in this article. Sometimes it's dx^i, sometimes it's dx^I? Randomblue (talk) 18:04, 2 February 2008 (UTC)

A form with capital indices is a k-form rather than a 1-form. This is explained in the article. Silly rabbit (talk) 18:12, 2 February 2008 (UTC)

When we integrate a function f over an m-dimensional subspace S of $\mathbb{R}^n$, we write it as

$\int_S f\,{\mathrm d}x^1 \cdots {\mathrm d}x^m.$

Consider ${\mathrm d}x^1$, ...,${\mathrm d}x^n$ for a moment as formal objects themselves

In the one case we go up to dx^m, and in the other dx^n. Is this an error? Randomblue (talk) 22:25, 2 March 2008 (UTC)

Yes, an error. It's so minor that I am not sure it's worth fixing it --- maybe somebody will be motivated by that to do some more substantial improvements :) 02:04, 27 March 2008 (UTC) —Preceding unsigned comment added by Commentor (talkcontribs)
But we are integrating here over S, not over Rn. S is a submanifold of dimension m. So one integrates m-forms over S, not n-forms. silly rabbit (talk) 02:20, 27 March 2008 (UTC)

## Stylistic elements

Umm... so is a math article supposed to make me laugh out loud multiple times? I think this page should be reworked stylistically.
Example: Consider dx1, ...,dxn for a moment as formal objects themselves, rather than tags appended to make integrals look like Riemann sums.
And worse: where dxI and friends represent basic k-forms
And "Gentle introduction" is a rather interesting section name... —Preceding unsigned comment added by 24.12.151.56 (talkcontribs)

So fix it. siℓℓy rabbit (talk) 13:14, 11 August 2008 (UTC)

## error in the lead paragraph

Not every differential form is a wedge product of exterior derivatives, contrary to the claim in the introduction attributed to Cartan. Since I am not sure what the editor's intention was here, I cannot correct it easily. Certainly Cartan did not say any such thing. Katzmik (talk) 07:33, 14 August 2008 (UTC)

Attributions aside, I'm not sure what the issue is here. On a smooth manifold the differential forms are defined locally as linear combinations of wedge products as in $\sum_if_i(\mathbf{x})dx_{i,1}\wedge\ldots dx_{i,k}$, where $k$ is some fixed nonnegative integer. Does my edit work for you? Orthografer (talk) 00:06, 17 August 2008 (UTC)

## exterior algebra and exterior derivative

The two exterior items mentioned in the lead are closely related but the use of the word "exterior" for both is actually very misleading. The former is essentially a concept in linear algebra, while the latter requires a differentiable structure. In particular, contrary to the claim in the lead paragraph, differential forms do not form an exterior algebra (the article on exterior algebra is purely linear-algebraic and deals with the finite dimensional case), but rather a differential algebra, or more precisely a differential graded associative algebra. Katzmik (talk) 07:40, 14 August 2008 (UTC)

This situation always arises with spaces of sections, and it is clearly better to avoid abuse of language whereever possible. There is no requirement in particular that exterior algebras must be finite dimensional, although I suppose I must assume some responsibility for the article Exterior algebra treating primarily the finite dimensional case. Nevertheless, a more accurate description would be that differential forms are sections of a sheaf of exterior algebras, or that they are sections of the exterior algebra of the cotangent bundle. siℓℓy rabbit (talk) 13:31, 14 August 2008 (UTC)

## Motivation

Here is a proposal for a different way to motivate the concept. Start with volume computation in $\mathbb R^n$. Observe that it is invariant under linear transformations with unit determinant. Therefore, the minimal structure needed to define volumes has less structure than the full coordinate system. Argue that volume is naturally a signed quantity. Then generalize to smooth manifolds. This gives the top dimensional differential form and justifies the anti-commutativity. To motivate the intermediate dimensional forms discuss signed lengths of curves invariantly. (Maybe this would be too long?) Opinions? Oded (talk) 16:11, 17 August 2008 (UTC)

This is nice, but somewhat non-standard. Ultimately we have to be able back up what we write by reliable secondary sources, so we have to take a fairly standard approach. Unfortunately most of this article is still rather weak: I expanded the vanilla motivation to draw attention to this weakness. I hope other editors who agree with my view that differential forms are rather important (!) will help to fix the basic shortcomings of the article. The relation to cohomology and homology needs massive expansion. The treatment of operations is utterly inadequate: what is the Lie derivative, the exterior derivative, Cartan's identity, etc. etc.? I'm looking forward to the luxury of fine-tuning the motivation and lead, but right now, this article has other issues. Geometry guy 20:40, 17 August 2008 (UTC)
The motivation in terms of determinants and more generally minors is done very nicely at exterior algebra (work by silly rabbit I believe and others). There is no need to duplicate it here. Katzmik (talk) 09:27, 20 August 2008 (UTC)

From my perspective, the main use of differential forms is in order to integrate them. They are the "right thing" to integrate, so to speak. Therefore, this seems like the most useful motivation. Of course, I realize there could be other perspectives as well. Oded (talk) 17:50, 20 August 2008 (UTC)

I agree. My first comment also was in agreement with your remarks, particularly concerning invariance under SL(n). I was merely pointing out that some of this discussion is available at the other page. You are welcome to copy some of it over to here if you like, or present your own perspective. Katzmik (talk) 08:29, 21 August 2008 (UTC)
Over compact oriented submanifolds. In the (more natural) cooriented situation, you need to use multivector densities. But in any case, it's not our job on Wikipedia to present our own perspective. Geometry guy 14:23, 21 August 2008 (UTC)

## error in definition of de Rham cohomology

In the definition of the exterior differential complex, it seems to me that the first term "R" should be deleted. Otherwise an interval has trivial 0-dimensional cohomology. Katzmik (talk) 09:00, 21 August 2008 (UTC)

Fixed. Geometry guy 14:00, 21 August 2008 (UTC)

## Unclearness in expression of 1-form

The expression $\mathrm d f = \sum_{i=1}^n \frac{\partial f}{\partial x^i} \mathrm{d} x^i$ used to introduce the 1-form is circular because it invokes the 1-form dxi. The circularity could be removed if the value of dxi had been given previously. The preceding paragraph seems to attempt to describe dxi but only gets as far as ∂xi / ∂xj which is not enough. I'm speaking as someone who has learned multivariable calculus but not yet differential forms. Halberdo (talk) 21:48, 12 January 2009 (UTC)

It seems that dxip(a) is the projection of a onto the xi axis. As I am new to this subject, I hesitate to add that explanation to the article, in case it is wrong; but if it is right, would someone please do so?Halberdo (talk) 22:17, 12 January 2009 (UTC)

Right, $\scriptstyle{dx^i}_p(a)=a^i$, where $\scriptstyle a=a^1e_1+\cdots+a^ne_n$ in case of euclidean forms. These $\scriptstyle{dx^i}_p$ are the generators of the dual space of the tangent space $\scriptstyle T_pM$, which in turn also is generated by the derivations $\scriptstyle\frac{\partial}{\partial x^i},$--kmath (talk) 23:39, 18 January 2009 (UTC)

## K-form redirect

It would be helpful if this article could be found by searching for "k-form". At he moment "k-form" redirects to "linear equation", an article in which the term "k-form" isn't even mentioned.Dependent Variable (talk) 04:03, 2 October 2009 (UTC)

Done--kmath (talk) 01:07, 4 October 2009 (UTC)

## Question for an expert

Comparison of vector algebra and geometric algebra states, "In advanced mathematics, particularly differential geometry, neither is widely used, with differential forms being far more widely used."

Could someone knowledgable say a few words on why differential forms are preferred (at least in some contexts) over vector algebra and geometric algebra?

SteelSoul (talk) 17:46, 16 February 2010 (UTC)

Vector algebra is specific to three dimensions. But there are many non-three-dimensional spaces out there: For instance, if I have a physical system consisting of two particles such as the Earth and the Moon, then I need six coordinates: three for the Earth and three for the Moon. Vector algebra does not work in this sort of space. Differential forms do.
Another perhaps less satisfying reason is that differential forms are more "natural" in a way that's hard to explain without a lot of abstract theory. The tangent bundle and cotangent bundle are interesting and canonically defined objects, and taking the exterior algebra is a pretty straightforward operation. From that, we get differential forms and all of their wonderful properties. Vector algebra, on the other hand, is messy: There's no apparent reason why we pick those formulas for div, grad, curl, and the cross product (and in fact the cross product has an implicit dependence on a metric—a metric is extra data). The only reason I know of to pick those formulas ultimately reduces down to facts about differential forms, where the formulas pop out without any fuss at all. Ozob (talk) 00:15, 17 February 2010 (UTC)

## Description of Clifford algebras is incorrect

Article has: 'The exterior product on a Clifford algebra differs from the exterior product of k-vectors (dual to the exterior product of k-forms) in that $v \wedge v = Q(v)$ in a Clifford algebra (the square of a vector is the quadratic form, applied to the vector), rather than 0; it is a non-anti-commutative ("quantum") deformation of the exterior algebra. This structure is used in geometric algebra.'

This is not correct, in the sense that the exterior product of a vector with itself is not the same as the square of a vector in a Clifford algebra, because the wedge product and the Clifford product are different operations with different results. Even when the underlying vector spaces are isomorphic, the two operations are generally distinct, and only coincide when the metric is totally degenerate (ie. 0). See [1] Penguian (talk) 11:25, 18 March 2010 (UTC)

Fixed. Ozob (talk) 00:18, 19 March 2010 (UTC)

## References

1. ^ P. Lounesto, Clifford Algebras and Spinors, Chapter 14.

## What is the example of a Differential 2-form?

I find the language under concept to be a bit confusing. In particular, when it's written "This is an example of a differential 2-form: the exterior derivative dα..." is it meant that the statement above is the differential 2-form or that the exterior derivative dα is the differential 2-form? — Preceding unsigned comment added by 46.65.200.69 (talk) 19:40, 26 February 2014 (UTC)

I've copyedited the article a little; it should be somewhat clearer now. Ozob (talk) 03:39, 27 February 2014 (UTC)
Ah! So the 2-form is dα and for there to exist f such that α=df, we require that this 2-form, that is, dα, is zero! Thank you, sir/madam, you are a scholar and a gentleman/lady! — Preceding unsigned comment added by 2001:630:12:2E1E:1DEC:488B:BD0C:C6A9 (talk) 11:58, 27 February 2014 (UTC)
You're welcome! Ozob (talk) 14:52, 27 February 2014 (UTC)