Talk:Exterior derivative

WHY DEFINE IT ANTISYMMETRIC???
Does anyone know where I can see a motivation for defining the exterior derivative antisymmetrically instead of just using the *normal* derivative of forms? I heard some time ago from a professor that it had to do with the manifold structure and not being transferred properly from chart to chart or something like that, unless we only take the antisymmetrization. I was not curious at that time so I didn't get a full explanation. any references for that? I always see that textbooks just define it and work with it.... (Arestes (talk) 17:57, 11 December 2011 (UTC)).


 * A "form" is by definition antisymmetric. It's defined that way because a form lives to be integrated; it gives you a way of linearly assigning a signed "volume" to a k-parallelepiped. If any two legs of the parallelepiped are the same, it's flat and the volume had better be 0.  From there and linearity, antisymmetry follows easily.  A somewhat deeper point is that you want the integral to be invariant in the sense that if f is a smooth diffeomorphism from a manifold M to a manifold N, $$\int_M f^*(\omega)=\int_N \omega$$. More generally, if  f is not surjective, but $$\omega$$ is 0 outside the image of M, then M could be $$R^n$$ and this is exactly what you need to have "independence of parameterization" by any one particular chart. This is equivalent to what your professor said about charts. If you think about the change-of-variables formula from multivariable calculus, it involves the determinant of the Jacobean; it is the antisymmetry of the forms that magically brings this determinant into being and thereby makes the integral independent of the parameterization.
 * Natkuhn (talk) 00:51, 18 December 2012 (UTC)


 * A slightly more down-to-earth if lazy edit: these forms are supposed to generalize determinants. Just as a determinant is antisymmetric (when you switch rows, the sign changes), so are these forms.   Tkuvho (talk) 10:38, 25 December 2012 (UTC)


 * Moreover, if you calculate, say, $$d(dx\wedge ydz)$$, by antisymmetry of forms this should be the same as $$-d(ydz\wedge dx)=-dy\wedge dz \wedge dx = -dx\wedge dy \wedge dz$$, whereas without the minus sign in the formula for d you would get the wrong sign in the answer. The skew nature of the exterior derivative is forced on you by the skew nature of the exterior forms.  Tkuvho (talk) 16:14, 25 December 2012 (UTC)

old edit
does the exterior derivative have a universal construction? does that even make sense to ask? -Lethe | Talk 06:49, Jan 21, 2005 (UTC)

What property would you want the exterior derivative to satisfy in the universal sense? I haven't heard of anyone treating this issue from a categorical perspective. Perhaps take a look at exterior algebra for more information on how this concept can be generalized. - Gauge 23:32, 7 Apr 2005 (UTC)

The theorem I know about is a characterisation as a natural differential operator (due I think to Palais). Charles Matthews 09:22, 8 Apr 2005 (UTC)

៛== codifferential ==

Should this page perhaps have something about the "codifferential"? i.e. the Hodge dual of the exterior derivative. --anon


 * There is something about this at Hodge_star. I don't know anything about Hodge-related stuff. If you feel like writing something in here, be bold! Oleg Alexandrov 17:45, 15 August 2005 (UTC)

Using the exterior derivative and the Hodge Star Hodge_star is the most economical way to present vector calculus for applications in curvilinear coordinates; here's a brief discussion with a few examples.

The Hodge Star in 3-dimensional euclidean space is given by:

$$ \displaystyle $$
 * 1=\sigma_1 \wedge \sigma_2 \wedge \sigma_3;

$$ $$
 * \sigma_1=\sigma_2 \wedge \sigma_3, *\sigma_2=\sigma_3 \wedge \sigma_1,*\sigma_3=\sigma_1 \wedge \sigma_2;

$$ $$
 * (\sigma_2 \wedge \sigma_3) =\sigma_1, *(\sigma_3 \wedge \sigma_1) =\sigma_2, *(\sigma_1 \wedge \sigma_2)=\sigma_3;

$$ $$
 * (\sigma_1 \wedge \sigma_2 \wedge \sigma_3)=1.

Expressions for Div, Grad, Curl and Laplacian Using the Hodge Star

If f is a scalar function and g is a vector function, we have:

$$ df=(\vec{\nabla} f)_i \sigma_i, $$

$$ d (g_i \sigma_i)=(\vec{\nabla} \times \vec{g})_i *\sigma_i, $$

$$ d (g_i *\sigma_i)=({\vec {\nabla}} \cdot {\vec{g}})*1, $$

$$ d * d f=\nabla^2 f *1. $$

where the sum is over repeated indices (the "einstein summation convention"). These expressions are coordinate independent, and this makes it easy to take div, grad, curl, in curvilinear coordinates.

Fundamental Theorem of Calculus

In classical vector analysis in three dimensions, the fundamental theorem of calculus reads:

$$ \displaystyle \int_P \vec{\nabla} f\cdot \vec{dl}=f(\vec{b})-f(\vec{a}) $$, where P is a parameter curve with endpoints $$\vec{a},\vec{b}$$.

$$ \displaystyle \int_{\partial A}\vec{g} \cdot \vec{dl}=\iint_{A}(\vec{\nabla} \times \vec{g})\cdot \vec{dA} $$, where A is an area with whose boundary is the closed loop $$\partial A$$.

$$ \displaystyle \iint_{\partial V} \vec{g} \cdot \vec{dS}=\iiint_{V} ({\vec {\nabla}} \cdot {\vec{g}})dV $$ where V is a closed volume enclosed by $$\partial V$$.

In term of the exterior derivative, these three expressions may be written in a unified form:

$$ \displaystyle \int_{C} dh =\int_{\partial C} h $$

where h is a form, C is a chain overwhich the form is being integrated, and $$\partial C$$ is the boundary of the chain.

'''Example 1. Canonical one-forms in Spherical Coordinates.'''

In spherical coordinates, a displacement by a small amout in the r direction results in a displacement dr, in the theta direction a displacement r dθ, and in the phi direction a displacement r sin θ dφ, so the canonical one-forms are given by:

$$ \displaystyle \sigma_r =dr, \sigma_\theta =r d\theta, \sigma_\phi =r \sin \theta d \phi. $$

Example 1.1: Gradient.

Consider a scalar function f(r,θ,φ) in spherical coordinates.

$$ df =\frac{\partial f}{\partial r} dr  +\frac{\partial f}{\partial \theta} d \theta +\frac{\partial f}{\partial \phi} d \phi $$

Rewriting the d's in terms of the canonical one-forms gives:

$$ df =\frac{\partial f}{\partial r} \sigma_r +\frac{1}{r}\frac{\partial f}{\partial \theta} \sigma_\theta +\frac{1}{r \sin \theta}\frac{\partial f}{\partial \phi} \sigma_\phi $$

The components of the σ's are the components of the gradient.

$$ \vec{\nabla} f=\frac{\partial f}{\partial r} \hat e_r +\frac{1}{r}\frac{\partial f}{\partial \theta} \hat e_\theta +\frac{1}{r \sin \theta}\frac{\partial f}{\partial \phi} \hat e_\phi $$

'''Example 1.2: Curl of a vector function in spherical coordinates. '''

Consider the (Dirac) vector potential of a magnetic monopole:

$$ \vec{A}=q \frac{(1-\cos \theta)}{\sin \theta} \hat e_\phi $$

To find the magetic field,

$$\vec{B}= \vec{\nabla} \times \vec{A}$$

we write the vector potential as a one-form:

$$ \vec{A}=q \frac{(1-\cos \theta)}{r \sin \theta} \sigma_\phi=q(1-\cos\theta) d \phi $$

and then take the exterior derivative:

$$ d (q (1-\cos\theta) d\phi) =\sin \theta d \theta \wedge d\phi=\frac{q}{r^{2}}*\sigma_r $$

The magnetic field is given by:

$$ \vec{B}= \frac{q}{r^{2}}\hat e_r $$

--Sfitzsi 02:11, 28 October 2007 (UTC)talk

definition
Maybe this is in relation with the first question, but regardless of lack of a universal definition or something, shouldn't it be pointed out that this definition is a bit risky as it uses coordinates. What is an important property is that it is independent of coordinates. Proof is probably too much to ask, but shouldn't this subtlety be mentionned?
 * Yes it should be mentioned. But actually, the invariant formula can be taken as a definition, which does not rely on local coordinates.  By the way, I did find an answer to my question above: the universal construction can be found in Hartshorne, and I'm going to add it some day. -lethe talk 23:17, 2 January 2006 (UTC)

differential topology vs differential geometry
Why does the intro say the exterior derivative operator of differential topology? In my experience, the exterior derivative is discussed in differential geometry texts. Differential topology deals with the topology of differential manifolds, such as cobordism. A glance at the table of contents in a typical book, like Differential Topology by Hirsch or Topology from the Differential point of view by Milnor will not turn up exterior derivative. Is it because of the deRham complex? This would be more likely to be covered in a book about algebraic topology. Rmilson 02:51, 27 April 2006


 * The two articles redirect to the same place, Differential geometry and topology, so your edit is meaningless, although I guess one day this conflation will be corrected, and then these links will be better. For what it's worth, in my mind, differential geometry is stuff that studies the local properties of extra stuff on manifolds like symplectic forms, volume forms, metric tensors, connections, almost compex structures, etc.  In this view, the exterior derivative belongs to the topology side.  And remember that the existence of a differential structure is definitely a topological question.  But I don't feel strongly enough about the issue to argue about it.  One day, someone will want to split that article, and we can argue and argue about which is which, and once we get it all straight, it will matter which way our links point, but until that day, it's pretty irrelevant. -lethe talk [ +] 03:31, 27 April 2006 (UTC)

No it was correct, roughly diff geometry if you have a type of curvature and diff.topology otherwise (for example, simplectic topology and Rieamnnian geometry)--Tosha 23:26, 1 May 2006 (UTC)


 * It would be good to have a source for this idea.  Differential topology is a pretty specialized subject.  Interestingly, Mathematics Subject Classification puts the exterior derivative in category 58AXX General theory of differentiable manifolds.  Differential geometry is 53-xx.  Differential topolgy is 57Rxx, part of 57-XX, Manifolds and cell complexes.

I seem to have stepped into this one before reading this talk page, sorry. But the last post supports my feeling: differential forms do not belong to either differential geometry (which is the land of Riemannian and I suppose Lorentz metrics) or differential topology (which has to do with things like transversality, cobordism, Morse theory, etc). So differentiable (or smooth) manifolds is the correct domain, and that's what my change reflected. Natkuhn (talk) 00:58, 18 December 2012 (UTC)

Typo?
I'm tempted to make the same edit as the one made by an anonymous user on 4 February and reverted by Oleg Alexandrov: shouldn't

"Note that if $$i = I$$ above then $$dx_i \wedge dx_I = 0$$"

in the section "Definition" say "\in I" instead of "= I"? i is an index, I is (as stated immediately previously in the definition) a multi-index, ie a subset of {1, ... n}. Spurts (talk) 08:40, 9 February 2008 (UTC)
 * Sorry. It was not clear what $$I$$ meant from the way things were defined. Note that given that $$I$$ is a multi-index, I am not sure you can apply the notation $$i\in I$$ which would only work if $$I$$ were a set. See the way I clarified the notation in the article. Comments and corrections welcome. Oleg Alexandrov (talk) 16:50, 10 February 2008 (UTC)
 * Yes, nice clear rephrasing. Spurts (talk) 18:36, 10 February 2008 (UTC)

Derivation or anti-derivation?
Hello, Isn't the Exterior derivation in fact an "antiderviation" and not a "derivation" (because of the minus sign?) I hope someone will be able to verify this... —Preceding unsigned comment added by 132.64.102.151 (talk) 14:50, 6 June 2010 (UTC)


 * The exterior derivation is a derivation in the graded world, an antiderivation in the ungraded world. It is correct to say derivation of degree 1. Simply to say derivation would indeed be wrong, or at least ambiguous.--345Kai (talk) 15:43, 13 August 2011 (UTC)

Einstein notational convention
Is the multi-index in the local coordinates section is being used to imply summation via Einstein notation convention? In the opinion of someone new to this topic (notably myself), the usage is unclear in the article. 99.19.84.64 (talk) 20:30, 5 August 2013 (UTC)
 * I agree – it seems like a curious mix of implicit and explicit summation. Somewhat confusing.  — Quondum 02:43, 7 August 2013 (UTC)
 * I'm also new to the topic, but I've seen the notation $$\omega = \sum_I f_I dx^I$$. Was it just a typo to leave off the sum in this line? — Carl (Seaplant (talk) 22:43, 4 December 2018 (UTC))

Different definitions
The second definition given here, in terms of local coordinates, seems to be the same as the one given by Spivak: CALCULUS ON MANIFOLDS W. A. Benjamin (1965), page 91

Spivak's next theorem, 4-10(2), then states

If \omega is a k-form and \eta is an l-form, then

d(\omega\wedge\eta) =  d\omega \wedge \eta  +  (-1)^{k l} \omega \wedge d\eta.

This conflicts with the first "axiomatic" definition on this web page. In it, using Spivak's notation, the sign of the second term is (-1)^k,  not  (-1)^{k l}.

I am not sure which of the first two definitions corresponds to the third "invariant formula" definition.

Possible (overlapping) sources of confusion:

- coordinates may be indexed from 1 to n, or from 0 to n-1

- the sign of the differential operator "d" may be reversed for odd-dimensional forms

- in a monomial with coordinates, the coefficient function may appear at the end: "d\x_1 \wedge ... dx_n \wedge df" rather than  "df \wedge dx_1 \wedge ... dx_n"

- in the 3rd definition, in terms of invariant formula, the sign of all terms containing a Lie bracket may be reversed.

Am I in a muddle, or is there real confusion here? AlHutchins (talk) 09:05, 14 May 2016 (UTC)