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Hi, actually I have several related questions, not worth opening different threads:

  1. What is the of the exterior derivative intuitively? What is its geometric meaning? A possible answer I know is, that it is dual to the boundary operator of singular homology. However I would prefer a more direct interpretation.

  2. What is a conceptually nice definition of the exterior derivative?

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Related: mathoverflow.net/questions/10574/… –  Qiaochu Yuan Apr 11 '10 at 19:11

7 Answers 7

up vote 28 down vote accepted

Many years back I wrote something about an intuitive way to look at differential forms here. In particular, figure 4 illustrates Stokes' theorem in a way that generalises to higher dimensions. Note that these are just sketches for intuition, and I've found them useful for illustrating various fields arising in physics, but they're not anything rigorous. They're also, in some sense, dual to the diagrams in Misner, Thorne and Wheeler. (There are some errors in that document, but I lost the source code many years ago...)

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Wow, this is just awesome :) I thought I already had good pictures in my mind of how a differential form looks, but I was wrong! –  Jan Weidner Apr 11 '10 at 19:49
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Very nice indeed. I strongly recommend reading the linked file. –  André Henriques Feb 19 '11 at 22:48

For 1-forms, you can get some intuition from Frobenius's theorem which states that a distribution D is integrable if and only if the ideal of differential forms that are annihilated by it is closed under exterior differentiation:

Let $\alpha$ be a 1-form on $M$. If $\alpha$ does not vanish, then ker $\alpha_x$ is a hyperplane in the tangent space to $M$ at $x$. Thus ker $\alpha$ is a hyperplane field in $TM$ (and is an example of a distribution). At every point in M, you should visualize a hyperplane passing through that point.

Frobenius's theorem gives conditions on whether this hyperplane field is integrable, that is, if one can fit the planes together to form hypersurfaces in $M$. In this case, it turns out that one can fit the planes together if and only if $d\alpha\equiv0$. (In the general case, where instead of $\alpha$ we have an algebraic ideal of 1-forms $\mathcal I$, this is $d\mathcal I\equiv 0\mod\mathcal I$).

Here's some more discussion: if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const. Similarly, if $d\alpha=0$ then it's clear that we can find such $f$ locally (not globally if $\alpha$ isn't exact).

Hence $d\alpha$ roughly measures how far this hyperplane field defined by ker $\alpha$ is from being tangent to hypersurfaces.

(I got the ideas from Appendix B of Ivey and Landsberg's book Cartan for Beginners).

Here's an example of a hyperplane field which is not tangent to any hypersurfaces. $\alpha = dz-y dx$ on $\mathbb R^3$ and $d\alpha = dx dy$:

standard contact structure on R^3

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Great answer! I wish I could accept both, your answer and sigfpe's. These two answers also fit nicely together. –  Jan Weidner Apr 11 '10 at 20:28
    
I wonder if there's a way to "see" the exterior derivative for k-forms with k>1 along these lines, see e.g. t3suji's comment here mathoverflow.net/questions/12266/… –  j.c. Apr 11 '10 at 20:34
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If M=R^2, the 1-forms w=dx and q=ydx determine the same (integrable) distribution of hyperplanes, but dq is not zero. Perhaps the proposition should be weaker: A 1-form w determines an integrable distribution iff a non vanishing real-valued function f exists such that d(fw)=0. (¿Is this true?) –  Marcos Cossarini Jul 7 '10 at 3:19
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Not quite right. Integrability of the kernel of $\alpha$ is measured by $\alpha \wedge d\alpha$, not by $d\alpha$. –  Ben McKay Feb 21 '13 at 21:28

I think that the best explanation is in Arnold's book "Mathematical methods of classical mechanics". Here it is: after fixing a chart on a manifold one can say that the value of $d\omega$ ($\omega$ is a n-form) on tangent vectors $(\xi_1, ...,\xi_{n+1})$ at point $x_0$ equals to the coefficient of the $(n+1)$-linear part of the function $F(\varepsilon)=\int_{\partial V(\varepsilon)} \omega$, where $V(\varepsilon)$ is a "curvilinear parallelepiped" with vertexes $x_0, x_0+\varepsilon \xi_1, ..., x_0+\varepsilon \xi_{n+1}$: $F(\varepsilon)=(d\omega)(x_0)(\xi_1, ...,\xi_{n+1})\varepsilon^{n+1}+o(\varepsilon^{n+1})$.

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This is a good explanation too! –  j.c. Apr 11 '10 at 20:29
    
In my opinion this is really "the" interpretation. Note that it treats functions and forms on an equal footing: the definition of the derivative of a function is exactly given by the above. Proving that $d\omega$ is in fact a form from this definition is also quite enlightening. Pedagogically, it is also great. It basically says "$d\omega$ is what makes stokes theorem true on 'infinitesimal' parallelepipeds". –  Steven Gubkin Dec 20 '13 at 22:13

There is a following (it seems to me it is not well-known but interesting) approach to differential forms. I'll try to reproduce it here. In this approach the exterior derivative is a very simple operation.

What is a differential k-form on a manifold $M$? Consider a (k+1)-product $V_{k+1}(M)=M\times...\times M$. Denote by $S_k(M)$ the space of all smooth skew-symmetric (with respect to a product structure) functions on $V_{k+1}$. Obviously any function from $S_k(M)$ equals to zero on the diagonal $\Delta=$ {$(x,x,...,x)| x\in M$}.

We define a subspace $L_k(M) \subset S_k(M)$ as follows: $L_k(M)$ consists of all elements of $S_k(M)$ of order smaller then $k$ along the $\Delta$. In other words $f\in S_k(M)$ if and only if for any smooth path $I(t)$ starting on the diagonal(i.e. $I(0)\in \Delta$) holds $I(t)=o(t^k)$.

Then one can identify the space of all k-forms $\Omega_k(M)$ with a quotient $S_k(M)/L_k(M)$.

What is the exterior derivative? Consider the following operator $\delta: S_k(M)\to S_{k+1}(M)$, $\delta f(x_1,...,x_{k+2}) =\sum (-1)^{i+1} f(x_1,..,\hat{x_i},...,x_{k+2})$. One can check that $\delta (L_k(M))\subset L_{k+1}(M)$ and that the induced operator $\Omega_k(M)=S_k(M)/L_k(M)\to S_{k+1}(M)/L_{k+1}(M)=\Omega_{k+1}(M)$ coincides with the exterior derivative $d$.

I know that approach from B.L. Feigin's lectures on multidimensional calculus (in russian here: http://ium.mccme.ru/f98/calcman.html).

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This is cool! Am I right in thinking that you can identify the set of k-vectors on M with the set of derivations on S_k, just as you can identify the set of 1-vectors on M with the set of derivations on S_1? –  Vectornaut Apr 12 '10 at 2:46
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This seems to me analogous to the way algebraic geometers define $\Omega^{1}$ as $\mathcal{I}_\Delta/\mathcal{I}_\Delta^{2}$, where $\mathcal{I}_\Delta$ is the ideal sheaf of the diagonal. –  Qfwfq Apr 12 '10 at 16:39

The exterior derivative is the unique (sequence of) linear map $d: \mathcal{A}^p (M) \to \mathcal{A}^{p+1}$, such that the following axioms hold:

  1. for a function $f$, $df$ is the total differential.
  2. For any function $f$ and any differential form $a$, the Leibniz rule $d(fa)= df \wedge a + f da$ holds.
  3. For any diffeomorphism $\phi: M \to N$, you have $\phi^{\ast} \circ d = d \circ \phi^{\ast}$.

I think that 3 is more natural or at least easier to motivate than the usual $dd=0$. But both properties are really equivalent.

Proof (of uniqueness): 2. implies locality, i.e. the value of $d a$ at a point $x \in M$ only depends on the value of $a$ in a neighborhood of $x$. This, together with the axiom 3, shows that it is enough to consider $M =\mathbb{R}^n$.

The group $\mathbb{R}^n$ acts by translations on $\mathbb{R}^n$. By axiom 3, for any translation-invariant form $a$ on $\mathbb{R}^n$, the form $da$ is again translation-invariant.

On the other hand, each nonzero $\lambda \in \mathbb{R}$ gives rise to the diffeomorphism $h_{\lambda}:x \mapsto \lambda x$ of $\mathbb{R}^n$. It is easy to check that it acts on translation-invariant $p$-forms by multiplication with $\lambda^p$. Thus for any translation-invariant $p$-form $a$, you get

$$\lambda^p d a = d (\lambda^p a) = d (h_{\lambda}^{\ast} a ) = h_{\lambda}^{\ast} d a = \lambda^{p+1} da,$$

which implies that any translation-invariant form is closed. Finally, note that any $p$-form on $\mathbb{R}^n$ can be written as a linear combination of translation-invariant form, with coefficients in $C^{\infty}(\mathbb{R}^n)$ (a basis for the translation-invariant forms is formed by the usual elements $dx_{i_1} \wedge \ldots \wedge x_{i_p}$).

From axioms 1 and 2, you now conclude that $d$ must be the exterior derivative that you knew before. This, of course, implies all the other properties of $d$.

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This is great! It's the first definition of exterior differentiation that ever really made sense to me. I think I'll be using this one from now on. –  Vectornaut Sep 14 '12 at 21:05

For 2: it is the unique extension of the total differential $d:C^\infty(M)\to\Omega^1(M)$ to a graded derivation of the algebra $\Omega^\bullet(M)$ of differential forms.

The map $d:C^\infty(M)\to\Omega^1(M)$ itself has a nice characterization as a universal derivation of the algebra $C^\infty(M)$ of functions satisfying certain rather reasonable conditions---this follows from Jaak Peetre's theorem.

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The definition of a graded derivation was originally just a natural generalisation of $d$, so this approach is almost circular, and I can't visualise it geometrically. –  Ben McKay Feb 21 '13 at 21:33
    
Well, my point is that you need only visualize the component in degree zero, as the rest is simply formalities. –  Mariano Suárez-Alvarez Feb 21 '13 at 23:27

Another conceptually nice definition of the exterior derivative is given in Bourbaki (Varietes differentielles et analytiques, Fascicule de resultats), (8.3.4) and (8.3.5). The idea is the following: if w is an exterior p-form on X, consider it as a section w: X to Omega^p(X) of the bundle Omega^p(X) of p-forms. It makes sense to take its derivative dw at each point x in X. Then one sees that dw corresponds to a p+1 exterior form.

By the way, a natural and simple definition of tangent vector on a smooth manifold is given in the same book in (5.5.1).

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So the derivative of $\omega \colon X \to \Omega^p(X)$ at $x \in X$ is I guess the tangent map $T_x(\omega) \colon T_x X \to T_{\omega(x)} \Omega^p(X)$. How do you get the $p+1$-form? –  Michael Murray Feb 21 '13 at 9:55

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