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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
up vote 38 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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Very nice indeed. I strongly recommend reading the linked file. – André Henriques Feb 19 '11 at 22:48
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Using these sorts of pictures, how do you see that the derivative of $dx + y\;dz$ is $dy \wedge dz$? Can you even draw a picture for $dx + y\;dz$? These pictures seem to characterize differential forms as "foliations of varying density," but I'm pretty sure that's only true for forms that are locally a function times an exact form, and there are lots of forms—for example, $dx + y\;dz$—that don't look like this. – Vectornaut Nov 6 '14 at 1:52
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By the way, in case you haven't seen it already, Gabriel Weinreich's Geometrical Vectors introduces differential forms and the exterior derivative from the perspective you describe. – Vectornaut Nov 6 '14 at 1:57
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@Vectornaut We could think of $dx + y\, dz$ as a formal sum of your "foliations", where we have a density of planes parallel to the $x$-axis and simultaneously a density of (half-)planes parallel to the $z$-axis. It would be somewhat analogous to how we can talk about the boundary of a linear combination of overlapping simplicies in homology. – epimorphic Jan 7 at 20:58
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The visualizations you refer to in Misner, Thorne, and Wheeler were originated by J.A. Schouten, and first presented in Ricci-Calculus: An Introduction to Tensor Analysis and its Geometrical Applications amazon.com/… . William L. Burke has given more concise and modern presentations, e.g., in Applied Differential Geometry. – Ben Crowell Jan 18 at 16:11

For 1-forms, you can get some intuition for exterior differentiation from how it shows up in 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 a foliation by hypersurfaces in $M$. For a hyperplane field defined by a single 1-form one can fit the planes together if and only if $d\alpha$ mod $\alpha$ is zero. This is usually expressed by the vanishing of $\alpha\wedge d\alpha$.

(In the general case, where instead of $\alpha$ we have a set of linearly independent 1-forms $\{\alpha_j\}_{j=1}^r$, the ideal in the algebra of differential forms on $M$ generated by $\{\alpha_j\}_{j=1}^r$ must be closed under exterior differentiation; equivalently $d\alpha_j\wedge\alpha_1\wedge\cdots\wedge\alpha_r=0$ for all $j$).

Two simple examples:

(1) if $\alpha=df$ then the field of hyperplanes ker $\alpha$ is actually tangent to the hypersurfaces $f=$const (and of course $d\alpha=0$).

(2) If $\alpha = g df$ for some non-vanishing function $g$, e.g. $\alpha=ydx$ in the upper half plane of $\mathbb{R}^2$, then this is just as good, since ker $\alpha$ is still tangent to $f=$const. Note that $d\alpha=dg\wedge df=(dg/g)\wedge\alpha$, which vanishes mod $\alpha$ and thus $\alpha\wedge d\alpha=0$.

Hence $\alpha\wedge d\alpha$, or $d\alpha$ mod $\alpha$ 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. Thanks to Marcos Cossarini and Ben McKay for pointing out in the comments that the original version of this was wrong!)

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

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
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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
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After 5 and a half years, I've finally fixed this. Thanks! – j.c. Sep 23 '15 at 13:19

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
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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
    
At the end of your third paragraph, should $I(t) = \mathrm o(t^k)$ be $f(I(t)) = \mathrm o(t^k)$? – L Spice Jan 19 at 15:19

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
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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
    
@Mariano Suárez-Alvarez Do you mean that for some subalgebra $A$ of $C^{\infty}(M)$ and $\Omega^1(M)$ truncated to be $A-$module $(A,\Omega^1(M))$ is Kähler differential? Could you give some refrences to your statement? – Fallen Apart May 26 '15 at 14:24
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@Mariano Suárez-Alvarez The sentence: "The map d:C∞(M)→Ω1(M) itself has a nice characterization as a universal derivation of the algebra C∞(M) of functions satisfying certain rather reasonable conditions" – Fallen Apart May 26 '15 at 15:40
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@FallenApart, ah. No , I do not mean that $\Omega^1(M)$ is the module of Kähler differentials of $C^\infty(M)$ (mostly, because it isn't! :) ) The operator $d:C^\infty(M)\to\Omega^1(M)$ can be characterized in terms of its functorial properties. This is surely done in detail in the book Natural Operations in Differential Geometry by Kolar, Michor and Slovak. – Mariano Suárez-Alvarez May 26 '15 at 16:14

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 $\omega$ is an exterior $p$-form on $X$, consider it as a section $\omega: X \to \Omega^p(X)$ of the bundle $\Omega^p(X)$ of $p$-forms. It makes sense to take its derivative $d\omega$ at each point $x \in X$. Then one sees that $d\omega$ 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

To start with 0-forms, $df$ codes how $f$ varies. In fact, it does this in a way that is, IMO, more natural than partial derivatives.

For example, if I want to know how $z = x^2 y$ varies with $x$ — actually that's a lie, I want to know how $z$ varies with $x$ as $y$ is held constant — then I compute an exterior derivative, setting $dy=0$:

$$ dz = 2xy \, dx + x^2 \, dy \equiv 2xy \, dx \pmod{dy} $$

Similarly, as (tangent) vectors are dual to one-forms, we can see that the exterior derivative is the thing you combine with a vector to get a directional derivative.

This is further supported by path integrals; if $\gamma$ is a path from $P$ to $Q$, then $\int_\gamma \, df = f(Q) - f(P)$; so again we see that $df$ is an encoding of how $f$ varies, and the path integral is how we accumulate the variation into a finite difference.

We can argue that $d(df)$ should be zero, ans the variation in the variation of $f$ is second derivative information, and differential forms are only intended to capture first derivative information. Similarly for $df \, df$.

Stokes' theorem expresses the analog of the fundamental theorem of calculus in higher dimensions, giving a way to see the exterior derivative of a differential form as encoding the higher degree variation.

Alternatively, we can appeal to Fubini's theorem to reduce to the one-dimensional case: here's a sketch.

Let x = $(x_1, \ldots, x_n)$ and $dx = dx_1 dx_2 \ldots dx_n$.

Suppose you wish to integrate $$ \int_X df \, dx $$ where $X$ is an $(n+1)$-dimensional region. If we let $X_x$ be the one-dimensional region defined by a constant value of $x$, then generalizing Fubini's theorem, we can write this as an iterated integral $$ \int_Y \left( \int_{X_x} df \right) dx $$ where $Y$ is some suitable $n$-dimensional space.

The integral $\int_{X_x} df$ is just the alternating sum of the values $f(P)$ where $P$ iterates over the endpoints of the curves comprising $X_x$, where the upper endpoints are added and the lower endpoints are subtracted. It's convenient to write this as an integral over a zero-dimensional surface: $\int_{\partial X_x} f$.

Consequently, the original integral can be written as $$ \int_Y \left( \int_{\partial X_x} f \right) dx $$ and again essentially by Fubini's theorem, we can identify this with $$ \int_{\partial X} f \, dx $$

Consequently, defining $d(f \, dx)$ as $df dx$ is exactly the right thing to do to generalize the fundamental theorem of calculus to get Stoke's theorem.

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The exterior derivative is an intrinsic way of talking about the gradient of a function. If you want to understand the intuitive meaning of the exterior derivative of $f$ you should make sure you understand $\nabla f$ properly. I am a little hesitant to post such an answer 5 years into the discussion but as I did not find any occurrence of the string "gradient" on the page I thought this might be useful. In the presence of a metric the relation between them is $\langle \nabla f, V\rangle=df(V)$ for tangent vectors $V$.

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To my mind the gradient only makes sense if you have a metric and may only be compared to $d$ on the level of $0$-forms. So it should not be necessary to understand $\nabla$ properly, before understanding $d$. – Michael Bächtold Jan 18 at 13:59
    
@MichaelBächtold, if one doesn't understand it at the level of $0$-forms one will certainly have trouble understanding it in general. Of course the gradient depends on the metric; that's why it needs to be replaced by the exterior derivative if one wants to work intrinsically. – Mikhail Katz Jan 18 at 14:04
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I mostly agree with your comment. But I'm not convinced that one should understand $\nabla f$ first. I have the impression that it may even hinder understanding, since $\nabla f$ is usually taught without explaining how it depends on more than $f$. Hence people have to forget hidden assumptions when they try to understand $df$. – Michael Bächtold Jan 18 at 14:47
    
There are some basic elementary geometric facts that form the foundation of understanding here, one of which being that a one-variable function is constant if its derivative is zero, that gradient has to do with direction of steepest descent, etc. Doing research in mathematics is not a formal game but rather involves understanding, without which it is difficult to do research. – Mikhail Katz Jan 18 at 14:50
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I'm not sure how to interpret your last comment. In case you are suggesting that without a metric the exterior differential necessarily carries no geometric meaning and reduces to its formal rules (Leibniz, $d^2=0$, naturality etc.) then I would disagree. Dan Piponis answer is an attempt to give a geometric understanding without a metric. Another quite elementary and intuitive approach can be found in Anders Kocks book Synthetic Geometry of Manifolds. His approach makes the statement "dual to the boundary operator of singular homology" directly accessible at the infinitesimal level. – Michael Bächtold Jan 18 at 15:18

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