What is the exterior derivative intuitively? - MathOverflow

What is the exterior derivative intuitively?

2010-04-11T19:05:58Z

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

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

Answer by Mariano Suárez-Alvarez for What is the exterior derivative intuitively?

2010-04-11T19:09:42Z

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.

Answer by Dan Piponi for What is the exterior derivative intuitively?

2010-04-11T19:28:25Z

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...)

Answer by Petya for What is the exterior derivative intuitively?

2010-04-11T20:11:36Z

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 $x_0, x_0+\varepsilon \xi_1, ..., x_0+\varepsilon \xi_{n+1}$: $F(\varepsilon)=(d\omega)(x_0)(\xi_1, ...,\xi_{n+1})\varepsilon^{n+1}+o(\varepsilon^{n+1})$.

Answer by jc for What is the exterior derivative intuitively?

2010-04-11T20:17:56Z

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 $d\alpha = dx dy$:

Answer by Petya for What is the exterior derivative intuitively?

2010-04-11T22:51:07Z

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).

Answer by Johannes Ebert for What is the exterior derivative intuitively?

2011-02-19T18:52:06Z

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:

for a function $f$, $df$ is the total differential.
For any function $f$ and any differential form $a$, the Leibniz rule $d(fa)= df \wedge a + f da$ holds.
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 $d$.

Answer by fcukier for What is the exterior derivative intuitively?

2011-02-19T19:36:30Z

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).