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One way to define the algebra of differential forms $\Omega(M)$ on a smooth manifold $M$ (as explained by John Baez's week287) is as the exterior algebra of the dual of the module of derivations on the algebra $C^{\infty}(M)$ of smooth functions $M \to \mathbb{R}$. Given that derivations are vector fields, 1-forms send vector fields to smooth functions, and some handwaving about area elements suggests that k-forms should be built from 1-forms in an anticommutative fashion, I am **almost** willing to accept this definition as properly motivated.

One can now define the exterior derivative $d : \Omega(M) \to \Omega(M)$ by defining $d(f dg_1\ \dots\ dg_k) = df\ dg_1\ \dots\ dg_k$ and extending by linearity. I am **almost** willing to accept this definition as properly motivated as well.

Now, the exterior derivative (together with the Hodge star and some fiddling) generalizes the three main operators of multivariable calculus: the divergence, the gradient, and the curl. My intuition about the definitions and properties of these operators comes mostly from basic E&M, and when I think about the special cases of Stokes' theorem for div, grad, and curl, I think about the "physicist's proofs." What I'm not sure how to do, though, is to relate this down-to-earth context with the high-concept algebraic context described above.

**Question:** How do I see conceptually that differential forms and the exterior derivative, as defined above, naturally have physical interpretations generalizing the "naive" physical interpretations of the divergence, the gradient, and the curl? (By "conceptually" I mean that it is very unsatisfying just to write down the definitions and compute.) And how do I gain physical intuition for the generalized Stokes' theorem?

(An answer in the form of a textbook that pays special attention to the relationship between the abstract stuff and the physical intuition would be fantastic.)

35A 1-form is a function which grows proportionally to how fast you are moving. Thus it doesn't matter how you parametrize the curve you are moving on - you either end up integrating a smaller function for a longer period of time, or a bigger function for a shorter period of time. This is why you can't integrate functions on manifolds - they have no intrinsic "unit speeds", because there are many choices of local coordinates - but you can still integrate differential forms. k-forms just generalize this to higher dimensions. – Steven Gubkin – 2011-01-27T20:25:22.107

Community wiki? – Harry Gindi – 2010-01-03T11:04:03.277

15Really? I would like to award reputation for good answers and I am not necessarily just looking for a list of recommendations; perhaps someone has a clear enough intuition that it can be described in a paragraph or two. – Qiaochu Yuan – 2010-01-03T11:30:06.320

2This is nitpicky, but it is traditional to denote the product structure on differential forms with a $\wedge$ ("wedge product"). – Pete L. Clark – 2010-01-03T12:22:09.277

9Have you seen

From Calculus to Cohomologyby Madsden and Tornehave? It's not really about physical intuition (which is why I'm making this a comment), but it might be helpful. – Akhil Mathew – 2010-01-03T15:28:11.9005I still think this should be community wiki because it's a sorted list. I didn't like an answer, and I'd like to vote it down, but not the user. – Harry Gindi – 2010-01-03T15:46:39.227

11+1 for "I freaking love this question". – B. Bischof – 2010-01-14T22:53:18.150

2@StevenGubkin I was waiting for this explanation for years! – Norbert – 2015-05-31T19:03:29.690

1I personally find multivariable calculus already better formulated in terms of covectors rather than vectors; IMO the exterior derivative is more natural than directional derivatives, especially in terms of symbolic calculation, and furthermore $\mathrm{d}(xy)$ is

always$y \mathrm{d}x + x \mathrm{d}y$, whereas there are several relevant ways in which $\frac{\partial}{\partial x}(xy)$ won't be $y$ (e.g. if $y$ depends on $x$, or you are working in different coordinates and mean to hold $x-y$ constant rather than holding $y$ constant). – Hurkyl – 2015-09-23T07:42:49.987