Why is the exterior differentiation operator sometimes visualized as the “boundary”?

I have been trying to get an intuitive grasp on exterior calculus and found this article particularly interesting :

http://homepage.mac.com/sigfpe/Mathematics/forms.pdf

I can understand why the exterior derivative of a function is visualized using the "level curves", since the derivative is supposed to show the growth of the function and the level curves are more tightly packed where the function grows faster. I can also understand why the wedge product is the intersection of the two sets of level curves, since it's pretty easy to see for the basis $dx \land dy$ and you can "expand" every 1-form in terms of dx and dy. (I find this one interesting since the wedge does look a little like an intersection symbol...)

Now my question comes with the "d" operator. In the link I gave the author says that it should be visualized as the "boundary" of the set of level curves, which makes sense when you try to interpret stoke's theorem, and also gives an intuitive sense to $d^2=0$, but I can't find an explanation as to why this would be a good visualization?

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$d$ is really a coboundary operator--the formal adjoint of $\partial$. This fact in the context of the de Rham complex is the content of Stokes' theorem. –  Steve Huntsman Mar 30 '11 at 12:09
You might learn from this earlier MO question: "Is the boundary $\partial S$ analogous to a derivative?" mathoverflow.net/questions/46252 . See especially Terry Tao's enlightening answer. –  Joseph O'Rourke Mar 30 '11 at 13:14

Just to give you some more intuition, let $M$ be an oriented smooth real manifold of real dimension $m$ and consider the space $^s\mathcal E^p(M)$, $s\in\mathbb N\cup\{+\infty\}$, of differential forms $C^s(M,\Lambda^p T^*_M)$ endowed with the topology of uniform convergence of all the derivatives of order $\le s$ of the coefficients on every compact subset (contained in a coordinate chart) of $M$. If $K\subset M$ is a compact subset, let $^s\mathcal D^p(K)$ denote the subspace of elements $u\in ^s\mathcal E^p(M)$ with support contained in $K$, together with the induced topology. Finally, put $$^s\mathcal D^p(M)=\bigcup_{K\subset\subset M} ^s\mathcal D^p(K).$$ The space $^s\mathcal D'_p(M)=^s\mathcal D'^{m-p}(M)$ is by definition the topological dual $(^s\mathcal D_p(M))'$ and is called the space of currents of dimension $p$ (or degree $m-p$) and order $s$ on $M$.

You should think of currents as a generalization of differential forms (locally they are differential forms with distributional coefficients), precisely how distributions generalize the concept of functions.

Example 1. If $f$ is a differential form of degree $q$ with $L^1_{\text{loc}}$ coefficients, we can associate to $f$ the current $T_f$ of dimension $m-q$ $$\langle T_f,u\rangle=\int_Mf\wedge u,\quad u\in ^0\mathcal D^{m-q}(M).$$ $T_f$ is of degree $q$ and of order $0$. The correspondence $f\mapsto T_f$ is injective.

Example 2. Let $Z\subset M$ be a closed oriented submanifold of $M$ of dimension $p$ and class $C^1$ (here $Z$ may have a boundary $\partial Z$). The current of integration over $Z$, denoted $[Z]$, is defined by $$\langle[Z],u\rangle=\int_Z u,\quad u\in ^0\mathcal D^p(M).$$ It is clear that $[Z]$ is a current of order $0$ on $M$ and that $\operatorname{Supp}(Z)=Z$ (the support of a current is the smallest closed subset $A\subset M$ such that the restriction of the current to forms with compact support contained in the complementary of $A$ is zero). Its dimension is $p=\dim Z$.

The exterior derivative $dT\in ^{s+1}\mathcal D'^{q+1}(M)=^{s+1}\mathcal D'_{m-q-1}$ of a current $T\in ^s\mathcal D'^q(M)=^s\mathcal D'_{m-q}$ is defined by $$\langle dT,u\rangle=(-1)^{q+1}\langle T,du\rangle,\quad u\in ^{s+1}\mathcal D^{m-q-1}(M).$$ For all forms $f\in ^1\mathcal E^q(M)$ and $u\in ^1\mathcal D^{m-q-1}(M)$, Stokes' formula implies $$0=\int_M d(f\wedge u)=\int_M df\wedge u+(-1)^qf\wedge du,$$ thus in our Example 1 we actually have $dT_f=T_{df}$ as it should be.

Finally, and this should clarify your question, in our Example $2$ another application of Stokes' formula gives $$\int_Z du=\int_{\partial Z}u,$$ therefore $\langle[Z],du\rangle=\langle[\partial Z],u\rangle$ and $d[Z]=(-1)^{m-p+1}[\partial Z]$.

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Hi diverietti, this is very nice stuff, can you recommend any book that develops these ideas? I want to learn some analysis on manifolds. –  Olivier Bégassat Mar 30 '11 at 14:19
@ Olivier Bégassat Mike Spivak's "Calculus on Manifolds" is a classic source. –  Dick Palais Mar 30 '11 at 14:25
Hi Olivier, it depends a lot on your kind of background. If you prefer a more geometrical/complex analytic flavor, then books as "Principle of algebraic geometry" by Griffiths and Harris, or "Complex Analytic and Differential Geometry" by Demailly are great. From a point of view of (real) differential geometry, Spivak's book is very well, but "Foundation of differentiable manifolds and Lie groups" by Warner is very good, too. –  diverietti Mar 30 '11 at 15:03
To see how this stuff is used in differential topology, look at the book by Bott and Tu. –  Deane Yang Mar 30 '11 at 15:47
Of course! Your are completely right, Deane! Bott and Tu is wonderful! –  diverietti Mar 30 '11 at 15:48

I'm not directly answering your question, but I would like to suggest two great books for you to learn the meaning of exterior calculus, integration and de Rham cohomology.

1. "Vector Analysis on Manifolds" by Janich. - I learned that stuff from this book and for me is still the best.

2. "Mathematical Methods in Classical Mechanics" by Arnold. - Well known for its great insight on the meaning of exterior derivative, Lie derivative, Stokes theorem, paralell transport, riemannian curvature, among other concepts.

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Whether a visualization is good really is a subjective thing. This visualization of $d$ works for the author, but it doesn't work for me. If you still find it perplexing after mulling it over, you're probably better off investing your energies into trying to understand $d$ directly rather than into trying to wrap your head around the visualization.

If you still seek visualizations, you could look to other properties of differential geometry to base them on instead of the idea of contour lines for a function.

One such property is that you can integrate an $n$-form over an $n$-dimensional region, so you can try and imagine $n$-forms as a way to measure things. If you can wrap your head around that, then, as in Steve's comment, you can try using Stokes' theorem visualize how $d$ must look.

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