Consider the Generalized Stokes Theorem:

\begin{equation} \int_Md\omega = \int_{\partial{M}} \omega \end{equation} Here, $\omega$ is a k-form defined on $R^n$, and $d\omega$ (a k+1 form defined on $R^n$) is the exterior derivative of $\omega$. Let M be a smooth k+1-manifold in $R^n$ and $\partial{M}$ (the boundary of M) be a smooth k manifold.

I know that the above theorem is simply a generalization of well-known vector calculus theorems. However, I am looking for the intuition behind the Generalized Stokes Theorem itself.

I started off by defining the exterior derivative at a point p in $R^n$ as: \begin{equation} d\omega_p =\lim_{|vol|\to 0}\frac{\int_{\partial{vol}} \omega}{|vol|} \end{equation}

In this case, "$vol$" represents a k+1 "parallelpiped" in $R^n$ that contains point p (with $|vol|$ being its "volume"). $\partial{vol}$ represents the boundary of this k+1 "parallelpiped", a k "parallelpiped" itself.

With this definition (assuming it is correct), can we say that $\omega$ represents an infinitesimal "flux" element through $\partial{vol}$ which would imply that $d\omega_p$ is simply the "flux density" at a point p?

If the above is true, can we take the idea that (when applying the Generalized Stokes Theorem) the interior "fluxes" through each $\partial{vol}$ within M cancel out leaving us with the total "flux" out of $\partial{M}$ as the intuition behind the Generalized Stokes Theorem?

Any help is much appreciated.

  • 1
    $\begingroup$ The divergence operator can be discovered by computing the flux of a vector field over the surface of a tiny cube. (Physicists often present this argument.) The divergence theorem then becomes intuitive, by thinking of a volume as being chopped up into tiny cubes. I suspect that, analogously, the $d$ operator can be discovered by computing the integral of a differential form over the boundary of a tiny parallelopiped. The generalized Stokes theorem would then become truly intuitive by thinking of a manifold as being chopped up into tiny parallelopipeds. $\endgroup$
    – littleO
    Commented Aug 8, 2019 at 18:10
  • $\begingroup$ I felt that defining dw in the way that I did made the most sense considering the definition of divergence. Hopefully, the analogy holds. $\endgroup$
    – JG123
    Commented Aug 8, 2019 at 18:19
  • 2
    $\begingroup$ How does your definition of exterior derivative make sense? $d\omega$ is a $(k+1)$-form, and yet you wrote down a real number on the right-hand side. Indeed, it only makes sense to integrate $\omega$ over the boundary of a volume when $\omega$ is an $(n-1)$-form. $\endgroup$ Commented Aug 8, 2019 at 18:33
  • $\begingroup$ That is a good point. Maybe we could refine the definition of dw by using a similar "flux density at a point" definition but making sure both sides of the equation are (k+1) forms? $\endgroup$
    – JG123
    Commented Aug 8, 2019 at 19:07
  • $\begingroup$ @Ted Shifrin Could we not just specify that dw is evaluated at the k+1 tangent vectors that parametrize the (k+1) parallelpiped? Both sides of the equation would have the same rank then, right? $\endgroup$
    – JG123
    Commented Aug 9, 2019 at 22:31

1 Answer 1


Yes, this is very good intuition for the theorem. All of this can be made precise.

In fact, it is made precise on pages 188-190 of the second edition of Arnold's Mathematical Methods of Classical Mechanics. There Arnold gives the following theorem, where $\omega$ is a given $k$-form on an $n$-dimensional manifold $M$.

First, the geometric setup. The idea is to construct a $(k+1)$-form by computing its value on a given list of $k+1$ vectors.

So let $\xi_1,\ldots,\xi_{k+1}$ be tangent vectors in $T_xM$, where $x\in M$. We can pull these back to $\mathbb{R}^n$ in the following way. Choose a coordinate system $\phi:U\to\mathbb{R}^n$ for $x\in U$ with $\phi(x)=0$. The preimages of the $\xi_i$ under the differential of $\phi^{-1}$ are tangent vectors $\xi_i^*$ in $T_0\mathbb{R}^n$. But we can naturally identify this tangent space with $\mathbb{R}^n$ itself. Let $\Pi^{*}$ be the parallelepiped in $\mathbb{R}^n$ spanned by these vectors. The map $\phi^{-1}$ carries this linear parallelepiped onto a "curvilinear parallelepiped" $\Pi$ in $M$ (this is like what you call "vol"), the boundary of which is a $k$-chain, $\partial\Pi$ (which is like what you call the boundary of "vol"). Define $F$ to be the integral of the given $k$-form $\omega$ on the boundary of this curvilinear parallelepiped:


Theorem. There is a unique $(k+1)$-form $\Omega$ on $T_xM$ which is the principal $(k+1)$-linear part at zero of $F$, i.e.


as $\epsilon\to0$. The form $\Omega$ does not depend on the choice of coordinates. And this unique form $\Omega$ is precisely $d\omega$ (in the usual calculation).

In this sense, $d\omega$ is indeed a kind of "flux density" of $\omega$.

The proof of the generalized Stokes theorem then follows, as littleO suggests in a comment, by making precise the idea of chopping up the manifold into little parallelepipeds. One just keeps track of all the orientations as one integrates the flux density over these little regions, $\int_{M}d\omega$. When you do so carefully, the interior fluxes cancel, leaving only the flux along the boundary, $\int_{\partial M}\omega$.


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