# The Unapologetic Mathematician

## Stokes’ Theorem (statement)

Sorry for the little hiatus. I’ve been busier than usual.

Anyway, now we come to Stokes’ theorem. You may remember something by this name if you took a good multivariable calculus course, but this is not quite the same thing. In fact, the Stokes’ theorem you remember is connected to but one special case of this theorem, which also subsumes Gauss’ theorem, Green’s theorem, and the fundamental theorem of calculus, all in one neat little package. The exact details of the connection, though, require us to move into the realm of differential geometry, so we’ll have to come back to them later.

But anyway, on to the theorem! We know how to integrate a differential $k$-form $\omega$ over a $k$-chain $c$. We also have a differential operator on differential forms and a boundary operator on chains. We can put these together in two ways: either we start with a $k-1$-form $\omega$, take its exterior derivative to get the $k$-form $d\omega$, then integrate that over the $k$-chain $c$; or we take the boundary of $c$ to get the $k-1$-chain $\partial c$, then integrate $\omega$ over that. What Stokes’ theorem asserts is that these two give the same answer. As a formula:

$\displaystyle\int\limits_cd\omega=\int\limits_{\partial c}\omega$

In a hand-wavy, conceptual way of putting it: integrating a differential form over the boundary of a region is the same as integrating its derivative over the interior. Indeed, if you look back over the results I mentioned above — even just the fundamental theorem of calculus — you can see this concept at work.

August 17, 2011