# 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 - Posted by | Differential Topology, Topology

1. We just learned this in my multivariable course, it’s at the very end of Marsden and Tromba. Very cool!

Comment by Jackson Walters | August 17, 2011 | Reply

2. I have a certain fondness for Green’s Theorem, because I grew up a few miles from where George Green used to live. His windmill (http://www.flickr.com/photos/njj4/4773588885) was restored to working condition during the 1980s, and a small science museum built next door. Down the hill, there’s a children’s playground with climbing frames, see-saws and slides in the shape of various mathematical symbols.

Comment by Nicholas | August 18, 2011 | Reply

3. […] We turn now to the proof of Stokes’ theorem. […]

Pingback by Stokes’ Theorem (proof part 1) « The Unapologetic Mathematician | August 18, 2011 | Reply

4. […] that we’ve proven Stokes’ theorem in the case of the standard cube, we can now tackle the general […]

Pingback by Stokes’ Theorem (proof part 2) « The Unapologetic Mathematician | August 20, 2011 | Reply

5. […] what conclusions can we draw from this? Well, Stokes’ theorem now tells us that the -form cannot be the differential of any -form — any function — […]

Pingback by An Example (part 3) « The Unapologetic Mathematician | August 24, 2011 | Reply

6. […] we come back to Stokes’ theorem, but in the context of manifolds with […]

Pingback by Stokes’ Theorem on Manifolds « The Unapologetic Mathematician | September 16, 2011 | Reply

7. […] John Armstrong: Integrals are Independent of Parametrization, Integration on Singular Cubes, Integration on the Standard Cube, Stokes’ Theorem (statement) […]