Stokes’ Theorem on Manifolds
If is such a manifold of dimension , and if is a compactly-supported -form, then as usual we can use a partition of unity to break up the form into pieces, each of which is supported within the image of an orientation-preserving singular -cube. For each singular cube , either the image is contained totally within the interior of , or it runs up against the boundary. In the latter case, without loss of generality, we can assume that is exactly the face of where the th coordinate is zero.
In the first case, our work is easy:
since is zero everywhere along the image of , and along .
In the other case, the vector fields — in order — give positively-oriented basss of the tangent spaces of the standard -cube. As is orientation, preserving, the ordered collection gives positively-oriented bases of the tangent spaces of the image of . The basis is positively-oriented if and only if is even, since we have to pull the th vector past others, picking up a negative sign for each one. But for a point with , we see that
for all . That is, these image vectors are all within the tangent space of the boundary, and in this order. And since is outward-pointing, this means that is orientation-preserving if and only if is even.
Now we can calculate
where we use the fact that integrals over orientation-reversing singular cubes pick up negative signs, along with the sign that comes attached to the face of a singular -cube to cancel each other off.
So in general we find
The last sum is finite, since on of the support of all but finitely many of the are constantly zero, meaning that their differentials are zero as well. Since the sum is (locally) finite, we have no problem pulling it all the way inside:
so the sum cancels off, leaving just the integral, as we’d expect. That is, under these circumstances,
which is Stokes’ theorem on manifolds.