## Stokes’ Theorem (proof part 1)

We turn now to the proof of Stokes’ theorem.

We start by considering the case where is the standard cube . Whipping out the definition of the boundary operator , the integral on the right proceeds as follows:

Now any -form can be written out as

where each term omits exactly one of the basic -forms. Since everything in sight — the differential operator and both integrals — is -linear, we can just use one of these terms. And so we can calculate the pullbacks:

It takes a bit of juggling with the definition of , but we can see that this determinant is if and otherwise. Roughly this is because takes the basis vector fields of and turns them into all of the basis vector fields of *except* the -th one. If then some basis -form has to line up against some basis vector field with a different index and everything goes to zero, while if then they can all pair off in exactly one way.

The upshot is that only the two faces of the cube in the direction contribute anything at all to the boundary integral, and we find

On the other side, we can calculate the differential of :

The tricky bit here is that when there’s nowhere to put this brand new , since it must collide with one of the other basic -forms in the wedge. But when then it can slip right into the “hole” where we’ve left out , at a cost of a factor of to pull the across the first terms in the wedge.

With this result in hand, we calculate the interior integral:

We can turn this into an iterated integral, which Fubini’s theorem tells us we can evaluate in any order we want:

which it should be clear is the same as our answer for the boundary integral above. Thus Stokes’ theorem holds for the standard cube.