First we write down the definition:
This actually isn’t that hard; there’s only the one basis vector to consider, and we find
We also have to calculate the composition
This lets us calculate
So, 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 — on . Why? Well, if we had , then we would find
which we now know not to be the case. Similarly, cannot be the boundary of any -chain, for if then
It turns out that there’s a deep connection between the two halves of this example. Further, in a sense every failure of a closed -form to be the differential of a -form and every failure of a closed -chain to be the boundary of a -chain comes in a pair like this one.
We follow yesterday’s example of an interesting differential form with a (much simpler) example of some -chains. Specifically, we’ll talk about circles!
More specifically, we consider the circle of radius around the origin in the “punctured” plane. I used this term yesterday, but I should define it now: a “punctured” space is a topological space with a point removed. There are also “twice-punctured” or “-times punctured” spaces, and as long as the space is a nice connected manifold it doesn’t really matter much which point is removed. But since we’re talking about the plane it comes with an identified point — the origin — and so it makes sense to “puncture” the plane there.
Now the circle of radius will be a singular -cube. That is, it’s a curve in the plane that never touches the origin. Specifically, we’ll parameterize it by:
so as ranges from to we traverse the whole circle. There are two -dimensional “faces”, which we get by setting and :
When we calculate the boundary of , these get different signs:
We must be very careful here; these are not vectors and the addition is not vector addition. These are merely points in the plane — -cubes — and the addition is purely formal. Still, the same point shows up once with a positive and once with a negative sign, so it cancels out to give zero. Thus the boundary of is empty.
On the other hand, we will see that this circle cannot be the boundary of any -chain. The obvious thing it might be the boundary of is the disk of radius , but this cannot work because there is a hole at the origin, and the disk cannot cross that hole. However this does not constitute a proof; maybe there is some weird chain that manages to have the circle as its boundary without crossing the origin. But the proof will have to wait.