An Example (part 3)
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.