## An Example (part 3)

Now we can take our differential form and our singular cube and put them together. That is, we can integrate the -form over the circle .

First we write down the definition:

to find this pullback of we must work out how to push forward vectors from . That is, we must work out the derivative of .

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.

You were right. This example was really illustrative. Thank-you. You have no idea how much your input has helped me, so keep posting. Tchuss!

Comment by Alice | April 27, 2015 |