## Path-Independent 1-Forms are Exact

Today we prove the assertion from last time: if is a -form on a manifold such that for every closed curve we have

then for some function . As we saw last time, the condition on is equivalent to the assertion that the line integral of over any curve depends only on the endpoints and , and not on the details of the path at all.

So, let’s define a function. In every connected component of , pick some base-point . As an aside, what we really want are the arc components of , but since is pretty topologically sweet the two concepts are the same. Anyway, if is in the same component as the selected base-point , we pick some curve from to and define

Remember here that the choice of doesn’t matter at all, since we’re assuming that is path-independent, so this gives a well-defined function given the choice of .

Incidentally, what would happen if we picked a different base-point ? Then we could pick a path from to and then always choose a path from to by composing and . Doing so, we find

So the only difference the choice of a base-point makes is an additive constant over the whole connected component in question, which will make no difference once we take their differentials.

Anyway, we need to verify that . And we will do this by choosing a vector field and checking that . So, given a point we may as well choose itself as the base-point. We know that we can choose an integral curve of through , and we also know that

for an integral curve. For any , we can get a curve from to by defining . And so we calculate (in full, gory detail):

So, having verified that at any point we have , we conclude that for the given function , and is thus exact.

[…] curve . But this means that every closed -form is path-independent, and path-independent -forms are exact. And so we conclude that , as […]

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