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.
Now, “conservative” is not meant in any political sense. To the contrary: integration is easy with conservative vector fields. Indeed, if we have a curve that starts at a point and ends at a point then fundamental theorem of line integrals makes it easy to calculate:
I didn’t go into this before, but the really interesting thing here is that this means that line integrals of conservative vector fields are independent of the path we integrate along. As a special case, the integral around any closed curve — where — is automatically zero. The application of such line integrals to calculating the change of energy of a point moving through a field of force explains the term “conservative”; the line integral gives the change of energy, and whenever we return to our starting point energy is unchanged — “conserved” — by a conservative force field.
This suggests that it might actually be more appropriate to say that a vector field is conservative if it satisfies this condition on closed loops; I say that this is actually the same thing as our original definition. That is, a vector field is conservative — the gradient of some function — if and only if its line integral around any closed curve is automatically zero.
As a first step back the other way, it’s easy to see that this condition implies path-independence: if and go between the same two points — if and — then
Indeed, the formal sum is a closed curve, since , and so
Of course, this also gives rise to a parallel — and equivalent — assertion about -forms: if the integral of around any closed -chain is always zero, then for some function . Since we can state this even in the general, non-Riemannian case, we will prove this one instead.