The Unapologetic Mathematician

Mathematics for the interested outsider

Path-Independent 1-Forms are Exact

Today we prove the assertion from last time: if \omega is a 1-form on a manifold such that for every closed curve c we have


then \omega=df for some function f. As we saw last time, the condition on \omega is equivalent to the assertion that the line integral of \omega over any curve c depends only on the endpoints c(0) and c(1), and not on the details of the path c at all.

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

\displaystyle f(x)=\int\limits_c\omega

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

Incidentally, what would happen if we picked a different base-point q? Then we could pick a path c' from q to p and then always choose a path d from q to x by composing c':q\to p and c:p\to x. Doing so, we find

\displaystyle g(x)=\int\limits_d\omega=\int\limits_{c'}\omega+\int\limits_c\omega=K+f(x)

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 df=\omega. And we will do this by choosing a vector field X and checking that X(f)=df(X)=\omega(X). So, given a point x we may as well choose x itself as the base-point. We know that we can choose an integral curve c of X through x, and we also know that


for an integral curve. For any t, we can get a curve c_t from x=c(0) to c(t) by defining c_t(s)=c(st). And so we calculate (in full, gory detail):


So, having verified that at any point x we have X(f)=df(X)=\omega(X), we conclude that \omega=df for the given function f, and is thus exact.

December 15, 2011 - Posted by | Differential Topology, Topology

1 Comment »

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

    Pingback by Simply-Connected Spaces and Cohomology « The Unapologetic Mathematician | December 17, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: