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

\displaystyle\int_c\omega=0

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

\displaystyle[X(f)](c(0))=\frac{d}{dt}f(c(t))\bigg\vert_{t=0}

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):

\displaystyle\begin{aligned}{}[X(f)](x)&=[X(f)](c(0))\\&=\frac{d}{dt}f(c(t))\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_{c_t}\omega\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_{[0,1]}c_t^*\omega\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[[c_t^*\omega](s)]\left(\frac{d}{ds}\bigg\vert_s\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c_t(s))]\left({c_t}_{*s}\frac{d}{ds}\bigg\vert_s\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(tc_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(c_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)t\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(c_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)\,d(st)\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^t[\omega(c(u))]\left(c_{*u}\frac{d}{dt}\bigg\vert_u\right)\,du\bigg\vert_{t=0}\\&=[\omega(c(t))]\left(c_{*t}\frac{d}{dt}\bigg\vert_t\right)\bigg\vert_{t=0}\\&=[\omega(c(0))]\left(c_{*0}\frac{d}{dt}\bigg\vert_0\right)\\&=[\omega(x)](X_x)\end{aligned}

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.

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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


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