# The Unapologetic Mathematician

## 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.