Let $\alpha$ be a closed $1$-form on a connected manifold $M$. We
assume that $M$ is simply connected, i.e. for all paths
$\gamma_0,\gamma_1$ with the same endpoints there exists a smooth
homotopy $h\colon[0,1]\times[0,1]\to M$ between $\gamma_0$ and
$\gamma_1$. Use the Poincare-Lemma to show that$$\int_{\gamma_0}\alpha = \int_0^1\gamma_0^*\alpha =
\int_0^1\gamma_1^*\alpha = \int_{\gamma_1}\alpha$$
Use this
observation to construct a smooth function $f$ so that $df = \alpha$.
I struggle to construct the desired function $f$.
So we are obviously looking for a $0$-form $f \in \Omega^0 M$ as the antiderivative $f = \int \alpha$ of our given $\alpha \in \Omega^1M$.
Since $\gamma_0,\gamma_1$ are homotopic, it does not depend over which $\gamma_t$ we integrate. Thus $$f = \int_{\gamma_1}\alpha = \int_0^1 \gamma^*\alpha$$
But how do i continue? I know the definition of the pullback of a differential form on a smooth manifold, but i do not know how to make use of it in this case.
Can anyone help me?
Thanks in advance for any advise!
Addendum:
For example, the solution to the problem starts with:
pick $\gamma_i(t)=tx^i$ in coordinates $x^i$
and then proceeds with
$ f = \int_\gamma \alpha = \int \alpha_{\gamma(t)}\frac{d\gamma}{dt}dt$
but i don't understand what's going on. I don't understand the coordinate-change nor the last integral.
Best Answer
Fix a point $x_0\in M$, and then for a given point $x\in M$, define $$f(x):=\int_\gamma \alpha$$ where $\gamma$ is any path from $x_0$ to $x$.