Problem: Define a differential $1$-form on the cylinder $$C = \{ (x,y,z) \in \mathbb{R}^3: x^2 + y^2 = 1 \} \subset \mathbb{R}^3 $$ which is closed but not exact.
My attempt (EDITED):
I am using $\omega = \frac{1}{x^2+y^2} (xdy - ydx) \in \Omega^1 (C)$. I showed that $\omega$ is closed by working out $d\omega = 0 $. My question is based on how to show that $\omega$ is not exact.
One sufficient condition that I know is that closed forms are exact if and only if the integral is zero if and only if the form vanishes at some point. However, this raises a few questions: How can I integrate over $C$ if it is not bounded? Furthermore, $C$ seems to be a two-dimensional manifold so integrating a $1$-form over $C$ doesn't seem to make quite sense.
Maybe the thing that makes more sense is to integrate over the curve, which I can choose to be $S^1 = \{ (x,y,0) \in \mathbb{R}^3: x^2 + y^2 = 1 \}$ because $\omega$ is a $1$-form so we can integrate around a curve that lives in $C$? (please correct me if I am wrong, this idea is not quite solidified for me)
So then by Stokes': $$\int_{S^1} \omega = \int_{S^1} xdy - ydx = \int_{D^2} d(xdy - ydx) = 2 \int_{D^2} dx \wedge dy = 2 \cdot \mathrm{vol}(D^2) = 2\pi$$ Can I integrate over $D^2$ even though the interior of $D^2$ does not intersect with $C$?
This implies that $\omega|_{S^1} \in \Omega^1(S^1)$ is a closed but not exact form. Consider the canonical projection map $\pi:C \rightarrow S^1$. Maybe there is some condition on $\pi$ such that $\pi^*(\omega|_{S^1}) = \omega \in \Omega^1(C)$ pullback to a closed but not exact $1$-form?
Thank you.