I tried this question, but I have no idea if I got it correctly.
On $\mathbb{R}^2$, let $\omega = (\sin^4 \pi x + \sin^2 \pi(x + y))dx - \cos^2 \pi(x + y)dy$.
Let $\eta$ be the unique $1$-form on the torus $T^2 = \mathbb{R}^2 / \mathbb{Z}^2$ such that $p^* \eta = \omega,$ where $p: \mathbb{R}^2 \to T^2$ is projection. The parametrized curve $\gamma: \mathbb{R} \to \mathbb{R}^2$ given by $\gamma(\theta) = (2\theta, -3\theta)$ is a line whose image $C \subset T^2$ is an oriented circle. Is $\eta$ closed? Exact?
First I showed that $\omega$ is closed: $$d\omega = \frac{\partial}{\partial y}(\sin^4 \pi x + \sin^2 \pi(x + y)) \wedge dx - \frac{\partial}{\partial x} \cos^2 \pi(x + y) \wedge dy = 0$$
And then I showed $\int_C \eta \neq 0$: $$\int_C \eta = \int_\gamma p^* \eta = \int_\gamma \omega = \int_\gamma (\sin^4 \pi x + \sin^2 \pi(x + y))dx - \cos^2 \pi(x + y)dy.$$ Carry out line integrals with $\gamma(\theta) = (2\theta, -3\theta):$ $$\int_\theta (\sin^4 \pi 2\theta + \sin^2 \pi \theta) 2d\theta + \cos^2 \pi \theta 3d \theta.$$
Hence, as long as $\theta \not\equiv 0$, $\int_C \eta \neq 0$, and it is positive.
Assume $\eta$ i s exact, then $\eta = d\alpha$ for some $\alpha$. By Stokes' theorem, its integral over $C$ would be: $$ \int_{C} \eta =\int_{C} d\alpha = \int_{\partial C} \alpha = 0 \ . $$ Since $\partial C = \emptyset$. This contradicts with the fact that $\int_{C} \eta \neq 0$.
$\eta$ is closed because $$d\eta = d^2 \int_C \eta = d^2 \int_\gamma \omega = d \omega = 0.$$
- Agustí Roig's answer checking if a 2-form is exact is very helpful for me to think of this question.
There is a standard embedding $i: T^2 \to S^1 \times S^1 \subset \mathbb{R}^2 \times \mathbb{R}^2$ (determined by the formula $i \circ p(x, y) = (\cos 2 \pi x, \sin 2 \pi x, \cos 2 \pi y, \sin 2\pi y))$. Is there a closed form $\xi$ on $\mathbb{R}$ for which $i^* \xi = \eta$?
My attempt following Amitesh Datta's hints:
Corollary. $\mathbf{H}^p(\mathbb{R}^k) = 0$ if $k > 0$.
Hence, assume that there is a closed form $\xi$ on $\mathbb{R}$, according to the corollary, $\mathbf{H}^1(\mathbb{R}^2 \times \mathbb{R}^2) = 0$. Hence, $\xi$ is exact. But $i^* \xi = \eta$ would also be exact, contradict.
Thank you very very much!
You might want a more elementary proof but here's one that comes to mind:
Theorem
Every closed form on $\mathbb{R}^n$ is exact.
If you're familiar with algebraic topology, then you might have seen this theorem already. If not, then can you prove it on your own? If so, then you've got an answer to your question (a corollary of the Theorem).
I hope this helps and I'm always happy to prove further hints if you encounter any difficulties in proving this Theorem!