# prove that a 1-form on the torus $T^2$ is closed, without computing its differential

I'm given he following closed form on $\Bbb R^2$ $$\omega = (\sin^4\pi x+\sin^2\pi(x+y))dx - \cos^2\pi(x+y)dy$$

and let $\eta$ be the unique $1$-form on the torus such that $p^*\eta=\omega$, where $p$ is the usual covering $\Bbb R^2\to T^2$.

Show whether $\eta$ is closed/exact or no.

I showed that $\eta$ is not exact since there is a loop s.t. $\int_C \eta \neq 0$. But I've no idea how to deal with closeness. Any suggestion?

• What is p? Also why not computethe differential? – lalala Aug 20 '17 at 21:10
• $p$ is the universal covering map from $\Bbb R^2$ to $T^2$. and concerning the differential, well you don't have an explicit expression for $\eta$ so don't know which differential you want to compute – Luigi M Aug 20 '17 at 21:22
• Notice that $p^*$ is an isomorphism and $d\omega = dp^*\eta = p^* d\eta$. So if we know that $d\omega = 0$ then this implies $d\eta = 0$ as well. Now you may write $$\omega = \sin^4 (\pi x) \, dx - dy + \sin^2 \pi(x+y) \, d(x+y)$$ to show that $\omega$ is closed. – Sangchul Lee Aug 20 '17 at 21:34
• @SangchulLee could you please elaborate on the fact that the map $p^*$ induces an isomorphism? – Luigi M Aug 20 '17 at 23:51
• Its injectiveness follows from the fact $p$ is a local diffeomorphism. I don't think it's surjective, but you don't need that. – Pedro Aug 21 '17 at 19:14