Differential forms on the circle $\mathbb{R}/\mathbb{Z}$

Question

Let $\pi:\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ be the projection.

1.- Let $t:\mathbb{R}\to\mathbb{R}$ be the identity function. Then there is a $1$-form $\eta\in\Omega^{1}(\mathbb{R}/\mathbb{Z})$ satisfying $\pi^{*}\eta=dt$.

2.- Let $F:\mathbb{R}\to\mathbb{R}$ be a smooth $1$-periodic function (i.e., $F(t+1)=F(t)$ for every $t$), and let $f:\mathbb{R}/\mathbb{Z}\to\mathbb{R}$ denote the function defined by $F$. Then $f\eta$ is closed.

3.- $f\eta$ is exact if and only if $\int_{0}^{1}F(t)dt=0$.

How can I prove the above statements?

I would appreciate any hint.

Answer 1

Define $\pi^* : \Omega^p(\mathbb{R/Z}) \to \Omega^p(\mathbb{R})$ (here the only itneresting values are $p=0,1$). You may try to show that this map is injective, and that its image is precisely the set of $\mathbb{Z}$-invariant elements (where $\mathbb{Z}$ acts on $\mathbb{R}$ with $(n,x) \mapsto x+n$).

To prove this, one direction is easy : $\pi^*\eta$ will always be $\mathbb{Z}$-invariant: can you see why ?

The other direction is a bit more subtle: you need to show that a $1$-form is always locally of the form $\pi^*\eta$, but use injectivity of $\pi^*$ (in fact the fact that $\pi$ is a local diffeomorphism) to show that this $\eta$ is unique locally (in $\mathbb{R}$). Gluing things back together will use the $\mathbb{Z}$-invariance: you need to show that the $\eta$ that you defined locally on $\mathbb{R}$ is actually well-defined locally in $\mathbb{R/Z}$.

To be more precise: (I hid it so you can think about it for yourself if you don't want a full solution)

Let $\omega$ be a $\mathbb{Z}$-invariant $1$-form and let $y\in \mathbb{R/Z}$, we wish to define $\tilde{\omega}_y(Y)$, $Y\in T_y(\mathbb{R/Z})$. Well take $x$ such that $\pi(x)= y$ and $X\in T_x\mathbb{R}$ such that $d_x\pi (X) = Y$ (there is only one such $X$, as $\pi$ is a local diffeomorphism) and put $\tilde{\omega}_y(Y)= \omega_x(X)$. Now this doesn't depend on the chosen $x$, because if I take another $x'$ then for some $n$, $x' = x+n= n\cdot x$ so $\omega_{x'}(X') = n^*\omega_x(X)= \omega_x(X)$ (by $\mathbb{Z}$-invariance; and because $X' = d_xn (X)$ where $n$ is seen as $t\mapsto t+n$).

This is also smooth because:

if you take a small enough neighbourhood of $y$, this is simply the pullback of the restriction of $\omega$ on some neighbourhood of $x$.

Note: this proof can be adapted almost verbatim to any discrete group $G$ acting on any manifold $M$ freely and properly.

How to apply this to your situation ? Well $dt$ is $\mathbb{Z}$-invariant ! Indeed, $dt_x(X) = X$ for any $x\in \mathbb{R}, X\in T_x\mathbb{R} (= \mathbb{R})$, so $dt_{x+n}(d_xn(X)) = dt_{x+n}(X) = X =dt_x(X)$. Hence by what we just did, $dt = \pi^*\eta$ for some $\eta$.

For 2., you just need to remember that $f\eta$ is a $1$-form. Are there many $2$-forms on a $1$-dimensional manifold ?

For 3., $\mathbb{R/Z}$ is a compact connected $1$-dimensional manifold. Have you seen De Rham cohomology ? If so, what is $H^1(\mathbb{R/Z})$ ?

If you haven't seen De Rham cohomology but have seen Stokes' theorem, if $f\eta$ is closed, that is $f\eta = d\omega$ for some $\omega$, what can you say about $\displaystyle\int f\eta$ ? Can you compute it in two different ways ?