Why is the general form of closed 1-forms on $\mathbb{R}^2 \setminus \{ 0 \}$ are as follows?

Question

Why is the general form of closed 1-forms on $\mathbb{R}^2 \setminus \{ 0 \}$ are given by $$ \omega = df + k d \theta, \qquad d \theta = \frac{-y dx + x dy}{x^2 + y^2}, \qquad k = \frac{ 1 }{ 2 \pi } \oint_{S^1} \omega. $$

Answer 1

Say we have a closed form $\omega$, with a given $k = \frac1{2\pi}\oint_{S^1}\omega$, and take the closed form $\sigma = \omega - k\,d\theta$. Now note that any integral of $\sigma$ around any loop is $0$, so by the gradient theorem, it is exact. Thus $\sigma = df$ for some $f:\Bbb R^2-\{0\}\to \Bbb R$, which makes $\omega = df + k\,d\theta$.

We can even construct such an $f$ directly. Pick an arbitrary point $p\in \Bbb R^2-\{0\}$, and an arbitrary value for $f(p)$, say $f(p) = 0$. Then we have $$ f(x) = \int_C\sigma $$ where $C$ is a curve from $p$ to $x$. This is well-defined because $\sigma$, as noted, is conservative.

The fact that the freedom you have to make non-exact, closed forms is captured entirely by the single real number $k$ is what, by definition, makes the de Rham cohomology of $\Bbb R^2-\{0\}$ be (isomorphic to) $\Bbb R$.