Differential forms on $S^1$ I'm reading this old question and there are some things I don't understand.
For example, why in the case of $S^1$ can every $1$-form be written in the form $f(\theta)d\theta=c d\theta+dg(\theta)$ where $g$ is differentiable and $c$ is some integral? What is special about $S^1$? The way the question is phrased suggests that on other manifolds there is no such representation of a $1$-form. 
 A: The general statement has to do with $n$ forms on a compact $n$-dimensional manifold $M$ without boundary.  It is a general fact that for such manifolds the $n$th de Rham cohomology group is $H^n_{dR}(M) = \mathbb R$.  Now if $\alpha$ is an $n$-form that represents a non-zero class in $H^n_{dR}(M)$ (i.e. $\alpha$ is not exact), then every other $n$th degree cohomology class is represented by a constant multiple of $\alpha$.  Now if $\mu$ is any $n$-form it is automatically closed and thus represents a class in $H^n_{dR}(M)$ and so is in the same class as $c \alpha$ for some $c \in \mathbb R$.  This means that $\mu$ and $c\alpha$ differ by an exact form, so $\mu = c\alpha + d\beta$ for some $n-1$-form $\beta$.
In your specific example, $\mu = f(\theta)d\theta, \alpha = d\theta$ and $\beta = g(\theta)$.
A: It means not every form on $S^1$ is necessarily an exact form, meaning it being the differential of a function on $S^1$. In this case, $fd\theta=cd\theta+dg$, being a closed form (you should see that the OP is talking about closed form), is not exact due to the existence of the term $cd\theta$ (note that closed form $\omega$ means $d\omega=0$ and therefore exact forms are automatically closd since $d^2=0$).
