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Why is $\theta \not \in C^{\infty}(S^1)$? I know that since $\int_{S^1} d\theta = 2\pi$ then $d\theta$ is not exact. Thus since $d(\theta)=d\theta$, $\theta$ must not be $C^{\infty}$ but it seems like it should since it is simply the identity function. Thanks for the help.

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To begin with, what is $\theta$? The fact that $d\theta$ is the notation used for a particular $1$-form does not mean that we have such a thing as "$\theta$". After all, $d\theta$ could be just that, a convenient piece of notation. But yes, it's actually more than that...

Suppose we have a family of functions $\theta_\alpha$, each of which is $C^\infty$ smooth on some open arc $U_\alpha\subsetneq S^1$. Suppose also that these arcs cover $S^1$, and that whenever $U_\alpha\cap U_\beta$ is nonempty, the difference $\theta_\alpha-\theta_\beta $ is locally constant on the intersection. Then we can define a $1$-form $d\theta$ by letting it be $d\theta_\alpha$ within each $U_\alpha$ and observing that they agree on overlaps. (Using a partition of unity, one can express this definition in more precise terms). All this applies to any manifold, not only $S^1$.

On $S^1$ we can consider the family of all proper subarcs $U_\alpha$, and pick a continuous (hence smooth) branch of the argument within each subarc. Or just use $U_j=S^1\setminus \{a_j\}$ for some distinct points $a_1,a_2$.

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Denoting by $\theta$ the argument of a point of $S^1$, then $\theta$ is not a function on $S^1$; it is a multivalued function. For this reason it is not smooth. In order to define $d\theta$, the approach indicated by @75064 is the correct one. In other words, one need to "cut" $S^1$, looking at it as a manifold with a chart consisting of open arcs with non trivial intersections.

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