a closed 1-form which is not exact

Question

On the unit circle S1 in the plane, let θ = arctan(y/x) be the usual polar coordinate. Show that dθ makes sense on S1 and is a closed 1-form which is not exact.

Answer 1

Ittay's answer is perfectly good. Here is a slightly more detailed version.

Your function $\theta(x,y)$ is $\mathcal C^\infty$ on $\mathbb R^2\setminus\{ x=0\}$. If you differentiate, you find $d\theta=\frac{xdy-ydx}{x^2+y^2}\cdot$ This expression now makes sense on $\mathbb R^2\setminus\{ (0,0)\}$; so $\omega=$"$d\theta$" (with the previous formula as a definition) makes sense in particular on $S^1$. Writing $\omega=Pdx+Qdy$, a direct computation of $\frac{\partial Q}{\partial x}$ and $\frac{\partial P}{\partial y}$ shows that $\omega$ is closed. If you compute the integral of $\omega$ on $S^1$ oriented counterclockwise (i.e. if you put $x=\cos\theta$, $y=\sin\theta$ and integrate from $0$ to $2\pi$), you will find $\int_{S^1}\omega=2\pi$, in accordance with the "winding number" interpretation. So $\omega=$"$d\theta$" is not exact since otherwise its integral along any closed curve would be $0$.

Answer 2

When you integrate that 1-form you get the winding number, which for $S^1$ is non-zero. Thus, the 1-form is not exact (since if it were, you could evaluate it by evaluating something at the endpoints, but $S^1$ is a closed curve). To show that it is closed, compute $d$ of it directly.