Let \(S^1\) be the unit circle as a 1-dimensional manifold. Let \(\omega\) be the 1-form on \(S^1\) defined by \[\omega=\frac{x+y}{x^2+y^2}\, \mathrm{d}x+\frac{y-x}{x^2+y^2}\, \mathrm{d}y,\] where \(x\) and \(y\) are the standard coordinates on \(S^1\). A simple calculation shows that this differential form is closed. A calculation also shows that when \(S^1\) is given the counterclockwise orientation, we have that \(\int_{S^1}{\omega}=\int_{0}^{2\pi}{-1\, \mathrm{d}\theta}=-2\pi\).

However, \(\omega\) clearly extends to a closed 1-form \(\Omega\) on the manifold \(M:=D^2\setminus\{(0,0)\}\), where \(D^2\) is the closed unit disk in \(\mathbb{R}^2\). Note that \(\partial M=S^1\). Stokes’ theorem then suggests that \[\int_{S^1}{\omega}=\int_{M}{\mathrm{d}\Omega}=\int_{M}{0}=0.\] What is the discrepancy?

The issue is that the precise statement of Stokes’ theorem calls for the support of the \(\Omega\) to be compact. Since our manifold \(M\) has a “hole” at the origin. Hence, Stokes’ theorem does not apply in this case. Of course, if we replace \(M\) with \(D^2\), then if \(\omega\) extends to a closed 1-form on \(D^2\), the support of this extension would indeed be compact so Stokes’ theorem would tell us that \(\int_{S^1}{\omega}=0\). Hence, it cannot be the case that \(\omega\) extends to a closed 1-form on \(D^2\).