Original motivation: How can I apply Stokes' Theorem to the annulus $1 < r < 2$ in $\mathbb{R}^2$?
Concerns:
- Since the annulus is a manifold without boundary, it would seem that Stokes' Theorem would always return an answer of $\int_M d\omega = \int_{\partial M} \omega = 0$ for compactly supported forms $\omega$. Is this correct?
- What about the annulus $1 < r \leq 2$? This seems like a manifold-with-boundary to me, yet an application of Stokes' Theorem will return a different answer. And what about $1 \leq r \leq 2$?
For instance, consider $\omega = -y\,dx + x\,dy$ on the annulus $1 < r \leq 2$, so that $d\omega = 2\,dx\,dy$. Then $$\int_M d\omega = 2\,\text{Area}(M) = 6\pi,$$
whereas $$\int_{\partial M} \omega = \int_0^{2\pi} 4\,dt = 8\pi,$$
where $\partial M$ is the circle $r = 2$.
What explains this discrepancy?
A friend of mine has suggested that this can be explained by the fact that $\omega = -y\,dx + x\,dy$ is not compactly supported on $1 < r \leq 2$, and hence Stokes' Theorem can't really be applied. Is this correct?
For reference, I am using the following version of the theorem:
Stokes' Theorem: Let $M$ be a smooth, oriented $n$-manifold with boundary, and let $\omega$ be a compactly supported smooth $(n-1)$-form on $M$. Then $$\int_M d\omega = \int_{\partial M} \omega.$$
Best Answer
Annulus with $1 < r < 2$ does not have a boundary and the form you pick is not compactly supported there. The form $\omega$ only vanishes at the origin so its support is in fact open in $\mathbb{R}^2$.