Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$\int_M d\omega = \int_{\partial M} \omega.$$

share|improve this question
4  
I completely agree with your friend. Stokes is only applicable to the form on the closed annulus $1 \leq r \leq 2$. Note also that the discrepancy of $2\pi$ is obtained by subtracting the integral of the form over the circle with radius $1$ -- you need to be careful about orientation! –  t.b. May 15 '11 at 22:13
1  
Well, the support of $\omega$ is the whole of your open annulus, which is not compact. So the Stokes theorem doesn't apply as stated. –  Zhen Lin May 15 '11 at 22:13
1  
OK, I understand where the discrepancy of $2\pi$ comes from, yes, but I suppose it just seems odd to me that the requirement of being compactly supported can carry such weight. Especially for forms like $\omega = -y\,dx + x\,dy$, which don't blow up at the boundary or anything... –  Jesse Madnick May 15 '11 at 22:17
5  
The point is not about blowing up at the boundary, but about the non-vanishing of the form at the boundary, which complicates the integration by parts. –  Henri May 15 '11 at 22:37
3  
Dear Jesse, "It just seems odd to me that the requirement ... can carry such weight". The example in your post shows exactly why this condition is important: otherwise there can be "missing" boundary contributions. The form $-y dx + x dy$, which is not compactly supported, "feels" the missing boundary along $r = 1$ in the way that a compactly supported form wouldn't. Regards, –  Matt E May 16 '11 at 3:24
show 2 more comments

1 Answer

up vote 1 down vote accepted

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 $\mathbb{R}^2$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.