In my differential geometry lecture, my professor stated Stokes' theorem as follows:

Let $M$ be a $n$-dimensional, piecewise continuously differentiable submanifold with boundary of $\mathbb R^m$. Then, for every continuously differentiable $(n-1)$-form $\omega$ with compact support on $M$, $$\int_M d\omega = \int_{\partial M}\omega$$

Now I want to find a counterexample for when $\omega$ is not compactly supported, but what is meant with "compact support" when talking about forms? For example, the form $$\omega = x \,dy \wedge dz + y \,dz \wedge dx + z \,dx \wedge dy$$ on $S^2$ was used as an application of Stokes' theorem, but I don't see how this has compact support? $x$ is unbounded on $\mathbb R^m$. On $S^2$, it of course is not, but how am I supposed to find something unbounded on a compact set?


Well, the support of a $k-$form $\omega\in \Omega^k(M)$ on a manifold $M$ is the set $$\text{supp}(\omega)=\overline{\{x\in M: \omega_x\ne 0\}}. $$ The support is said to be compact when it is compact as a topological space. In your particular example we have that any form $\omega$ on a compact manifold $M=S^2$ is compactly supported, since $\text{supp}(\omega)$ is a closed set, and a closed subset of a compact topological space is compact.

  • $\begingroup$ So I can not find a counterexample? $\endgroup$ – Staki42 Jan 26 '18 at 19:47
  • $\begingroup$ It's more that on a compact manifold any form is compactly supported. $\endgroup$ – Alekos Robotis Jan 26 '18 at 19:48
  • $\begingroup$ So I could just omit the "compactly supported" part in the statement of the theorem, right? I guess I a counterexample is only to be found when I also drop the $M$ compact condition. $\endgroup$ – Staki42 Jan 26 '18 at 19:51
  • $\begingroup$ Well, no you shouldn't omit that. We don't require that $M$ be compact, unless I've misread. It's just that in the case of $S^2$ the manifold is compact. $\endgroup$ – Alekos Robotis Jan 26 '18 at 19:52
  • 1
    $\begingroup$ No problem. Good luck! $\endgroup$ – Alekos Robotis Jan 26 '18 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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