# Compact support of a p-form?

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.
• 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. – Staki42 Jan 26 '18 at 19:51
• 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. – Alekos Robotis Jan 26 '18 at 19:52