Exterior derivative of $n-1$ form vanishes somewhere

Question

I'm trying to show that on a compact orientable manifold without boundary, the exterior derivative of any $n-1$ form vanishes at some point.

I think I should be using the Stokes' theorem but don't see how. Any hints?

Answer 1

Yes,if $\omega$ is a such a form and $d\omega$ does not vanish, $d\omega$ is a volume form and $\int_Md\omega\neq 0$ by Stokes. contradiction since $\int_Md\omega=\int_{\partial M}\omega=0$, since the boundary $\partial M$ of $M$ is empty.

Answer 2

The other answer is less complete than I would like so I'm going to write my own, even if this is a pretty old post. Let $\{U_i\}$ be an open cover of $M$ by connected open subsets and consider $\{\lambda_i\}$ a partition of unity subordinate to it. I claim $\lambda_i \rm{d} \omega$ has the same sign for every $i$. Indeed, if it were otherwise, then we would have a partition of disjoint open sets$$M = \left(\bigcup_{i \in I_1} U_i \right) \bigcup \left(\bigcup_{j \in I_2} U_j \right) $$ where $\{U_i\}_{i \in I_1}$ are the sets where $\lambda_i \rm{d} \omega$ has positive sign and similarly for $I_2$. This cannot happen because $M$ is connected. And since:

$$0 = \int_{M} \rm{d} \omega = \sum_{i \in I} \int_{M} \lambda_i \mathrm{d} \omega$$ we then get $$\int_{M} \lambda_i \mathrm{d} \omega = 0$$ for all $i$. By writing the local coordinate representation of $ \lambda_i \mathrm{d} \omega $ with respect to each $U_i$, we see this implies $\lambda_i \mathrm{d} w$ has a zero at each $U_i$. Let one of them be at a $p \in M$ where $\lambda_i(p) \neq 0$ (which exists since $\sum_{i} \lambda_i(p) = 1)$. Evidently $\mathrm{d} \omega(p) = 0$.