# If $\omega \in \Omega^{n-1}(M)$ then $d_p\omega = 0$ at some $p \in M$.

Let $M$ be a compact, oriented, smooth manifold of dimension $n$.

I have to show that if $\omega \in \Omega^{n-1}(M)$ then $d_p\omega = 0$ at some $p \in M$.

My first attempt was to use the following proposition:

"If $M$ is an oriented smooth manifold and $\omega \in \Omega_c^{n-1}(M)$ then $$\int_M d\omega = 0,$$ where $\Omega_c^p(M)$ are the p-forms on $M$ with compact support."

If $\omega \in \Omega^{n-1}(M)$ has compact support is sufficient to apply the proposition and take a chart $(U,h)$ so that $supp(\omega) \subset U$, then $$0 = \int_M d\omega = \int_{h(U)} (h^{-1})^* d\omega,$$ so exists $p \in h(U)$ so that $((h^{-1})^*d\omega)_p = 0$, and it implies that $d_p\omega = 0$ because $(h^{-1})^*$ is an isomorphism.

Is this proof right? How can I generalize it to the general case?

Thank you so much for your help.

• $M$ is compact... but how do you choose such a $U$ ?and how do you go from a zero integral to $0$ at some point ? – Max May 13 '18 at 17:32

Your proof works indeed if the support of $\omega$ is contained in a coordinate neighborhood (compact support does not imply this). The idea for the general case is basically the same: if the integral of a continuous function on a connected open set is $0$, then that function has a zero.
When there is no such neigborhood, we can try to cover $M$ by coordinate neighborhoods $U_i$. There are some problems: the $U_i$ may overlap. But okay, we can probably work around this by taking them disjoint and not integrating over a measure $0$ set (the boundaries of the $U_i$). But the problem is still not solved: if the sum of the integrals of $d\omega$ over disjoint $U_1$ and $U_2$ is $0$, it does not imply that $d\omega$ is $0$ at some point of $U_1 \cup U_2$: what could happen is that $d\omega$ is strictly negative in one coordinate neighborhood and strictly positive in the other. In fact, it is the only problem that could occur! (The sign of a differential form $\alpha$ in a chart $(U, \phi)$ means, the sign of the function $g$ such that $\phi_*\alpha = g dx_1 \cdots dx_n$.)
The key is that $M$ is oriented: there exists a cover by charts $(U_i, \phi_i)$ such that $\det(D(\phi_i\phi_j^{-1}))>0$. Now suppose $d\omega$ is nowhere $0$. Then (assuming $M$ is connected) each $\phi_*d\omega$ has the same sign, wlog they are of the form $f_i dx_1 \cdots dx_n$ with each $f_i > 0$. Now replace the $U_i$ by smaller $V_i$ (they need not be open) so that we obtain a disjoint cover of $M$. Then: $$\int_M d\omega =\sum _i\int_{\phi_i(V_i)}\phi_*d\omega >0$$ a contradiction!
More details on how to choose the $V_i$: take a finite cover by coordinate neighborhoods $U_1, \ldots, U_n$, and let $V_i =U_i \setminus\bigcup_{j<i} U_j$. Then $M$ is the disjoint union of the $V_i$, the $\phi_i(V_i)$ are Borel sets of $\mathbb R^n$, and the equality $$\int_M d\omega =\sum _i\int_{\phi_i(V_i)}\phi_*d\omega >0$$ follows by additivity of the integral. (If we want to dig deeper, it's going to depend on how you define $\int$.)
• Can we always choose coordinate neighborhoods $V_i$ so that $M = \cup_i V_i \cup A$, where $m(A) = 0$? – Javier González May 14 '18 at 13:04
• I changed the argument a bit, without talking about measure $0$ sets. But to be honest, I'm not very confortable with the formalism, and I'd have to think how one defines $\int_M\omega$ again. – punctured dusk May 14 '18 at 16:02