Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M$ be a compact connected $n$-manifold without boundary. Let $\mu\in\Omega^{n-1}(M)$, show that there exists a point $p\in M$ such that $d\mu(p)=0$.

share|cite|improve this question
You can typeset mathematics as in LaTeX on this website: enclose mathematics in $ or $$ as appropriate. But please consider phrasing your question as a question, not as an order. – Zhen Lin Aug 5 '11 at 10:15

First, assume that $M$ is oriented. Then, by Stokes's theorem, we have $\int_M d\mu = \int_{\partial M} \mu = 0$ since $\partial M = \emptyset$. This can't happen if $d\mu$ is never $0$.

If $M$ is nonorientable, let $N$ by any orientable cover of $M$ (say, the universal cover, or the orientation covering). Let $\pi:N\rightarrow M$ be the covering map.

Then by the previous case, $d(\pi^*\mu)$ is $0$ on some point $p$ of $N$. But $0 = d(\pi^*\mu) = \pi^* (d\mu)$. Since $\pi$ is a local diffeomorphism, $\pi^*$ induces an isomorphism from $\Omega(M)_{\pi(p)}\rightarrow \Omega(N)_{p}$. Since the image of $d\mu$ under this isomorphism is $0$, we must have $d\mu(\pi(p)) = 0$.

share|cite|improve this answer
On second thought, one should definitely use the orientation covering as the universal cover may not be compact. – Jason DeVito Sep 2 '11 at 3:59

Your Answer


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.