So the problem I have is:

Let $\theta$ be a closed 1-form on a compact Manifold M without boundary. Further suppose that $\theta \neq 0$ at each point of M. Prove that $H^{1}_{dR}(M)\neq 0$.

The only approach I see to doing this is finding a closed loop in M such that the integral over it is non-zero. In local coordinates, writing $\omega=f_{1}(x)dx^{1}+...+f_{n}(x)dx^{n}$ and using the fact that $f_{i}(x)\neq 0$ in some neighborhood of $x$ I can integrate along that segment to get a nonzero integral. Then, switching to different coordinates if needed, if $f_{i}(x_{0})=0$, then there is some other $f_{j}$ so that it is nonzero around that point, and picking the segment with right orientation, I can integrate along this path to increase the value of the integral.

Continuing this indefinitely, I get a constantly increasing value. However, I don't think this is the right approach since I can't guarantee that the curve closes. Though, I could imagine the following. If at some point this continuation crosses itself, then I'm done. If not, then eventually this curve must somehow become dense on the manifold, maybe arguing by compactness (this is just my intuition and is not rigorous, I am imagining a curve on the sphere or torus wrapping around without intersecting). Then my starting point will be close to some other point on the curve, so that when I connect them, the integral along the segment will be potentially negative, but not so negative to cancel the contribution from the other segments.

However, I think there should be a much simpler and rigorous solution to this. So, any thoughts and input would be appreaciated!


1 Answer 1


Suppose there exists $f: M\rightarrow \mathbb{R}$ such that $df = \theta$. What happens at a maximum of $f$?

  • 1
    $\begingroup$ Doi, I am an idiot. Now that is an efficient proof! $\endgroup$ Dec 29, 2014 at 4:20
  • $\begingroup$ How can you say $f$ has a maximum? $\endgroup$
    – Heisenberg
    May 7, 2017 at 6:48
  • 1
    $\begingroup$ @Heisenberg Compactness $\endgroup$ Jul 25, 2018 at 16:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .