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Say that $N$ is an oriented, compact, connected manifold without border. If $\operatorname{dim}(N) = k$, does it always exists some $k$-form $\omega$ such that $\int_N \omega \neq 0$?

I know how to proceed in some particular cases (e.g., $S^k$), but I have no idea how to prove the general case, if it is even true. In the literature I've checked (mostly Lee's Introduction to Smooth Manifolds, 2nd edition, and Guillemin/Pollack Differential Topology) there is nothing, also - but I may have missed something.

Mtguy
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2 Answers2

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Let $(U, \varphi)$ be an oriented local chart of $M$. On $\varphi (U) \subset \mathbb R^k$ you have a $k$-form $\omega_0 = dx^1 \wedge \cdots \wedge dx^k$ and a bump function $b : \varphi(U) \to \mathbb R$ (i.e. a smooth function with compact support in $\varphi(U)$), then $$\varphi^*(b \omega_0)$$ is a smooth $k$-form on $M$ (by extending to zero outside of $U$). By definition of the integration,

$$ \int_M \omega = \int_{\varphi (U)} b\omega_0 = \int_{\varphi (U)} b(x) dx^1\cdots dx^k.$$

One can choose this to be non-zero (e.g. choose $b\ge 0$ and $b>0$ on an open subset)

Arctic Char
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From Lee's Introduction to Smooth Manifolds, 2nd edition:

Theorem 17.31 If $M$ is a compact connected orientable smooth $n$-manifold, then $H_{\text{dR}}^{n}(M)$ is $1$-dimensional, and is spanned by the cohomology class of any smooth orientation form.

Here, $H_{\text{dR}}^{n}(M)$ is the standard de Rham cohomology group, that is, $$H_{\text{dR}}^{n}(M)=\{\text{closed $n$-forms on $M$}\}/\{\text{exact $n$-forms on $M$}\}.$$

If you see the proof of Theorem 17.30, you can see that the isomorphism is given by \begin{align*} H_{\text{dR}}^{n}(M) & \to \mathbb{R}, \\ \omega & \mapsto \int_{M} \omega. \end{align*}

Hence, the answer to your question is affirmative.

Sebathon
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  • This is such an overkill..... – Arctic Char Dec 05 '21 at 14:51
  • Yes. I just wanted to point out that this is, indeed, explained in Lee's book. – Sebathon Dec 05 '21 at 15:15
  • Even then, this is arguably not the way to go about it, since this is a non-trivial theorem, whose content is the injectivity of the map, not the surjectivity. The surjectivity is much more elementary and also explained in Lee's book. Indeed, just looking at the proof of Theorem of 17.30 (from which 17.31 is a corollary) shows that he gives precisely the argument in Arctic Char's answer there. Even without that, Lee already explains the correspondence between orientations and orientation forms in Prop. 15.5 and proves the integral of an orientation form is positive in Prop. 16.6(c). – Thorgott Dec 05 '21 at 16:24
  • @Thorgott you are absolutely right. I wrote this as a way to say "look, here is a proof for what you want in the book that you mentioned. Maybe it would be instructive if you go to the book and look again". I also know what you mentioned about the other chapters of Lee's book, but my idea was that the author could figure that out after a second search. – Sebathon Dec 05 '21 at 16:44