One half of De Rham's theorem says that if $\omega$ is a closed $p$-form on a smooth manifold that integrates to zero against all $p$-cycles, then $\omega$ must be exact.

My question is slightly different: whether for a fixed $\omega$ the integrates-to-zero-against-all-cycles condition, in itself, implies that $\omega$ is closed. Call the integrates-to-zero hypothesis the "vanishing condition."

Certainly this statement is true for $1$-forms on an open subset $U$ of $\mathbb{R}^n$. In that case, the vanishing of $\omega$ against all $1$-cycles implies that $\int_{\gamma_1}\omega=\int_{\gamma_2}\omega$ when $\gamma_1$ and $\gamma_2$ are $1$-chains with the same endpoints; therefore we can define an explicit primitive of $\omega$ on each connected component of $U$ by $f(x)=\int_y^x\omega$ for some fixed point $y$. Thus for this special case the vanishing condition implies not only that $\omega$ is closed but also that $\omega$ is exact.

To what extent does the conclusion that $\omega$ is closed generalize to $p>1$ and/or beyond open subsets of Euclidean space? I know I'm not thinking very clearly about this. My guess is that the result won't generalize; if it did, we wouldn't need to assume the closure of $\omega$ in the hypothesis of the half of de Rham's theorem I stated above. But I can't think of a counterexample.


Yes. Pick a point $p$ and a coordinate chart around it; we may as well work in $\Bbb R^n$ around zero. Fix a $(p+1)$-plane $H$ and let $S(\varepsilon)$ be the $\varepsilon$-sphere in it, and $D(\varepsilon)$ the $\varepsilon$-ball. Stokes' theorem says that $\int_{S(\varepsilon)} \omega = \int_{D(\varepsilon)} d\omega$. If $d\omega$ evaluates nontrivially on some $(p+1)$-plane, choose that to be $H$. Then the latter integral must be nonzero for sufficiently small $\varepsilon$ - it's the integral of a nowhere zero function - and thus the first integral is nonzero. So if $\omega$ evaluates to zero against every $p$-cycle it is closed.

  • $\begingroup$ Both you and the OP should find another variable name than $p.$ You use it both as a point and a dimension, and he and you both use it for both the point and $p$-cycles. May I suggest $x$ for the point? $\endgroup$ – hkr Dec 31 '16 at 20:55
  • $\begingroup$ Feel free to edit both the OP and this post with notation changes; StackExchange is communal. $\endgroup$ – user98602 Dec 31 '16 at 20:55
  • $\begingroup$ Ah, yes, of course. So the hypothesis of de Rham's first theorem ("If $\omega$ is a closed form such that $\int_\gamma\omega$ vanishes for all cycles $\gamma$, then $\omega$ is exact") really is redundant. But I get why it is formulated that way: the goal is to show that integration gives an isomorphism from $H_{DR}^p(M)$ to $\text{Hom}(H_p(M;\mathbb{Z}),\mathbb{R})$, and there we need the closure conditions. $\endgroup$ – symplectomorphic Dec 31 '16 at 23:45
  • $\begingroup$ Also, FWIW, I just found this in Conlon, Corollary 8.2.10; I've never seen it made explicit elsewhere. The real point is that $\omega$ is closed iff its integral against every boundary vanishes. $\endgroup$ – symplectomorphic Dec 31 '16 at 23:47
  • $\begingroup$ @symplectomorphic Right. My answer is equivalent to that because you can write every boundary as a finite union of the boundary smooth simplices (simplices cancelling with orientation, as is usual). $\endgroup$ – user98602 Dec 31 '16 at 23:50

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