Cross-posted from MSE.

I know that De Rham cohomology reveal some properties of the topology of smooth manifolds by finding closed differential $k$-forms $\mathsf{d}\omega=0$ that are not exact $\omega\neq\mathsf{d}\eta$. I wonder

Question: Why non closed differential forms do not play important role for the topology of a manifold?

One can say that because the closed forms work pretty well i.e. the PDE approach to closed and exact forms that asks for solvability of a system of differential equations, then Poincare Lemma enters. All goal of the solvability is based on closedness and exactness of system of differential equations. The other reason is that the set of all non-closed forms, is not a vector space. But these are not a good and convincing reason for non-closed forms not affecting on the topology of the manifold.

Here are some of MSE users comments which seems to be useful but has been posted with some doubts:

  • There is a theory going back to Sullivan that derives the real homotopy type of a smooth manifold from its algebra of differential forms.
  • In a word (well a few), in this book, Postnikov towers and spectral sequences are showing up, and with those spectral sequences, non-closed differential forms.
  • 11
    $\begingroup$ I'm not sure I agree with the premise of the question, but one answer could be that when you integrate a closed form over a submanifold, the answer doesn't change when you perturb the submanifold, whereas it does depend for non-closed forms. $\endgroup$ Jun 6 at 18:23
  • 7
    $\begingroup$ However, non-closed forms do appear in topology: e.g. contact forms. Just not in de Rham cohomology. $\endgroup$ Jun 6 at 18:24
  • 5
    $\begingroup$ To me it seems that in order to take this question seriously you have to view manifold theory purely through the lens of the De Rham theorem. This seems a rather artificial perspective. $\endgroup$ Jun 6 at 19:08
  • 5
    $\begingroup$ The question seems to implicitly work from the pov that "topology of manifold = cohomology". Now once you take that pov, the answer is tautological: by definition cohomology only cares about closed forms. If you change your pov and no longer consider the cohomology as the only interesting topological property of a manifold, the question doesn't make sense, because then the whole algebra of forms contains interesting information, as you point out yourself. $\endgroup$ Jun 6 at 20:57
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    $\begingroup$ @user2520938: By "as you point out yourself" you meant the firs itemized case? i.e. the works of Sullivan? $\endgroup$
    – C.F.G
    Jun 7 at 5:10

Actually, the non-closed forms on manifolds play an essential role in the definition of Massey products, which are 'higher cohomology' operations.

Another place where they make an essential appearance is in the construction of the Hopf invariant for smooth maps $f:S^{2n-1}\to S^n$. Recall that if $f:S^{2n-1}\to S^n$ is smooth and $n>1$, one can construct a quadratic map $\eta_f:H^n_{dR}(S^n)\to H^{2n-1}_{dR}(S^{2n-1})$ as follows: Let $\alpha\in\Omega^n(S^n)$ be a top degree form. Then $f^*\alpha\in\Omega^n(S^{2n-1})$ is not only closed, but is exact, i.e., $f^*\alpha = \mathrm{d}\beta$ for some $(n{-}1)$-form $\beta$ on $S^{2n-1}$. Then let $\eta_f([\alpha]) = [\beta\wedge\alpha]\in H^{2n-1}(S^{2n-1})$.

It is easy to show that this is a well-defined map $\eta_f:H^n_{dR}(S^n)\to H^{2n-1}_{dR}(S^{2n-1})$ of degree $2$, constructed using the structure of the full algebra of differential forms, not just the closed forms.

There are many other examples where the full algebra of differential forms is used to define topological invariants of smooth manifolds, for example, Gelfand-Fuks cohomlogy, secondary characteristic classes, etc.


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