When dealing with a top degree differential form $\mu$ in a manifold $M$, a way of "computing" its cohomology class is integrating it through the whole manifold. For instance, if the integral $ \int_M \mu=0 $ then the form is exact.

I've been trying to find a similar condition (computing an integral or something like this) when dealing with the relative cohomology class of $\mu$ with respect to a submanifold. For example, let $Z$ be an hypersurface in $M$ and $H(M;Z)$ its relative cohomology group.

I'm working with the definition given by Godbillon, the group is given by closed and exact forms that vanish at $Z$. But I've found very little references or information on these topics. Any ideas or references, other than Godbillon or Bott-Tu? Any help is appreciated.



Assume that the pair $Z,M$ is oriented. There are two cases: If $Z$ represent a non-vanishing $n-1$-cycle, then the map $H^{n-1}(M)\to H^{n-1}(Z)\stackrel{\int_Z}{\to} \mathbb{R}$ is onto by Poincare duality (the restriction of a cocycle to $Z$ and integrating is like cupping with the poincare dual of $Z$ which can not anihilate the cup pairing).

In that case the long exact sequence of the pair $(Z,M)$ gives us an isomorphism $H^n(M,Z)\cong H^n(M)$ and the invariant is the integral.

In the case where $0=[Z]\in H_{n-1}(M)$ we can find a domain $U$ in $M$ such that $Z=\partial U$. Then, we can consider the invariant $\mu\mapsto \int_U \mu$. Note that this is well defined only on forms with total integral $0$ because we can replace $U$ by its complement, but once we have chosen $U$ we can consider the map $\alpha \mapsto (\int_U \alpha, \int_M \alpha)$ which gives us a map $H^n(M,Z)\to \mathbb{R}^2$.

Note that the first is well defined because it vanishes in forms of type $d\beta$ with $\beta|_Z=0$ by Stokes formula. Moreover, the idea to consider this functional is by formally interpret the long exact sequence of the pair: We take a form $\alpha$ with total integral $0$ and write it as $d\beta$ for $\beta$ not nessesarily $0$ on $Z$. Then its class in $H^{n-1}(Z)$ is exactly the preimage of $\alpha$ in $H^{n-1}(Z)$, and this is well defined only modulo $H^{n-1}(M)$ since we could add a closed form to $\beta$. But the class of $\beta$ in $H^{n-1}(Z)$ is represented by its integral, which by stokes theorem is exactly the integral of $\alpha$ on $U$, so we don't really need to choose $\beta$ to compute it!

  • $\begingroup$ Thank you a lot! I've got two doubts about this. First, this works for any $U$? I mean just fixing one, any, it's enough? And secondly, let me check if I understood this well. If I want to check if $\alpha$ is exact in the relative cohomology I have two cases. If the class of $Z$ is non-vanishing, I just need to check that $\int_Z \alpha = 0$ and $\alpha$ exact in the usual cohomology. If the class is vanishing, then for this $U$ I need that $\int_U \alpha=0$ as well as $\alpha$ exact in the usual cohomology. Am I right? $\endgroup$ – Reb Sep 18 '18 at 8:28
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    $\begingroup$ Correct, except that I guess you meant $\int_M \alpha$ and not $\int_Z$. And in this story we need to assume that $Z$ is connected, then there are only two options for $U$, but their sum with the correct sign is $M$ so its redundant to consider both. $\endgroup$ – S. carmeli Sep 18 '18 at 9:16
  • $\begingroup$ How much of this is still correct if $Z$ contains different connected components? I'm specially interested in the case where $M$ is a compact surface and $Z$ is a union of disjoint closed curves. Do you have any reference on this? $\endgroup$ – Reb Sep 26 '18 at 14:33
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    $\begingroup$ I have no reference, your question is the first time I thought about this issue. In general, the complement of $Z$ have components $U_i$ and the integrals over them exhaust all the functionals on the relative cohomology. To see this, note that the relative cohomology is just the cohomology with compact support of the complement, since $M$ is a compactification of $M\backslash Z$. Then, for open orientable manifold, top cohomology with compact support is isomorphic to 0-th homology. $\endgroup$ – S. carmeli Sep 27 '18 at 22:54

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