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Hello,

this may be a trivial question, but I am not very familiar with the topic. Let (M,g) be a Riemannian Manifold. (In fact, we don't need the metric here.)

What exactly does it take for two k-submanifolds $S$ and $S'$ to lie in the same homology class? And why does that imply $~ \int_S ~ \omega = \int_{S'} ~ \omega ~$, where $\omega$ is a closed differential form on M.

This must be somehow a consequence of Stokes' Theorem. I would be pleased with an answer for Dummies. :)

$~$

Greetings,

Henry

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    $\begingroup$ I believe your questions are all answered on the De Rham Cohomology Wikipedia page: en.wikipedia.org/wiki/De_Rham_cohomology In particular $\int_S \omega =\int_{S'} \omega$ for all closed forms $\omega$ if and only if $S$ and $S'$ are homologous (modulo torsion, or with real or rational coefficients, take your choice). $\endgroup$
    – Ryan Budney
    Commented Oct 3, 2010 at 15:24
  • $\begingroup$ This question is more appropriate for math.stackexchange.com $\endgroup$
    – Deane Yang
    Commented Oct 4, 2010 at 0:17

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By triangulating each manifold $S$ and $S'$, you can regard them as sums of singular simplices. (A singular simplex is a continuous map $f\colon \Delta^n\to M$.) So the very simple answer is that $S$ and $S'$ are the same homology class if and only if they are the boundary of some sum of singular simplices. This can happen if, for example, there is a map of an oriented manifold of one higher dimension into $M$ whose boundary is the disjoint union $S\cup -S'$. If all the singular simplices are smooth maps, then Stokes theorem indeed tells you that the integral along the boundary of each singular simplex is zero, which then implies that the integral $\int_{S\cup-S'}\omega=0$, or $\int_S\omega=\int_{S'}\omega$. Someone else can probably comment on why you can pass from the topological to the smooth category.

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  • $\begingroup$ Thanks, this helps a lot. Concerning your last sentence, it may make things easier to mention the fact that, for my purposes, we can assume S and S' to be smooth. $\endgroup$
    – Henry Wegener
    Commented Oct 3, 2010 at 16:22
  • $\begingroup$ so you may be interested in the notion of cobordism (see e.g. en.wikipedia.org/wiki/Cobordism and references) $\endgroup$
    – Pietro Majer
    Commented Oct 3, 2010 at 16:51

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