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Can anyone explain to me what it is said in the following article : http://indico.ictp.it/event/a06114/material/0/0.pdf , page : $3$, by Mr. Aroldo Kaplan :

The paragraph says :

Stokes : $$ \int_M d \omega = \int_{ \partial M }\omega $$ implies : $$ d \omega = 0 \ \ \Longleftrightarrow \ \ \int_{ \mathrm{boundary} } \omega = 0 \ \ \ \ \ \ \ \ \ \mathrm{and} \ \ \ \ \ \ \ \ \partial M = 0 \ \ \Longleftrightarrow \ \ \int_M \mathrm{exact} = 0 $$ so, defining :

$ H^k ( X ) = \dfrac{ \{ \omega \in \bigwedge^k \ : \ d \omega = 0 \} }{ \{ \omega \in \bigwedge^{k} \ : \ \omega = d \phi \} } = \dfrac{ \mathrm{closed} }{ \mathrm{exact} } $ , $ \ H_k ( X ) = \dfrac{ \{ M \subset X \ : \ \partial M = 0 \} }{ \{ M \subset X \ : \ M = \partial N \} } = \dfrac{ \mathrm{cycles} }{ \mathrm{boundaries} } $

the bilinear (function?): $$ H^k ( X ) \times H_k ( X ) \to \mathbb{R}, \ \ \ \ \ \ \ \ \ \ \ \ ( [ \omega ] , [M] ) = \int_M \omega $$ ( = flux of $ \omega $ through $ M $ ) is well defined.

So, I still do not understand why: $$ H^k ( X ) \times H_k ( X ) \to \mathbb{R}, \ \ \ \ \ \ \ \ \ \ \ \ ( [ \omega ] , [M] ) = \int_M \omega $$ is well defined. Can you explain to me that please ?

Thanks in advance for your help.

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    $\begingroup$ What are you unclear about? Integrating a closed form or over a boundary gives $0$, so the integral $\int_M \omega$ depends only on $[M]$ and $[\omega]$. $\endgroup$ – anomaly Aug 8 '16 at 21:12
  • $\begingroup$ @anomaly : When is the bilinear well defined ? When we don't integrate a closed form as well as when we don't integrate over a boundary ? In this case : $ [ \omega ] \not \in H^k ( X ) $ and $ [ M ] \not \in H_k (X) $. I don't really understand that clearly. :-) $\endgroup$ – Lina45 Aug 8 '16 at 21:19
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    $\begingroup$ Bah, I meant exact rather than closed above. $\endgroup$ – anomaly Aug 8 '16 at 21:26
  • $\begingroup$ I don't know what you mean, but it sounds like you don't understand the definition of $H_*$ and $H^*$. $\endgroup$ – anomaly Aug 8 '16 at 21:27
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    $\begingroup$ A boundary still has a homology class- the zero class. Likewise the zero cohomology class is the equivalence class of all exact forms. You can say e.g. that a representative for a nonzero cohomology class is closed but not exact. $\endgroup$ – Max Aug 8 '16 at 21:52
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let's prove that $([\omega],[M])=([\omega+d\phi],[M])$.

indeed, $([\omega+d\phi],[M])-([\omega],[M])=\int_M d\phi=\int_{\partial M}\phi=0$, because $M$ is a cycle.

secondly, let's prove that $([\omega],[M])=([\omega],[M+\partial N])$.

indeed, $([\omega],[M+\partial N])-([\omega],[M])=\int_{\partial N}\omega=\int_N d\omega=0$, because $\omega$ is closed.

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  • $\begingroup$ Thank you very much SIr. :) $\endgroup$ – Lina45 Aug 9 '16 at 12:45

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