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I am not sure how this integral that is being calculated using Stoke's theorem shows that the $n$th de Rham cohomology group with compact supports of $\mathbb R^n$ is not trivial.

How does the fact that the integral of $\omega$ over $\mathbb R^n$ is zero show that $H_c^n(\mathbb R^n) \ne 0$?


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Since $\mathbf R^n$ has dimension $n$ and $\omega$ is an $n$-form, we have that $\omega$ has to be closed (as there are no nontrivial $(n+1)$-forms on $\mathbf R^n$). Hence $\omega\in Z_c^n(\mathbf R^n)$. But $\omega\not\in B_c^n(\mathbf R^n)$ because we'd would have $\eta$ with compact support such that $\omega=d\eta$, and that would imply that $\int_{\mathbf R^n} \omega =0$ as Spivak claims. This is a contradiction because $\omega$ is non-negative and positive at some point (hence on a neighborhood of that point by continuity). Hence $\omega$ is a nonzero element of $H_c^n(\mathbf R^n)$.

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  • $\begingroup$ Why is $\int_{\partial \mathbb R^n} \eta =0$? $\endgroup$ – Al Jebr Dec 11 '18 at 0:34
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    $\begingroup$ @AlJebr It's because $\partial \mathbf R^n = \emptyset$. $\endgroup$ – Zircht Dec 11 '18 at 0:41

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