# How does this example from Spivak that $H_c^n(\mathbb R^n) \ne 0$?

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 $$H_c^n(\mathbb R^n) \ne 0$$?

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