# de Rham cohomology with compact support isomorphic to $\mathbb{R}$

The $$k$$-dimensional de Rham cohomology with compact support is defined as $$H_c^k(M)=\frac{Z_c^k(M)}{B_c^k(M)}$$ where $$Z_c^k(M)$$ is the vector space of all closed $$k$$-forms with compact support on $$M$$ and $$B_c^k(M)$$ is the vector space of all $$k$$-forms $$d\eta$$ where $$\eta$$ is a $$(k-1)$$-form with compact support on $$M$$. A Comprehensive Introduction To Differential Geometry states that for a connected orientable manifold $$M$$ (I think $$M$$ is considered as with $$\partial M=\emptyset$$) we have $$H_c^n(M)\simeq \mathbb{R} \quad \quad \quad (*)$$ Then he says that this assertion is equivalent to two other assertion, which I list below:

1. Given a fixed $$\omega$$ such that $$\int_M\omega\neq0$$, then for any $$\omega'$$ n-form with compact support there exists a real number $$\lambda$$ such that $$\omega' - \lambda \omega$$ is exact (It is not mentioned, but I assume both $$\omega$$ and $$\omega'$$ are supposed to be closed)

2. An closed form $$\omega$$ is the differential of another form with compact support if $$\int_M \omega=0$$

I wanted to prove this (the equivalence of this assertions), but I cannot see how this is true.

Any help is welcome

• @EricWofsey by a form of this sort I mean closed and compact supported.... On the other hand I thought about your second comment, yes it looks like I am trying to prove 2, that was not my intention. I am really more interested in first understanding why this statements are equivalent to each other Apr 8 '20 at 5:37
• Note that $\omega$ and $\omega'$ being closed is automatic: any form whose degree is the same as the dimension of the manifold is closed. Apr 8 '20 at 13:53

First, there is an error in your statement (1): it should say $$\omega'-\lambda\omega$$ is exact, not $$\omega-\lambda\omega'$$. (Also, $$n$$ should presumably be the same as the $$k$$ you mention elsewhere, the dimension of $$M$$.)

Now there is a linear map $$I:H^k_c(M)\to\mathbb{R}$$ which maps $$[\omega]$$ to $$\int_M\omega$$ (note that this is well-defined because the integral of an exact form is $$0$$). Since there exists a compactly supported $$k$$-form with nonzero integral, $$I$$ is surjective. Since $$\mathbb{R}$$ is a 1-dimensional vector space, the following are then equivalent:

• (a) $$I$$ is an isomorphism.
• (b) $$H^k_c(M)$$ is 1-dimensional, i.e. $$H^k_c(M)\cong \mathbb{R}$$
• (c) $$\ker I$$ is trivial.

But (c) is exactly the same as your statement (2), since an element of $$\ker I$$ is $$[\omega]$$ such that $$\int\omega=0$$ and to say $$[\omega]=0$$ means that $$\omega$$ is the differential of a form with compact support. And of course (b) is the same as your statement (*).

It remains to be shown that your statement (1) is also equivalent to the others. Note first that (1) says exactly that $$[\omega]$$ spans $$H^k_c(M)$$, since it says that for any $$[\omega']\in H^k_c(M)$$, there is some scalar $$\lambda$$ such that $$[\omega']=[\omega]$$. So, (1) implies $$H^k_c(M)$$ is 1-dimensional. Conversely, assume statement (2). Fix $$\omega$$ such that $$\int_M \omega\neq 0$$ and take any other $$\omega'$$, and let $$\lambda=\frac{\int_M\omega'}{\int_M\omega}$$. Then $$\int_M \omega'-\lambda\omega=0$$, so $$\omega'-\lambda\omega$$ is exact.