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

Thanks in advance

  • $\begingroup$ @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 $\endgroup$ Apr 8 '20 at 5:37
  • $\begingroup$ Note that $\omega$ and $\omega'$ being closed is automatic: any form whose degree is the same as the dimension of the manifold is closed. $\endgroup$ 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.


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