If a smooth, compact, connected manifold $M$ is not orientable then its top De Rham cohmology is zero I know how to prove that if a (smooth, compact, connected) manifold of dimension $n$ is orientable then $H^n_{dR}(M) = \mathbb{R}.$ I was wondering why the converse is true, i.e., if the top comology isn't zero....how can I construct a never-vanishing n-form? I get that I can use an n-form $w$ that is closed but not exact...but does that imply that it is never-vanishing?
Thanks!
 A: Let $M$ be closed connected non-orientable and $\omega$ a top-dimensional form. Pull this back to the orientable double cover $\tilde M$; because the nontrivial deck transformation $\iota$ is orientation reversing, we have $$-\int \tilde \omega = \int \iota^*\tilde \omega,$$ and so $\int \tilde \omega = 0$. Therefore there is a form with $d\eta = \tilde \omega$. You can replace $\eta$ with $\frac 12(\eta + \iota^* \eta)$ to get an $\iota^*$-invariant form whose derivative is $\tilde \omega$; thus it descends to an antiderivative of $\omega$ on $M$, making every top form exact. 
This is not the constructive approach you're hoping for, I don't think. I think I have such an approach but it's tedious and not insightful. 
A: I don't know if this is preferable or not, but you can argue directly that if $M$ is an $n$-dimensional manifold, either non-compact or non-orientable, then $H^n_{\text{dR}}(M)=0$. I believe Spivak writes this out in detail in Volume I of his five-volume opus. But here's a sketch of an argument. I invite you to write out all the details.
The fundamental lemma is that if $\psi$ is a compactly supported $n$-form on $\Bbb R^n$ with integral $0$, then $\psi=d\xi$ for compactly supported $(n-1)$-form $\xi$. (You can prove this directly from the fact that $H^{n-1}_{\text{dR}}(S^{n-1}) \cong \Bbb R$.)
Suppose $M$ is non-compact. Cover $M$ with a locally finite collection of charts $U_i$ so that $U_i\cap U_{i+1}\ne\emptyset$ for all $i$. Let $\omega$ be an $n$-form. By a partition of unity argument, you can assume $\text{supp}(\omega)\subset U_1$. Choose forms $\omega_i$ with $\text{supp}(\omega_i)\subset U_i\cap U_{i+1}$ and with $\int_M\omega_i\ne 0$. Then by the lemma there are forms $\eta_i$, supported in $U_i$, with $\omega-c_1\omega_1=d\eta_1$ on $U_1$, and $\omega_i - c_{i+1}\omega_{i+1} = d\eta_{i+1}$ on $U_{i+1}$, $i=1,2,\dots$.
So $\omega=d\eta_1+c_1\omega_1=d(\eta_1+c_1\eta_2)+c_1c_2\omega_2 = \dots = d(\eta_1+c_1\eta_2+c_1c_2\eta_3+\dots)$, where the sum on the right is locally finite and so makes sense.
If, on the other hand, $M$ is non-orientable, there is an orientation-reversing loop in $M$, which we cover with charts $U=U_1,U_2,\dots, U_s, U_{s+1}=U$ so that $U_i\cap U_{i+1}\ne\emptyset$ and $x_{i+1}\circ x_i^{-1}$ is orientation preserving for $i=1,\dots,s-1$, and $x_{s+1}\circ x_s^{-1}$ is orientation-reversing. Start with an $n$-form $\omega_0$ supported in $U$ with $\int_M \omega \ne 0$. As before, create $n$-forms $\omega_i$ supported in $U_i\cap U_{i+1}$ with $\int_M \omega_i>0$, $i=1,2,\dots,s-1$. [We're using the charts $x_i$ to define the orientations here.] Set $\omega=-\omega_0$. Then, proceeding as before, we'll get $\omega = c\omega_0 + d\eta$ for appropriate $c>0$ and $\eta$. This means that $-(1+c)\omega_0=d\eta$, so $\omega_0$ is exact. (Again, a partition of unity argument reduces to this case.)
