# 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!

• I have read that you can see that the tangent bundle must be disconnected...maybe you can see that $H_{dR}^0(TM)$ has dimension 2? Jun 26, 2017 at 15:37
• What do you mean when you say the tangent bundle is disconnected? Jun 26, 2017 at 20:18
• @AmitaiYuval: To hazard a guess, perhaps the OP meant the bundle of $n$-forms is disconnected? Jun 26, 2017 at 20:45
• @JasonDeVito I'm assuming you mean $\Lambda^n(M)$, which is a line bundle over $M$, and is always connected. So I'll assume further you mean this bundle with the zero section removed. This is actually clearly related to the existence of a volume form, and so, I think you answered my question. Jun 26, 2017 at 21:23
• I guess you can call it the orientation bundle of $M$. If $M$ is non-orientable, it is also called the orientable double cover. And yes, this double cover is connected $\Leftrightarrow M$ is non-orientable. Jun 27, 2017 at 19:28

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.

• Thanks! I'll wait for a couple of days in case there is a constructive, intuitive proof and if not I'll mark your answer as accepted! Jun 27, 2017 at 18:47

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.)

• I asked my topology professor about your claim that "Cover M with a locally finite collection of charts U_i so that U_i intersect U_{i+1} is not empty for each i." After about a week of thinking about it, the prof told me that it's NOT possible (for an arbitrary manifold M). It is the locally finite condition that cannot be guaranteed. Can you please elaborate on the proof of your statement? Nov 10, 2017 at 16:41
• @Curiosity All my manifolds are paracompact. Aren't yours? Nov 10, 2017 at 16:45
• Yes, they are paracompact, the only issue was the existence of a cover by charts satisfying both conditions 1) locally finite and 2) U_i intersect U_{i+1} is not empty for each i. The problem was that if you have a property 2 for a cover, then you pass to a locally finite subcover, you may break the chain condition 2. Nov 10, 2017 at 17:17
• @Curiosity: I don't see a problem. The key lemma is this: Given a finite covering of a compact set by coordinate charts, we can renumber them (repeating indices if necessary a finite number of times) so that $U_i\cap U_{i+1}\ne\emptyset$. (This is easily proved by induction.) Now cover $M$ by an increasing sequence $\{K_j\}$ of compact sets and apply the lemma to $K_j-\text{int}(K_{j-1})$, paying attention to choose an overlapping pair of sets as we proceed from $j$ to $j+1$. Nov 10, 2017 at 17:38
• I remember someone proposed an argument like this, but then someone mentioned what would happen if the sets had to go "back and forth" an infinite number of times crossing the set K_1 for example. Would this violate the local finiteness condition? (Being compact, one can get an infinite number of points in K_1, which would then have a convergent subsequence to a point, which then would belong to infinitely many of the U_i's? This would imply the U_i's constructed like this would not be locally finite. Nov 10, 2017 at 18:22