Is it true that a compact non-orientable manifold $M^n$ must have $H_{dR}^n(M)=0$?

Question

I have seen this statement assumed to be true several times --- I just can't find a reference, and now I'm starting to suspect there can be catch somewhere.

Answer 1

Try to see: "Differential form in algebraic geometry" wrote by Bott and Tu.

I'm not going to give you an actually formal proof, but a way to think about this.

Basically if $M$ is an orientable manifold, the choise of an orientation for it is equivalent to choose a basis for $H^n(M, \Bbb R)\equiv \Bbb R$, so up to positive multiplier a basis for $\Bbb R$ are $\pm 1$, infact, supposing $M$ connected, there are two different orientations. I am almost sure that this equivalence is describe in the book above.

If $M$ is non-orientable, then you cannot choose an orientation for it, this means that there not exists a basis for the vector space $H^n(M, \Bbb R)$, this imply that $H^n(M)$ must be zero.