1

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.

Francisco
  • 363
  • Do you know any reference for this, if you assume integer coefficient,i.e how to prove this using algebraic topology??for example see Hatcher. Then the rest will be just change of coefficient from integer to real. – Anubhav Mukherjee Jul 14 '16 at 15:24
  • 1
    Let $\tilde{M}$ be the oriented double cover of $M$. $H^n_{dR}(\tilde{M}) \simeq \mathbb{R}$, assuming $M$ is connected. Moreover $H^n_{dR}(M)$ injects into $H^n_{dR}(\tilde{M})$ and consists of elements which are invariant under sign change (corresponding to a change in orientation). Thus $H^n_{dR}(M) = 0$. I think though if you take cohomology with $\mathbb{Z}_2$ coefficients, then you should get $H^n(M, \mathbb{Z}_2) \simeq \mathbb{Z}_2$. Please check with experts though. – Malkoun Jul 14 '16 at 15:37
  • @Malkoun: Since $M$ is $\mathbb{Z}_2$-orientable, $H^n(M, \mathbb{Z}_2) = H_0^c(M, \mathbb{Z}_2) = H_0^c(M) \otimes \mathbb{Z}_2 = 0$ for $M$ noncompact. – anomaly Jul 14 '16 at 15:43
  • @Malkoun Thank you and yes, the last statement can be found in the first answer of http://math.stackexchange.com/questions/1742467/non-orientable-manifolds-and-mod-2-homology – Francisco Jul 14 '16 at 15:46
  • See also Chapter 8 of Volume 1 of Spivak's Comprehensive Intro to Differential Geometry. – Ted Shifrin Jul 15 '16 at 22:40

1 Answers1

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.

InsideOut
  • 6,883
  • In Corollary 5.8, they state that if $M$ es compact, orientable and connected, then $H_{dR}^n(M)\cong\mathbb{R}$. I am not sure if they say anything at all when $M$ is not orientable.

    What I was thinking is that if $M$ was not orientable, then you might pick an $n$-form that is not a volume form, but still not a boundary. Anyway, Malkoun already explained that this cannot happen.

     – Francisco Jul 14 '16 at 18:36