3
$\begingroup$

I am trying to find if there is a orientable compact manifold $M$ of dimension 10 with the 5th cohomology group of De Rham $H^5_{DR}(M)\cong \mathbb R$.

But, I can find such an example or prove that it is not possible. Anyone have some ideas?

Thank you.

$\endgroup$
  • $\begingroup$ Do you know about Poincaré duality? $\endgroup$ – Najib Idrissi Dec 9 '14 at 18:21
3
$\begingroup$

It's not possible. There is a nondegenerate antisymmetric pairing $H^5\otimes H^5\to \mathbb R$. This means that $H^5$ can't be $1$ dimensional.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ ...because in fact it must be even-dimensional. More generally if $5$ is replaced by any odd number. $\endgroup$ – Qiaochu Yuan Dec 9 '14 at 19:39
  • $\begingroup$ @QiaochuYuan Nice! But how to show it? $\endgroup$ – Chieh LIU Dec 9 '14 at 22:06
  • $\begingroup$ @Chieh: it follows by Poincare duality. On a closed orientable $4k+2$-manifold there is a nondegenerate antisymmetric pairing $H^{2k+1} \times H^{2k+1} \to H^{4k+2}$, and such things only exist on even-dimensional vector spaces (en.wikipedia.org/wiki/Symplectic_vector_space). $\endgroup$ – Qiaochu Yuan Dec 9 '14 at 22:11
  • $\begingroup$ @QiaochuYuan Got it. Thank you! $\endgroup$ – Chieh LIU Dec 9 '14 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.