Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any direct way to prove that $n$-manifold is orientable? In AT we can just calculate $n$'th homology group and check whether it's $\mathbb Z$ or $0$. But I want a geometric method, using differential forms. Thanks!

share|cite|improve this question
Are you intending to do any of your homework yourself or have you posted it all? – Alex B. Jun 24 '11 at 10:38
Sorry,I have thought these problems for a long time,but I can't find any idea, so I just have to find help... – henry Jun 25 '11 at 12:20
Interesting question, btw (one in the title, I mean). – Grigory M Jun 26 '11 at 12:11

If you have a nowhere vanishing n-form, then the manifold is orientable.

share|cite|improve this answer
But how to construct this n-form?I have an idea:for any deck transformation ,it's free, so degree is 1 ,i.e. keep orientable, then we can construct n-form of X from local n-form of S^(2n+1). Is it right? – henry Jun 27 '11 at 10:22
@henry "so degree is 1" -- why not $-1$? (And are you using that sphere is odd, not even dimensional?) – Grigory M Jun 29 '11 at 20:33
@ Grigory :Yes ,here I use this condition. – henry Feb 23 '12 at 13:37

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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