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!
If $S^{2n+1}$ is covering space of $X$, then $X$ is orientable.
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differential-topology
differential-forms
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3Are you intending to do any of your homework yourself or have you posted it all? – 2011-06-24
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0Sorry,I have thought these problems for a long time,but I can't find any idea, so I just have to find help... – 2011-06-25
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0Interesting question, btw (one in the title, I mean). – 2011-06-26
1 Answers
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If you have a nowhere vanishing n-form, then the manifold is orientable.
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0But 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? – 2011-06-27
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0@henry "so degree is 1" -- why not $-1$? (And are you using that sphere is odd, not even dimensional?) – 2011-06-29
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0@ Grigory :Yes ,here I use this condition. – 2012-02-23