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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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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0@ Gr$i$gory :Yes ,here I use th$i$s co$n$d$i$tion. – 2012-02-23