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