# If $S^{2n+1}$ is covering space of $X$, then $X$ is orientable.

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!

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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.

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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