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If $M$ is a contractible differentiable manifold, then $M$ is orientable.

Asaf Karagila
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user58930
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  • You haven't given any of your thoughts, @user – Rustyn Jan 25 '13 at 00:15
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    By the Poincare lemma, all closed forms are exact. Then by the natural projection $\pi: TM \to M$.., compute the homology groups and show isomorphic to $\mathbb{Z}. –  Jan 25 '13 at 00:22
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    @user- Please see FAQ page on how to ask a good question. –  Jan 25 '13 at 00:38
  • @Jaivir Baweja, I don't understand your hint about Poincare lemma. What homology groups are you computing? –  Jan 25 '13 at 01:08

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If $M$ is contractible, then any fibre bundle $E\rightarrow M$ is trivial. In particular, it means that the tangent space of $M$ and its cotangent space are trivial bundles.

Then, one can choose a global $n$-frame $(\xi_1,\cdots,\xi_n)$ of $T^\star M\rightarrow M$ and look at the $n$-form on $M$ given by $\omega = \xi_1\wedge \cdots \wedge \xi_n$.

It is a nowhere vanishing $n$-form, so it induces on orientation of $M$.

Bebop
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