There is another question on this site which confirms that for an orientable hypersurface of an orientable manifold, the normal bundle is trivial. However, I was wondering if this result generalises to arbitrary orientable submanifolds.

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    $\begingroup$ What is true in that case is that the normal bundle is always orientable. For one-dimensional real vector bundles (such as the normal bundle of a hypersurface), orientability is equivalent to triviality. $\endgroup$
    – Jack Lee
    Aug 2, 2016 at 21:58

1 Answer 1


It is not true. For example, take a nontrivial vector bundle $E \to X$, then the zero section $Z \subset E$ has normal bundle isomorphic to $E$.

One can also find example when the ambient space is $\mathbb R^M$, since there are orientable manifolds which cannot be embedded in the euclidean space with trivial normal bundle. Together with Whitney's embedding theorem, there are orientable submanifolds in euclidean spaces with nontrivial normal bundle.

  • $\begingroup$ So now I need to know that there exists a nontrivial vector bundle which is orientable and has orientable zero section. It's not obvious to me that such a thing could exist. $\endgroup$
    – gj255
    Aug 2, 2016 at 22:22
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    $\begingroup$ The simplest example would be the tangent bundle $TX=E$ of $X =\mathbb S^2$. The tangent bundle is always orientable, and $\mathbb S^2$ is also orientable. But it's tangent bundle is nontrivial. @gj255 $\endgroup$
    – user99914
    Aug 2, 2016 at 22:26
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    $\begingroup$ I saw the example in this link. But the even dimensional real projective space is non-orientable. $\endgroup$
    – maplemaple
    Jan 31, 2018 at 21:15
  • $\begingroup$ Do you have some idea in this question? Thanks! math.stackexchange.com/questions/2630328/… $\endgroup$
    – maplemaple
    Jan 31, 2018 at 21:35
  • $\begingroup$ math.stackexchange.com/a/734712/481713 $\endgroup$
    – maplemaple
    Jan 31, 2018 at 21:37

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