Suppose $N^{n+1}$ is a orientable smooth manifold, $M^n$ is a smoothly embedded hypersurface in of $N^{n+1}$. Is the normal bundle of $M^n$ trivial?


1 Answer 1


This is true if and only if $M$ is also orientable. Call the normal bundle $\xi$. Then $TM \oplus \xi = i^*TN$, where $i$ is the inclusion map. So $TM \oplus \xi$ is orientable. The only orientable line bundle is the trivial line bundle (proof: use the orientation to construct a nonvanishing section). So if $\xi$ is trivial, then $TM$ must be orientable. Conversely if $TM$ is orientable then you can use the orientation on it and $TN$ to construct an orientation on $\xi$.

For concrete counterexamples take $\Bbb{RP}^n\hookrightarrow \Bbb{RP}^{n+1}$ for $n>0$ even.

  • $\begingroup$ I understand. Thank you very much for helping me again! A further question: If $N$ is simply connected or $H_1(N)$ is torsion free, will any embedded closed hypersurface of $N$ orientable? $\endgroup$
    – 王文龙
    Oct 25, 2015 at 6:24

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