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Consider $\Sigma$ a non-orientable $n$-dimensional manifold immersed in a $n+1$-dimensional manifold M, such that the normal bundle $N\Sigma$ is not trivial, $\tilde \Sigma$ its orientable double cover and $\pi:\tilde \Sigma \to \Sigma$ the covering map.

Is is true that the vector bundle $\pi ^* N\Sigma$ given by the pullback $N\Sigma$ by the map $\pi$ is trivial?

Some references would be nice, thanks.

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Cheap answer: No, Take non-orientable $\Sigma$ with $b_1\Sigma>1$. By looking at rational cohomology we have that the transfer map of a finite covering $\bar \Sigma$ gives an isomorphism: $$ H_i(\Sigma;\mathbb Q) \hookrightarrow H_i(\bar \Sigma ;\mathbb Q) \to H_i(\Sigma;\mathbb Q), $$ hence, as indicated, the first map injects and $b_1\bar \Sigma >1$. In particular the map $H^1(\Sigma;\mathbb Z/2) \to H^1(\Sigma;\mathbb Z/2)$ is non-trivial, i.e. there exists $0 \neq \omega$, s.t. $\pi^*\omega \neq 0$. There is an isomorphism (as explained in this answer: https://math.stackexchange.com/a/1277969/156302 ) $\omega_1 :\{\text{line bundles over }X\} \to H^1(X;\mathbb Z/2)$, so $\omega$ gives us a line bundle $L\to \Sigma$. Note that this is a non-orientable bundle and pulls back to the non-orientable line bundle corresponding to $\pi^*\omega$. We get a codimension 1 embedding $\Sigma \to L$ by the zero section which does the trick. If $\Sigma$ is compact and you want to imbed into a compact space take the disk bundle $\Sigma \to D(L)$.

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  • $\begingroup$ However note that some refinements of your question have a positive answer. $\endgroup$ – Daniel Valenzuela Sep 21 '15 at 10:09
  • $\begingroup$ Which particular refinements? Thank you. $\endgroup$ – Lawen Sep 21 '15 at 14:36

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