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I am very new to differential geometry. I am familiar with fibered manifolds, fibered bundles (i.e fibered manifold with local trivialization) and sections. No I want to motivate the existence of local sections and for that I am looking for an example of a fibered bundle that admits no global section. Can someone tell me in which book I can find such an example. It would be best if it is very basic. I now there is the example of the slit tangent bundle of $S^2$ but I haven't introduced tangent bundles and I haven't found a good one anywhere else.

Thanks!

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The mapping $\pi :\mathbb S^1 \to \mathbb S^1$, $\pi(z) = z^2$ defines a $\mathbb Z_2$ fiber bundle over $\mathbb S^1$ which has no continuous section.

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  • $\begingroup$ What is a $\mathbb{Z}_2$ fiber bundle and why is there no global section? $\endgroup$ – JDoe Jul 13 '16 at 13:53
  • $\begingroup$ A $\mathbb Z_2$ fiber bundle is just a fiber bundle with each fiber having two elements. $\endgroup$ – user99914 Jul 13 '16 at 14:02
  • $\begingroup$ But if I take for example $\phi = z^{\frac{1}{2}}$ and choose z = $(-1,0) \in S^1$ then $(\pi \circ \phi)(-1,0) = \pi(i, 0) = (-1,0)$. And this should be valid for all $z \in S^1$. I don't see why there is no global section. $\endgroup$ – JDoe Jul 13 '16 at 14:15
  • $\begingroup$ $z^{\frac 12}$ is not a continuous mapping @JDoe $\endgroup$ – user99914 Jul 13 '16 at 14:46
  • $\begingroup$ Okay I see. That's new to me. In Saunders "The geometry of jet bundles" he says a section is simply a map from base space to total space such that $\pi \circ \phi = $ Id on the base space. Thanks! $\endgroup$ – JDoe Jul 13 '16 at 20:01

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