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Global Section of Fiber Bundles with Contractible Fiber

  1. May 4, 2010 #1
    Is it true in complete generality that every fiber bundle with contractible fiber have a global section? Or do some sort of restrictions on the bundle need to be made? I ran across a mention of this fact in Guillemin and Sternberg's "Supersymmetry..." and I'm not sure how to prove it.
  2. jcsd
  3. May 5, 2010 #2
    Do you know any obstruction theory? With that in hand, it's trivial. I'm not sure how far outside CW complexes obstruction theory extends, however.

    If you don't know obstruction theory, here's a rough idea of how the argument works. Suppose we have a CW-structure on our base space, and assume inductively we've defined a section over the (k-1)-skeleton. We would like to extend the map over the k-skeleton. The potential problem with this is that we're attempting to extend a map defined over a sphere to the disk it bounds. The "obstruction" to being able to extend it is precisely that the map may be homotopically nontrivial. (think of the section as a map D->D x F, x -> (x,s(x))). This, of course, is not a problem at all if the fiber is contractible.
  4. May 6, 2010 #3


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    I would think that over local trivializations of the bundle the contraction mapping would make a new section that is constant. I think you can piece these constant maps together over all local trivializations because the contraction can be done to any point in the fiber.

    haven't checked this though.
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