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Pullback of fibre bundle

  1. May 12, 2007 #1
    Chris Isham said (Modern Differential Geometry for Physicists) it's fairly easy to show that the pullback bundles of a (locally trivial) fibre bundle by two homotopic maps are isomorphic. But I haven't had a clue. Can somebody tell me how to prove?
     
  2. jcsd
  3. May 12, 2007 #2

    mathwonk

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    Ok I think I know how to do it. Its tricky because the isomorphism does not seem to be at all unique. that makes it not really stick out for you.

    you have

    an open cover by Ui, and on each overlap you have two clutching data, sij and tij, which are maps into the group of automorphisms of the fiber F.

    then you need isomorphisms, i.e. homeomorphisms fi:Ui-->Ui, such that

    on overlaps, sij o fj = fi o tij. so you take f1 to be anything, then you need f2 = sij^(-1) 0 f1 o tij on Ui meet Uj. so define it that way on Ui meet Uj and use an extension lemma like urysohn, to extend f2 to all of U2.

    continue with f3,......

    it seems compactness is needed. you could look it up in atiyah's book on k theory, or husemollers book on bundles, or steenrod, or any other book on bundles, like hatcher's algebraic topology, which is free online.
     
  4. May 16, 2007 #3

    mathwonk

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    i seem to have neglected to make f2 a homeomorphism, simply by using any old extension in fact i cant understand any of it myself. what the heck is sij^(-1)?
     
  5. May 16, 2007 #4

    mathwonk

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    here is a free reference that uses parallel transport to assign an isomorphism of pullbacks to a homotopy for principal fiber bundles.

    http://arxiv.org/pdf/math/0105161
     
  6. May 16, 2007 #5

    mathwonk

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    well i looked in bott tu p. 58 for the vector bundle case, and it is knid of like what i said.

    one shows the isomorphism class of a bundle on YxI, restricted to Yx{0} cannot change locally near 0.

    i.e. the homotopy gives a pull back bundle E on YxI, and we look at its restriction to Yx{0} and we just cross that restriction with I to get a family F of bundles on YxI, which have isomorphic restrictions to every Yxt.

    So we have an isomorphism of these bundles E,F over 0 and want to extend the isomorphism say to Yxe where e is a small interval containing 0.

    To do this view the isomorphism as a section defined over Yx{0}, of the subbundle Iso(E,F) of Hom(E,F).

    Then since this last bundle has Eucklidean space fibers, we can extend, again by a Urysohn type lemma.

    then near t=0 the section lies in Iso(E,F), and using compactness of Y and I we get our theorem over YxI.

    It can be extended to paracompact Y. This argumenT uses that the bundle of isomorphisms of two vector bundies is itself a fiber bundle contained in a euclidean space bundle, which seems a little special for all fiber bundles..

    to go further i guess we need to know your definition of a fiber bundle. but i will probably leave it here.
     
  7. Jun 4, 2007 #6

    mathwonk

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    was this useful?
     
  8. Jul 14, 2007 #7
    At my present level, I can't comprehend what you said. I have bookmarked this page, in future I will refer back to it. Thank you mathwonk, all I know now is that it's not fairly easy.
     
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