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Is an associated bundle isomorphic to the principal bundle?

  1. Nov 29, 2009 #1

    Fredrik

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    Edit: Never mind. I should have looked more carefully at the examples in my book. The obvious right action of Z2 on S1 gives us a principal bundle that we can use to construct many different associated bundles, including the cylinder, the Möbius strip and the Klein bottle. Those spaces aren't even homeomorphic to each other, and they're even more obviously not homeomorphic to the total space of the original principal bundle.

    Let [itex]\pi:E\rightarrow M[/itex] be the projection of a principal bundle with structure group G. (I'm taking the group action of G on E to be a right action). If there's a left action of G on a space F, then we can define an "associated" fiber bundle by defining the total space to be the set of orbits of the right action of G on [itex]E\times F[/itex], and the define the following projection

    [tex]\pi_F:\frac{E\times F}{G}\rightarrow M[/tex]

    [tex]\pi_F([p,f])=\pi(p)[/tex]

    Here [p,g] is the orbit through (p,f). The right action of G on [itex]E\times F[/itex] that I mentioned above is defined by

    [tex](p,f)g=(pg,g^{-1}f)[/tex]

    Is this associated bundle supposed to be isomorphic to the original fiber bundle? The associated bundle isn't a principal bundle, but there could exist a fiber bundle isomorphism between them. If so, how is that function defined? For each [itex]f\in F[/itex], the map

    [tex]\phi_f:E\rightarrow\frac{E\times F}{G}[/tex]

    defined by

    [tex]\phi_f(p)=[p,f][/tex]

    could be a candidate, but I don't see a way to prove that it's bijective, at least not without additional assumptions. Are there assumptions missing in what I wrote above, or is the associated bundle not supposed to be fiber bundle isomorphic to the principal bundle that was used to define it?

    My failure to prove that the function is injective goes like this:

    [tex]\phi_f(p)=\phi_f(p')\implies [p,f]=[p',f']\implies (p',f')\in [p,f]\implies \exists g\in G\quad (p',f')=(p,f)g=(pg,g^{-1}f)[/tex]

    [tex]\implies\exists g\in G\quad p'=pg\ \mbox{and}\ f'=g^{-1}f[/tex]

    But to say that there exists a g such that the last equation is satisfied is to say that the group action on F is transitive, and we didn't assume that it was. I have similar problems when I try to prove that the function is surjective. I have also tried the map [itex][p,f]\mapsto p[/itex]. It's obviously surjective, but I can't prove that it's injective without assuming that the group acttion on F is transitive.
     
    Last edited: Nov 29, 2009
  2. jcsd
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