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On local trivializations and transition functions of fibre bundles

  1. Aug 27, 2013 #1
    Hi everyone!

    I would like to ask you some clarifications on an explicit example of local trivializations and transition functions of fibre bundles: namely on the [tex][-1,1]\hookrightarrow E\rightarrow S^1[/tex] bundle (which I guess is the simplest possible example).

    Following Nakahara (chapter 9, example 9.1) we pick [tex]U_1=(0,2\pi)\,\,U_2=(-\pi,\pi)[/tex] as an open covering for the base space [tex]S^1[/tex] and label [tex]A=(0,\pi)\,\,B=(\pi,2\pi)[/tex] the intersection [tex]U_1\cap U_2[/tex].

    Now Nakahara takes as local trivialization on A
    [tex]\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,t)[/tex]
    for [tex]\theta\in A \,\,t\in [-1,1][/tex], then he says that on the B section we have two possible choices, namely

    [tex]1)\,\,\,\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,t)[/tex]
    [tex]2)\,\,\,\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,-t)[/tex].

    Now, my questions:

    1) Shouldn't we have both this possibilities for the A sector as well? Does Nakahara simply not state them because they would not add anything to the example (you end up with either a cylinder or the moebius strip anyway :) )

    2) Why, disregarding the specific sector (A or B), are we limited to the two choices aboce, i.e.
    [tex]\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,\pm t)[/tex]
    ?

    I mean, couldn't I pick the choice

    [tex]3)\,\,\,\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,\frac{t}{a})[/tex]
    with [tex]a\in\mathcal R/\{0\}[/tex]??
    In this way we should have for the transition function [tex]t_{12}(\theta):t\rightarrow \frac{t}{a}[/tex], so that with the final choice of

    [tex]1)\,\,\,\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,t)[/tex]
    on the A sector and
    [tex]3)\,\,\,\phi_1^{-1}(u)=(\theta ,t)\,\,\text{and}\,\,\phi_2^{-1}(u)=(\theta ,\frac{t}{a})[/tex]
    on the B sector the bundle would have the structure group [tex]G=\{e,1/a\}[/tex] which is something different from the cylinder or the Moebious strip!!



    Where am I mistaking??

    Thanks to all of you for your time!
     
  2. jcsd
  3. Sep 19, 2013 #2

    lavinia

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    The transition functions must be isomorphisms of the fiber. Multiplication by a real is not even a homeomorphism of the unit interval

    But you can take the whole real line as the fiber and then multiplication by a real will be a linear isomorphism of the fiber thus making the bundle a one dimensional vector bundle.

    That said, you are correct that different structure groups will give the same total space of the bundle but different bundles because they have different structure groups.

    However for a line bundle over the circle, the structure group can be reduced to either the trivial group or to Z/2Z. In the first case the bundle is trivial and the total space is a cylinder. In the second case, the bundle is non-orientable and the linear isomorphism in Z/2Z is just multiplication by -1. This is the Mobius band.

    Try going through the cases you pose and see which ones can be reduced to the trivial group and which can be reduced to Z/2Z but no further.

    Another instructive example is the twisted torus.

    Take a cylinder and identify the opposite boundary circles by a 180 degree rotation. If you consider the structure group to be the whole circle then this bundle is trivial. That is, its structure group can be reduced to the identity map. If you consider the structure group to be Z/2Z ( the square of a 180 degree rotation is the identity) then this bundle is not trivial.
     
    Last edited: Sep 19, 2013
  4. Sep 22, 2013 #3
    Thank you for your kind answer lavinia! But I'm afraid I'm gonna need a little more help to get to the bottom of this :)

    Is it now clear that i cannot take [itex]a\in\mathbb R/0[/itex], but what is I restrict to [itex]1\leq |a|<\infty [/itex]? This gives indeed a homomorphism beween the fiber! I really cannot see how such transition function can give a structure group ONLY reducible to [itex]e[/itex] and [itex]\mathbb Z_2[/itex]

    Thanks again for your help!
     
  5. Sep 27, 2013 #4

    lavinia

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    Numbers with absolute value greater than 1 still don't work. Multiplication moves the end points either shrinking them or expanding them.
     
  6. Oct 1, 2013 #5
    Ok, I totally messed up-- I meant [itex]|a|\leq 1, a\neq 0[/itex].

    What about this mapping?

    Sorry for the inconvenience :) :) :)
     
  7. Oct 1, 2013 #6

    lavinia

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    Multiplying by any number other than 1 or -1 can not be a homeomorphism of a finite interval.
     
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