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On transition functions of fiber bundle

  1. Oct 26, 2011 #1
    I don't understand why the constructed fiber bundle E have g_{\alpha \beta} as its transition function.

    The problem is in the pdf file,thank you!

    Attached Files:

  2. jcsd
  3. Oct 26, 2011 #2


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    It's almost a tautology once you know the little bit of implied information lying behind that statement. Because so far you have defined E. But what about the bundle projection pr:E-->B? Well, it's the natural choice: send [x,y] to x. So now, why is this an F-bundle with structure group G? Well, because pr:E-->B admits local trivialisations over the family {U_\alpha}. Indeed, pr-1(U_\alpha) is homeomorphic to U_\alpha x F via [x,y]-->(x,y). And for U_\beta another guy in that family, what is the transition function [itex]U_{\alpha}\cap U_{\beta} \times F \rightarrow U_{\alpha}\cap U_{\beta} \times F[/itex]? Indeed, it is just (x,y)-->(x,g_{\alpha \beta}y)!
  4. Oct 27, 2011 #3
    Thank you quasar,but I'm sorry I still can't understand the connection between the way we construct E and the transition function[tex]g_{\alpha \beta}[/tex].In E, (x,y) is equivalent to (x,g_{\alpha \beta} by definition.But how does this equivalent relation induce that g_{\alpha \beta} is the transition function?Why there is an equivalent relation?

    In order to prove g_{\alpha \beta} is the transition function,we need to find fiber-preserving homeomorphisms [tex]\psi_\alpha[/tex] and [tex]\psi_\beta[/tex] for [tex]g_{\alpha \beta}(x)=\psi_\alpha \psi_\beta^{-1}[/tex]

    Thank you very much!
  5. Oct 27, 2011 #4


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    Mmh, maybe your problem stems from the technicalities of the definition. Because strictly speaking it is not true that in E, (x,y) is equivalent to [itex](x,g_{\alpha \beta}y)[/itex]. Because E is obtained by quotienting a disjoint union over the index alpha. This means that the elements of E are actually equivalences classes of elements of the form [itex]((x,\alpha),y)[/itex] where [itex]((x',\beta),y')[/itex] is identified to [itex]((x,\alpha),y)[/itex] iff x'=x and [itex]y'=g_{\alpha \beta}y[/itex].

    So what we're doing here is we're constructing E by taking a covering {U_\alpha} of the base B and considering the trivial bundles [itex]U_{\alpha}\times F[/itex] over each U_\alpha. Then we glue all of these trivial bundles along fibers over the points where they "intersect" (i.e. over the intersections [itex]U_{\alpha}\cap U_{\beta}[/itex]) by using homeomorphisms coming from the action [itex]G\rightarrow \mathrm{Homeo}(F)[/itex] of the group G on F. This induces a potential "twisting" in the bundle.

    For instance, the Mobius bundle can be constructed in this way using G=Z/2Z-->{┬▒IdR} and a covering of S1 of only two open sets {U1,U2}. Then the intersection of U1 and U2 has two connected components. On the first, glue along the fibers following IdR, and on the second, glue along the fibers following -IdR.

    With that, define now a projection map pr:E-->B that sends [itex][((x,\alpha),y)][/itex] to x. This is obviously independent of the class so it is well defined. Now E is a fiber bundle because it is trivializable over the U_\alpha's by the map [itex]\Phi_{\alpha}:pr^{-1}(U_{\alpha})\rightarrow U_{\alpha}\times F[/itex] that says "for a class in [itex]pr^{-1}(U_{\alpha})[/itex], pick the representative that belongs to [itex]U_{\alpha}\times F[/itex], say [itex]((x,\alpha),y)[/itex], and send it to (x,y)". Now suppose \beta is another index. How does [itex]\Phi_{\beta}:pr^{-1}(U_{\beta})\rightarrow U_{\beta}\times F[/itex] acts on the same element [itex][((x,\alpha),y)][/itex] of E? Well, it says "pick the representative that belongs to [itex]U_{\beta}\times F[/itex]... well that's [itex]((x,\beta),g_{\alpha\beta}y)[/itex] by definition of the equivalence relation! So send it to [itex](x,g_{\alpha\beta}y)[/itex]."

    So you see, the transition function associated with the trivialisations [itex]\Phi_{\alpha}[/itex] and [itex]\Phi_{\beta}[/itex] is [itex]g_{\alpha\beta}[/itex], in the sense that [itex]\Phi_{\beta}\circ\Phi_{\alpha}^{-1}(x,y)=(x,g_{\alpha\beta}y)[/itex].
    Last edited: Oct 27, 2011
  6. Oct 29, 2011 #5
    Thank you very much again quasar!Now I really understand it,haha:rofl:
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