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- Thread starter kakarotyjn
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quasar987

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kakarotyjn

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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!

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quasar987

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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!

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

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-->{±Id

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].

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kakarotyjn

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Thank you very much again quasar!Now I really understand it,haha:rofl:

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