On transition functions of fiber bundle

[kakarotyjn]

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!

 

[quasar987]

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 U_{\alpha}\cap U_{\beta} \times F \rightarrow U_{\alpha}\cap U_{\beta} \times F? Indeed, it is just (x,y)-->(x,g_{\alpha \beta}y)!

 

[kakarotyjn]

Thank you quasar,but I'm sorry I still can't understand the connection between the way we construct E and the transition functiong_{\alpha \beta}.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 \psi_\alpha and \psi_\beta for g_{\alpha \beta}(x)=\psi_\alpha \psi_\beta^{-1}

Thank you very much!

 

[quasar987]

Thank you quasar,but I'm sorry I still can't understand the connection between the way we construct E and the transition functiong_{\alpha \beta}.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 \psi_\alpha and \psi_\beta for g_{\alpha \beta}(x)=\psi_\alpha \psi_\beta^{-1}

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 (x,g_{\alpha \beta}y). 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 ((x,\alpha),y) where ((x',\beta),y') is identified to ((x,\alpha),y) iff x'=x and y'=g_{\alpha \beta}y.

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 U_{\alpha}\times F 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 U_{\alpha}\cap U_{\beta}) by using homeomorphisms coming from the action G\rightarrow \mathrm{Homeo}(F) 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_R} 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 Id_R, and on the second, glue along the fibers following -Id_R.

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

So you see, the transition function associated with the trivialisations \Phi_{\alpha} and \Phi_{\beta} is g_{\alpha\beta}, in the sense that \Phi_{\beta}\circ\Phi_{\alpha}^{-1}(x,y)=(x,g_{\alpha\beta}y).

 

[kakarotyjn]

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