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!

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quasar987

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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 $U_{\alpha}\cap U_{\beta} \times F \rightarrow U_{\alpha}\cap U_{\beta} \times F$? 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 function$$g_{\alpha \beta}$$.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

Homework Helper
Gold Member
Thank you quasar,but I'm sorry I still can't understand the connection between the way we construct E and the transition function$$g_{\alpha \beta}$$.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-->{±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 $[((x,\alpha),y)]$ 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 $\Phi_{\beta}\circ\Phi_{\alpha}^{-1}(x,y)=(x,g_{\alpha\beta}y)$.

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kakarotyjn

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

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