On transition functions of fiber bundle

1. Oct 26, 2011

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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2. Oct 26, 2011

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)!

3. Oct 27, 2011

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!

4. Oct 27, 2011

quasar987

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 $\Phi_{\beta}\circ\Phi_{\alpha}^{-1}(x,y)=(x,g_{\alpha\beta}y)$.

Last edited: Oct 27, 2011
5. Oct 29, 2011

kakarotyjn

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