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in the definition of a fiber bundle associated to a principal bundle,we have this: (P*F)~ s.t. (p,f)~(p.g,g-1.f)

where P is principal bundle and F is a Manifold. such way of gluing the points of P*F together gives us a quotient topological space such that satisfies in conditions of a fiber bundle with fiber F and a certain set of transition functions that comes from action of group G (fiber of P) of F. a natural question raise here: is this unique way of gluing?is the relation ~ that defined above plays a crucial role in construction of bundle? can we propose another relation that gives same bundle structure? i guess it can be like that: (p,f)~(p.g,g.f) if the first relation is crucial element of construction,how?show me it is unique and not all considerations could be satisfied by another relation. if it is not true,why this way introduces in textbooks without any discussion about general case? and what is general case?

mja
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  • The question title seems to presume a fibre bundle is associated to (i) a principal $G$-bundle $P$ and (ii) a fibre $F$. That's not the whole story: You also have to specify an action $\rho$ of the structure group $G$ on the fibre $F$. The expression "$g^{-1}f$" refers to $\rho(g^{-1})f$. That aside, it's a little difficult to make sense of what you're asking. Perhaps what bothers you is the implicit freedom (or rather, the necessity) to choose the action $\rho$...? – Andrew D. Hwang Jul 09 '17 at 17:44
  • i know that.suppose we fix a principal g bundle and fiber F and an action.my question is about first equivalence relation.such definition is somehow unnatural and non-necessary for bundle construction.is it possible that define equivalence relation ~ differently and reach the same bundle structure? – mja Jul 10 '17 at 00:52
  • I think, even if you find another way of gluing the patches together you will end up the same thing. Here I am looking at the theorem 8.2 in Steenrod's Topology of Fiber bundles, it says " two bundles having the same base space, fibre and group are equivalent iff their associated principal bundles are equivalent. We know that this associated principal bundle totally determined by local trivializations and translation maps. What I am trying to say is if you want to define a "associated" bundle from principal one with the same translation maps it looks like this is the way. – muratguner Jul 10 '17 at 03:37
  • thank you and i going to look up the theorem you mentioned.but what you said is the equivalence relation is not important,right?if we choose different equivalence relation,we will end up the different quotient topology.for example consider this:(p,f)~(p.g,f).i guess it give us the trivial bundle.anyway,i will be very thankful, if you could provide me an proof that says bundle structure is free of choosing the equivalence relation. – mja Jul 11 '17 at 10:11

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