Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The definition of fiber bundle can be found from here: Definition of Fiber Bundle

Then Bredon defines Principal bundle in the exercise as follows:

enter image description here

I am not able to show how K acts naturally on X and the rest of the exercise.

My try was-

We need to get a map $X \times K \rightarrow X$. Let $(k,x) \in K\times X$.

Then $p(x)\in B$ and by the trivial fibration condition there exists $U \ni p(x)$ such that there is homeomorphism $\varphi :U\times F \rightarrow p^{-1}(U)$ .

Send $(k,x)$ to $\varphi(p(x),k)$.

Note: By definition of Principal Bundle F and K are same.

Am I correct?

How to prove the next part of the exercise?

share|cite|improve this question

1 Answer 1

I dont think you have defined an action yet. If we take $g \in K$ then we use the map $\varphi$ to define the action where if $(u,h) \in U\times K$ then $$g(u,h)=(u,gh)$$ now you need to use the fact that $K$ is the structural group to show this this is well defined.

The orbit space is $B$ since $K$ acts transitively.

If you have a section, set the values equal to the group identity element and use this to define a homeomorphism with the trivial bundle.

Both these last statement need to have the details supplied.

share|cite|improve this answer
Given an element $(x,k)\in X\times K$ we have to define an element of $X$. How does another element $g\in K$ come into picture? –  Babai Jul 16 '14 at 15:59
I didnt understand your notation, you are still not defining an action. How should $K$ act n a fibre ? It should act by group multiplication. In your definition $k$ send the entire fibre to the element $k$, and this is not invariant under change of coordinates. –  Rene Schipperus Jul 16 '14 at 16:13
K acts on F by right translation, yes that is correct. But then what is the action of K on X ? And K and F are same by definition. –  Babai Jul 16 '14 at 16:20
I changed my notation to avoid confusion. –  Rene Schipperus Jul 16 '14 at 16:20
I guess you want to say this- for $(x,g)\in X\times K$ consider the fibration $\varphi$ over $U$ such that $p(x) \in U$ So $x\in p^{-1}(U)$. Let $(u,h)=\varphi ^{-1}(x)$ So,send $(x,g)$ to $\varphi (x,gh)$. Am I correct? –  Babai Jul 16 '14 at 16:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.