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Principal Bundle

Find the definition of a fiber bundle here- Definition of Fiber Bundle

I am having difficulty in proving that the natural action of $K$ on $X$ is well-defined:

Let us recall how does K acts on X $(x,g)\in X \times K$ consider the fibration $\varphi_{U}$ over $U$ such that $p(x)\in U$ So $x\in p^{-1}(U)$. Let $(p(x),h)=\varphi_{U}^{-1}(x)$ So,send $(x,g)$ to $\varphi_{U}(p(x),gh)$.

Let $x\in V$ also. Let $(p(x),h_1)=\varphi_{V}^{-1}(x)$.

How do we show that- $\varphi_{U}(p(x),gh)=\varphi_{V}(p(x),gh_1)$?

Also can someone give any hints how to show the orbit space is isomorphic to B.

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up vote 1 down vote accepted

By definition of fiber bundle, there exists $\theta_{U,V}:U\cap V\to K$ such that $\phi_V(b,k)=\phi_U(b,k\cdot\theta_{U,V}(b))$ on $p^{-1}(U\cap V)$. Evaluating this equality on $(p(x),h_1)$, we get: $$\phi_U(p(x),h)=x=\phi_V(p(x),h_1)=\phi_U(p(x),h_1\cdot\theta(p(x)))$$

This means that $\theta_{U,V}(p(x))=h_1^{-1}\cdot h$. In particular,


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Can you please give some hints on how yo show the orbit space is isomorphic to the base space B – Babai Jul 17 '14 at 20:52
Uh, sorry, I forgot it. Well, you have a $K$ invariant surjective map $p:X\to B$, i.e. if $g\in K$ and $x\in X$, $p(gx)=p(x)$. This induces a surjective map $\psi:X/K\to B$. We only need to show that $\psi$ is injective. If we take $x,y\in X$ such that $\psi([x])=\psi([y])$, since $\psi([x])=p(x)$ we have $p(x)=p(y)$. But the action of $K$ on the fibers is transitive, hence there exists $g$ such that $gx=y$, and this implies $[x]=[y]$. – Giulio Bresciani Jul 17 '14 at 20:59
Thank you so much! – Babai Jul 17 '14 at 21:15

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