# Difference between various type of bundles having a group as fibre

I am trying to understand the difference between these three objects:

1- a fiber bundle in which the fiber is a group $G$

2- a fiber bundle in which the fiber is a group $G$ and the structure group is also $G$

3- a principal $G$-bundle

Denote by $\{U_i,\phi_i \}$ a local trivialization of the bundle and consider the transition function $\phi_i\circ\phi^{-1}_j:U_i\cap U_j\times G\rightarrow U_i\cap U_j\times G$, $\phi_i\circ\phi^{-1}_j(x,g)=(x,t_{ij}(x,g))$, with $t_{ij}(x,\cdot):G\rightarrow G$ a diffeomorphism. As far as I have understood, the difference between 1 and 2 is that in 1 the transition functions need not to be a left $G$-action, while in 2. they need to, i.e.~I can write $t_{ij}(x,g)=\rho(\tau_{ij}(x),g)$, with $\tau_{ij}:U_i\cap U_j\rightarrow G$ and $\rho: G\times G \rightarrow G$ a left action of $G$ on itself. The difference between 2 and 3 is that while in 2 any action will do, in 3 the action must be left multiplication. As a consequence only in 3 it is possible to globally define a right action of $G$ on the total space of the bundle. Is all of this correct?

To better understand the theory I am trying to think of some examples. Let us consider bundles over $S^1$ with fiber $S^1$ (which can be thought as the group $U(1)$). Cover $S^1$ with two open sets $U_1=\{\exp(ix):-\epsilon<x<\pi+\epsilon\}$, $U_2=\{\exp(ix):\pi-\epsilon<x<2\pi+\epsilon\}$. The overlap consist of two open sets. Let us take the identity as transition functions over one of them and consider transition functions over the other, $t:V\rightarrow G$ with $V=\{\exp(ix):-\epsilon<x<\epsilon\}$.

As an example of 1 I thought of taking $t(x,g)=g^{-1}$, which is not an action of the fiber on itself (basically it is mapping each point on the fibre to its conjugate). By the way is this the Klein bottle?

An example of 2 but not 3 would be to take $t(x,g)=\rho(\tau(x),g)$ with $\tau(x)=\exp(ix)$ and $\rho(\exp(ix),g)=\exp(2ix)g$. (I am not sure about this one because defining instead $\tau(x)=\exp(2ix)$, $\rho$ left translation the map $t$ that one obtains is identical) In the non-Abelian case another example would be to take $\tau(x)=h\in G$ and $\rho(h,g)=hgh^{-1}$.

Finally an example of 3 would be given by $\tau(x)=\exp(ix)$ and $\rho$ left multiplication.

Again, is what I wrote correct?

Thanks for the help and sorry for the long question!

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Your question is really helpful to understand things in a better way. Thanks a ton @GFR. –  Babai Jul 19 '14 at 19:59
I am glad to hear that! –  GFR Jul 21 '14 at 10:50