A **smooth vector bundle** of rank $n$ is usually defined as a smooth map $p: E \longrightarrow B$ together with **a real vector space structure on each fiber** $E\_b := p^{-1}(b)$ such that:


* **(Local Triviality):** There is an open covering $\{U\_i\}\_i$ of $B$ and trivialisations $\{\phi\_i\}\_i$ where
$\phi\_i: p^{-1}(U\_i)\longrightarrow U\_i\times \mathbb{R}^n$ is a diffeomorphism satisfying $p =pr\_1 \circ \phi\_i$ and $\phi\_{ib}: E\_b\longrightarrow \mathbb{R}^n$ is a linear isomorphism for every $b\in B$.


Alternatively, some authors (**eg.** S. Lang, Amiya Mukherjee, Michor P.W) define the notion of a vector bundle atlas.


A vector bundle chart is a diffeomorphism
$\phi: p^{-1}(U)\longrightarrow U\times \mathbb{R}^n$ such that $p = pr\_1\circ \phi$.


Two vector bundle charts $\phi\_i, \phi\_j$ are said to be compatible if there exist a smooth map $\phi\_{ij}:U\_i\cap U\_j\longrightarrow GL(n, \mathbb{R})$ such that $\phi\_j\circ\phi\_i^{-1}(x, v) = (x, \phi\_{ij}(x)\cdot v)$


Finally, a vector bundle atlas is a set $\{(U\_i, \phi\_i)\}\_i$ of compatible vector bundle charts where $\{U\_i\}\_i$ is an open cover of $B$. Two vector bundle atlases are equivalent if their union is again a vector bundle atlas.


Now, one can define a **smooth vector bundle** of rank $n$ as a smooth map $p: E \longrightarrow B$ together with **an equivalent class of vector bundle atlases.**
The two definitions are equivalent since:


* The local triviality axiom provide an (equivalent class of) atlas.
* An atlas provide a real vector space structure on each fiber: the only ones such that $\phi\_{ib}: E\_b\longrightarrow \mathbb{R}^n$ is an linear isomophism for every $i$ and $b\in B$.


A natural question is: can we have two non-equivalent atlases giving the same real vector space structure on the fibers? The answer is no, and that why the two definitions are equivalent.


Now consider a smooth map $p: E \longrightarrow B$ together with a Lie group $G$ acting on a smooth manifold $F$.


We can define the notion of a $G$-bundle atlas (with typical fiber $F$) as before by replacing $\mathbb{R}^n$ with $F$ and $GL(n, \mathbb{R})$ with $G$. An equivalent class of $G$-bundle atlas is a $G$-bundle structure and given a $G$-bundle structure we have a $G$-bundle. Thus a vector bundle is just a $GL(n, \mathbb{R})$-bundle.


Here are my questions:


* Can we, instead of giving a $G$-bundle atlas, give a structure on each fiber as in the case $G = GL(n, \mathbb{R})$?
* Is there a simple example of a data $p: E \longrightarrow B$ together with a Lie group $G$ acting on a smooth manifold $F$ with two non-equivalent $G$-bundle atlas on it? If so, could you explain the differences between the two examples.
* Could you explain why it is important to choose a $G$-bundle atlas; what is the meaning of this structure?


In the case of $G = GL(n, \mathbb{R})$ I am satisfied because it's equivalent with the "vector space structure on each fiber" definition but I am confused with the general case.


Thank you for helping!


Paul.



Vector Bundle Structure
Consider the torus $T^2$, its tangent bundle $p:TT^2\rightarrow T^2$ is an $R^2$-differentiable bundle.You can also see this bundle as an $O(n)$-bundle by defining differentiable metrics on $T^2$, here $G=O(2)$ acts on $R^2$. There exist flat and non flat differentiable metrics on the torus which define non equivalent $O(2)$-reductions of the tangent bundle.


