It is well known that a (real) vector bundle $\pi : E\to B$ over a topological space (or manifold) $B$ is a fibre bundle whose fibres
$$F=\pi^{-1}(x), \ \ \ x\in B $$
over any $x\in B$, are diffeomorphic to a vector space $V$. On the other hand, a principal $G$-bundle is a fibre bundle $\pi : P\to B$ over $B$ with a right free action of a Lie group $G$ on $P$ such that for any open set $U\subset B$, the locally trivial fibrations defined by:
$$
\Phi_{U} : \pi^{-1}(U)\to U\times G, \ \ \Phi_{U}(p)=(\pi(p), \varphi_{U}(p)).
$$
Here
$$
\ \varphi_{U} :\pi^{-1}(U)\to G
$$
is a $G$-equivariant map, that is $\varphi_{U}(pg)=\varphi_{U}(p)g$, for all $p\in\pi^{-1}(U)$ and $g\in G$. In the last case the fibers are submanifolds of $P$ which are always diffeomorphic with the structure group $G$.
Although for any vector bundle $\pi : E\to B$, its fibers $F\cong V$ can be considered as Lie group with operation the vector addition, in general we do not include the vector bundles
as examples of principal $G$-bundles (Although, to every vector bundle we can associate the frame bundle which is a ${\rm GL}_{n}\mathbb{R}$-principal bundle, but I don't speak here about associated bundles).
My question is about a good explanation about the fact that IN GENERAL vector bundles (themselves) do not give examples of principal $G$-bundles. For example, the tangent bundle $TM$ of a smooth manifold is a prototype example of a vector bundle, but itself it cannot be considered as a principal bundle for a Lie group $G$, is this true?
Thus I ask:
Which is the basic difference between a vector bundle an a $G$-bundle, which does not allows us (almost always??), to consider the vector bundles themselves as examples of $G$-bundles?
For example, a $G$-bundle is trivial (isomorphic to the product bundle), if and only if it admits a global section, but I think that this is not true for vector bundles.
A second question is about examples of vector bundles which can be considered the same time as $G$-bundles for some Lie group (I think that such an example is a cylinder)
Best Answer
The difference is that, for a vector bundle, there is usually no natural Lie group action on the total space that acts transitively on the fibers. The fact that all of the fibers are, individually Lie groups, doesn't mean that there is a Lie group that acts on the whole space, restricting to each fiber to be a simply transitive action. The simplest example of this is the nontrivial line bundle over the circle. Another example is the tangent bundle of $S^2$.