Can bundles be trivial if the structure group is not? I am confused about a basic concept and any help to clear up my misunderstanding is appreciated:
In https://arxiv.org/abs/hep-th/0611201, for example, it is stated that a fibre bundle is trivial if and only if all transition functions can be chosen to be identity maps. If this is not the case, the fibres are ‘twisted’  in a way that the bundle  cannot be simply described by a product manifold. This non-trivial aspect is encoded in a non-trivial structure group S of the bundle. As a first example often the struture group of the  Mobius strip versus a cylinder  is discussed. So far so good!
My confusion arises when considering a principle bundle $ P= P(M,G) $ over some base manifold $M$ where the fibre $G$ is the structured group $S$ itself, e.g. $G \simeq S $. This is because I can choose $M \simeq \mathbb{R^4}$ which makes the fibre bundle trivial no matter how "complicated" one chooses $G$ (because  $\mathbb{R^4}$ can be contracted to a point). This seems like a contradiction to me.
Furthermore image I start out with  $ P= P(M,G) $ and systematically "test out" different base manifolds $M$ while keeping $G$ fixed. Now as soon as the topology of $M$ changes ( say $M$ cannot be contracted to a point unlike before) one needs to expects the structure group of the bundle to change as well; but it was fixed to be $G$ in the first place? How can this be?
One way out I see if one replaces $G \simeq S $ by $G \supseteq S$ requiring that the structured group only needs to be a subset of $G$. So in the trivial case from before $ P = P(\mathbb{R^4},G)$ the structured group would only pick out the trivial sub-group $S=\{e\}$ from G. However even in this case, in my second example of "testing out" different base-manifolds for a given $G$, which theorem guarantees me that one always stays within $G \supseteq S$ for an arbitrary choice of $M$?
many thanks!
 A: The problem is that "the structure group of a bundle" is more a matter of choice than an intrinsic object. As you observe that structure group of a principal $G$-bundle always is $G$. However, there is the concept of reduction of structure group, and triviality of a bundle is related to the question of the smallest structure group that the bundle can be reduced to. For example, the structure group of a (real) vector bundle of rank $n$ initially is $GL(n,\mathbb R)$. The fact that any bundle admits a bundle metric is equivalent to the fact that the structure group can always be reduced to $O(n)\subset GL(n,\mathbb R)$. Further reductions depend on additional structures on the bundle which may not exist in general. Finally, there exists (almost by definition) a reduction of structure group to the trivial group $\{e\}$ if and only if the bundle is trivial (and such a reduction corresponds to a choice of trivialization). 
In the language of principal $G$-bundles, the question for a given principal $G$-bundle $P$ takes the following form: What is the smallest subgroup $H\subset G$ such that there is a principal $H$-bundle $\tilde P$ for which $P$ can be realized as the space of orbits of the $H$-action on $\tilde P\times G$ defined by $(u,g)\cdot h=(u\cdot h,h^{-1}g)$. For the trivial group this means that $\tilde P$ has to be the base manifold $M$ and $P\cong \tilde P\times G$. 
