# Vector bundles and principal $G$-bundles

I am trying to understand the notion of a principal $G$-bundle versus a vector bundle. Here $G$ is a Lie group.

Supposedly, principal $G$-bundles are a generalization of vector bundles. My problem here is that most sources, for example the wikipedia page, talks about bundles over $GL_n(\mathbb{R})$ or some other such matrix group. But the fibers of vector bundles are of the form $\mathbb R^n$ and not $GL_n(\mathbb{R})$. So, how are principal bundles the generalization of vector bundles?

On the other hand, is there some kind of correspondence between vector bundles and $GL_n(\mathbb{R})$-bundles, so that principal bundles are in some indirect way a generalization?

On the other hand, is there some kind of correspondence between vector bundles and $\text{GL}_n(\mathbb{R})$-bundles, so that principal bundles are in some indirect way a generalization?
Yes. Given a principal $G$-bundle and a linear representation $\rho : G \to \text{Aut}(V)$, you get an associated vector bundle whose fibers look like $V$ instead of $G$. This gives you a functor from principal $\text{GL}_n(\mathbb{R})$-bundles to $n$-dimensional vector bundles (taking the standard $n$-dimensional representation) which is an equivalence of categories (the inverse functor is given by taking the frame bundle).
• For the benefit of the OP: Given a principal $G$-bundle and any (say, smooth) $G$ action on a manifold $M$, you get an associated bundle with fiber $M$. Commented Feb 27, 2013 at 19:54
• For a general G is (not necessarily $GL_n(\mathbb R)$) is it a equivalence of categories? What if we take G is a subgroup of $GL_n(\mathbb R)$? Can we give a reference of the proof of the equivalence in case of G=$GL_n(\mathbb R)$? Commented Jul 19, 2014 at 20:05