# Principal bundles and fibre bundles

Let $$\pi_P:P\rightarrow M$$ a principal $$G$$ (right action) bundle. Let $$F$$ be a manifold with a left action of $$G$$. Then we have the notion of associated fibre bundle over $$M$$ whose fibre is $$F$$. I do not know if there is any connection in the other way which is an inverse for the above construction.

Suppose we start with a $$Gl(n,\mathbb{R})$$ bundle. We have associated vector bundle whose fibre is $$\mathbb{R}^n$$ which is a vector bundle.

Suppose we start with a vector bundle of rank $$n$$, we have the notion of what is called frame bundle which is a principal $$Gl(n,\mathbb{R})$$ bundle.

This says that there is a correspondence between vector bundle of rank $$n$$ and principal $$Gl(n,\mathbb{R})$$ bundle.

Do we have such correspondence in case of fibre bundles? Given a fibre bundle, $$\pi:E\rightarrow M$$, with fibre $$F$$, can I produce a principal $$G$$ bundle with an action of $$G$$ on $$F$$ such that associated fibre bundle is precisely the one I started with?

• You could look at Steenrod, The Topology of Fiber Bundles, for the theory on topological spaces. – Ben McKay Mar 13 at 16:39
• @BenMcKay ok ok. I will see :) – Praphulla Koushik Mar 14 at 2:58

Yes. There is a bundle $$\mathrm{Fr}(E)\to M$$ whose fibre at $$m\in M$$ is the space $$\mathrm{Iso}(F,E_m)$$, where $$E_m$$ is the fibre of $$E$$ over $$m$$. Then $$G=\mathrm{Aut}(F)$$ acts on the right of $$\mathrm{Fr}(E)$$ by composition. That this is a locally trivial bundle follows from the fact $$E$$ is locally trivial. Moreover this action makes $$\mathrm{Fr}(E)$$ a principal $$G$$-bundle. The canonical action of $$G$$ on $$F$$ means we can form the associated bundle $$\mathrm{Fr}(E)\times_G F$$, and this is (IIRC) isomorphic to $$E$$.
I didn't specify what kind of isomorphisms should be used. In the case that $$E$$ is a vector bundle, we take linear isomorphisms, and this is just the usual frame bundle, with $$\mathrm{Aut}(\mathbb{R}^n) = GL(n,\mathbb{R})$$. For an arbitrary continuous fibre bundle, we take homeomorphisms, and then we have $$G=\mathrm{Homeo}(F)$$. For a smooth fibre bundle we take diffeomorphisms (but then note that one lands in infinite-dimensional manifold territory, and one has to be a little bit more careful if the fibre $$F$$ is a non-compact manifold).
• i might be misunderstanding something,,, isn’t it the case that $F$ is same as $E_m$? We are taking automorphisms group of fibre.. that is a Lie group.. corresponding notion is what you are calling as $Fr(E)\rightarrow M$.. is it the case?? Is there any specific reason to write $E_m$ and $F$ both and not just say $\text{Aut}(E_m,E_m)$?? In case of vector bundle, where $E_m$ is an $n$ dimensional real vector space, we have $\text{Aut}(E_m,E_m)=Gl(n,\mathbb{R})$... – Praphulla Koushik Mar 13 at 11:10
• They are isomorphic, but not canonically. This isomorphism arises from the local trivialisations, which can be nontrivial. If $F=E_m$ for all $m$, then $E_m = E_{m'}$ for arbitrary pairs. The whole point of connections and parallel transport is that such canonical identifications do not exist. Automorphisms of the fibre are not a Lie group when you are in the topological category, which I left as a possible case covered by this construction. – David Roberts Mar 13 at 11:34
• That is true, it is not canonical isomorphism... :) :) i understand why you write $\text{ISO}(F,E_m)$.... – Praphulla Koushik Mar 13 at 11:37
• @David Roberts, in the topological context, I think that this works only if $G = Homeo(F)$ is a topological group, i.e. if $F$ is locally compact and locally connexe for example. – ychemama Mar 14 at 9:53