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?

  • $\begingroup$ You could look at Steenrod, The Topology of Fiber Bundles, for the theory on topological spaces. $\endgroup$
    – Ben McKay
    Mar 13, 2019 at 16:39
  • $\begingroup$ @BenMcKay ok ok. I will see :) $\endgroup$ Mar 14, 2019 at 2:58

1 Answer 1


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).

  • $\begingroup$ 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 $\text{Aut}(E_m,E_m)=Gl(n,\mathbb{R})$... $\endgroup$ Mar 13, 2019 at 11:10
  • $\begingroup$ 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. $\endgroup$
    – David Roberts
    Mar 13, 2019 at 11:34
  • $\begingroup$ That is true, it is not canonical isomorphism... :) :) i understand why you write $\text{ISO}(F,E_m)$.... $\endgroup$ Mar 13, 2019 at 11:37
  • $\begingroup$ @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. $\endgroup$
    – ychemama
    Mar 14, 2019 at 9:53
  • 1
    $\begingroup$ @Rei the topological group of self-diffeomorphisms of a non-compact manifold is not very nice, compared with the infinite-dimensional (I think Fréchet) Lie group in the compact case. I can't recall offhand if there is a manifold structure, but if there is it might be something rather bad eg not complete, possibly only locally convex at best... $\endgroup$
    – David Roberts
    Feb 20 at 20:16

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