I am reading Parallel transport on principal bundles over stacks. I quote from their paper :

Let $G$ be a Lie group and $M$ a $C^{\infty}$ manifold. Recall that a choice of a connection $1$-form $A\in \Omega^1(P,\mathfrak{g})^G$ on a principal $G$-bundle $P$ over the manifold $M$ and a choice of base point $x\in M$ gives rise to the holonomy map $$\Omega(M,x)\rightarrow \text{Aut (fiber of $P$ at $x$)}\cong G,$$ where $\Omega(M,x)$ is the set of smooth loops at $x$ in $M$.

I do not have a probelm with above set up, except that I do not know how they are identifying $\text{Aut (fiber of $P$ at $x$)}\cong G$. I think along with fixing a point $x\in M$, they are also fixing a point $u\in \pi^{-1}(x)$, then, I know what is the map $\Omega(M,x)\rightarrow G$. If some one can clarify about it, it is good. But that is not the question. Question is from another line in the paper :

For a connected manifold $M$ holonomy map uniquely determines the connection $A$, and infact the bundle $P$ itself.

The reference they gave is

Shoshichi Kobayashi. La connexion des varietes fibrees. I, II. C. R. Acad. Sci. Paris, 238: 318–319, 443–444, 1954.

It is in a language that I can not read.

Can some one give an English reference where this is proved or a sketch of the proof is given or can some one write a sketch of the proof here?

Edit : As pointed in comments, there is an obvious way to identify $\text{Aut (fiber of $P$ at $x$)}\cong G$ if $G$ is abelian. In case $G$ is non abelian, can some one point me to a reference where this identification is explained.

  • 2
    $\begingroup$ The group of G-equivariant automorphism of a G-space with a free and transitive action is G itself, because any such automorphism is determined on any given point, and then given the original point and the point it is sent to, there is a unique element of G taking one to the other. Given any other starting point, it gives rise to the same group element. Moreover, composition of these automorphisms corresponds to group multiplication. $\endgroup$ Aug 18, 2019 at 9:14
  • $\begingroup$ @DavidRoberts That is absolutely correct... I did not realise about the restriction of automorphism group to equivariant automorphism group... So, there was some confusion.. As you said, if a Lie group $G$ acts freely, transitivley on a manifold $M$, the equivariant automorphism group is identified with the Lie group $G$... Here, $G$ acts on $\pi^{-1}(x)$ freely and transitively and the maps $\pi^{-1}(x)\rightarrow \pi^{-1}(x)$ are actually $G$-equivariant.. So, this is clear,, $\endgroup$ Aug 18, 2019 at 9:36
  • 1
    $\begingroup$ If you're happy with uniqueness up to homotopy: By taking classifying spaces, the map $\Omega M \to G$ gives us a map $B(\Omega M) = M \to BG$, which is just the classifying map of the bundle, and so determines the bundle up to homotopy. Note that $B(\Omega M) = M$ requires $M$ connected. $\endgroup$ Aug 18, 2019 at 23:13
  • $\begingroup$ Oh, ok.. I don’t know much about this $B(\Omega(M)=M$.. I can not even convince myself that this is possible.. can you give so:e reference.. Assuming this, I got your point... Holonomy map is a map $\Omega(M,x)\rightarrow G$... See it as $\Omega(M)\rightarrow G$.. Then “take” the classifying space to get the map $M\rightarrow BG$.. then pullback the principal bundle $EG\rightarrow BG$ along this $M\rightarrow BG$ to get a principal bundle $P\rightarrow M$... what you said is mostly clear except that result..can you give some reference... $\endgroup$ Aug 19, 2019 at 3:39
  • $\begingroup$ @DavidRoberts Is it not necessary for $G$ to be an abelian group? I see that the map $\phi:X\rightarrow X$ given by $\phi(x)=xg$ is not a $G$-equivariant map.. $\phi(x.h)=xgh\neq (xg).h=\phi(x).h$ unless $G$ is abelian.. Am I misunderstanding something here? $\endgroup$ Sep 15, 2019 at 15:55

1 Answer 1


See Group of loops, holonomy maps, path bundle and path connection by Jerzy Lewandowski and related question on MO.

  • $\begingroup$ Thank you. I do not have access to that article in my home. Once I go to my institute, I will check that and respond. thank you.. $\endgroup$ Aug 19, 2019 at 3:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.