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.

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    $\begingroup$ 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. $\endgroup$
    – David Roberts
    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
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    $\begingroup$ 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

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