Let $M$ be a manifold. Let $G$ be a Lie group and $\mathfrak{g}$ be its Lie algebra.

Let $P(M,G)$ be a principal bundle. Recall that, a connection on $P(M,G)$ is a distribution $\mathcal{H}\subseteq TP$ satisfying certain conditions. Equivalently, a $\mathfrak{g}$-valued differential forms on $P$ satisfying certain conditions.

Question : Is there a characterization of connections on $P(M,G)$; in the sense, a one-one correspondence between the set of connections on $P(M,G)$ and some "well-described set"?

Question : Can we say something if we restrict to the case of integrable connections?

Questions : Can we say something when the bundle is trivial bundle? A characterization for integrable connections on trivial principal bundle?

Edit : May be it is already clear but, I want to specify that I am seeing two connections $\omega, \omega’$ to be same if there exists an isomorphism $\varphi :P\rightarrow P’$ Of vector bundles such that $\varphi^*\omega’=\omega$, that is the pullback of $\omega’$ is equal to $\omega$.

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    $\begingroup$ Connections are affine over $\Omega^1(M;\mathfrak g)$, and in particular form an affine space. Integrable connections are the same thing as flat conns, which are the zeroes of the curvature map $F_A$. Such a map gives rise to a holonomy map $\pi_1 M \to G$. The space of all flat connections on $P$ carries the action of the automorphism group of $P$, and the space of homomorphisms to $G$, the action of conjugation. After quotienting this descends to a bijection from the space of flat conns to the subset of $\text{Hom}(\pi_1 M, G)$ so that $\tilde M \times_{\pi_1 M} G \cong P$ as G-bundles. $\endgroup$
    – mme
    May 3, 2020 at 17:53
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    $\begingroup$ If your question is how to tell what the topological type of the bundle $\tilde M \times_{\pi_1 M} G$ is given the homomorphism $\rho: \pi_1 M \to G$, you will probably first want to know how to determine the isomorphism classes of principal G-bundles over M, and then to determine it by hand for your representation $\rho$... in particular it's not hard if $G = U(1)$! $\endgroup$
    – mme
    May 3, 2020 at 17:56
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    $\begingroup$ Atleast for trivial Principal $G$ bundles a partial answer to your question is mentioned in the Theorem 1 in arxiv.org/pdf/1003.4485.pdf. Also for non trivial Principal $G$ bundles a partial answer can be found in the section "Idea" in ncatlab.org/nlab/show/connection+on+a+bundle where you may find a one-one correspondence(not sure about 1 direction) between connections on Principal $G$ bundles over $M$ and an appropriate subset of functors from Path groupoid of $M$ to the Atiyah Lie Groupoid of the principal $G$ bundle. $\endgroup$ May 6, 2020 at 21:17
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    $\begingroup$ @AdittyaChaudhuri +1 for Theorem $1$ that gives a one-one correspondence between $\mathfrak{g}$-valued $1$-forms on $M$ and the set of connections on the trivial principal bundle $M\times G\rightarrow M$... I do not understand the other half of the theorem that says there is a one one correspondence between the above mentioned set and the set of "smooth" functors $\mathcal{P}_1(M)\rightarrow G$.. I never understood what is smooth structure on $\mathcal{P}_1(M)$.. Do you want to shre your thoughts on smoothness of these functors? $\endgroup$ May 7, 2020 at 3:51
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    $\begingroup$ :) See if you can make it as an answer by adding as many details as you can.. $\endgroup$ May 7, 2020 at 5:04

1 Answer 1


This is not an answer. This is in response to Mike Miller's comment.

Let $M$ be a manifold, $\tilde{M}$ to be its associated universal cover (a simply connected covering space over $M$). I do not understand why they are not assuming $M$ to be connected. I am assuming $M$ is a connected manifold.

The following result is from the book Differential Geometry - Bundles, Connections, Metrics and Curvature by Clifford Henry Taubes.

Theorem $13.2$ (classification theorem for flat connections) says that, there is a bijection between the sets $\mathcal{F}_{M,G}$ and the set $\text{Hom}(\pi_1(M),G)/G$ where,

  1. $\mathcal{F}_{M,G}$ denote the set of equivalence classes of pairs $(P,A)$, where $P\rightarrow M$ is a principal $G$ bundle, and $A$ is a flat connection on $P(M,G)$. Two pairs $(P,A)$ and $(P',A')$ are equivalent, if there is an isomorphism of principal $G$-bundles $(\varphi,1_M):(P,M)\rightarrow (P',M)$ such that $\varphi^*A'=A$ (pullback of connection $A'$ is the connection $A$).
  2. $\text{Hom}(\pi_1(M),G)/G$ denote the set of equivalence classes of group homomorphisms $\pi_1(M)\rightarrow G$. Two morphisms $f_1:\pi_1(M)\rightarrow G$ and $f_2:\pi_1(M)\rightarrow G$ are equivalent if there exists $g\in G$ such that $f_1=gf_2g^{-1}:\pi_1(M)\rightarrow G$.

The bijection $\mathcal{F}_{M,G}\rightarrow \text{Hom}(\pi_1(M),G)/G$ is given as follows:

  • given a principal bundle $P(M,G)$ with flat connection $A$, we get a group homomorphism $\pi(M)\rightarrow G$. Its equivalence class gives an element in $\text{Hom}(\pi_1(M),G)/G$.
  • Let $\rho:\pi_1(M)\rightarrow G$ be a representative of an element in $\text{Hom}(\pi_1(M),G)/G$. Consider the trivial principal $G$-bundle $\tilde{M}\times G\rightarrow \tilde{M}$. The map $\rho:\pi_1(M)\rightarrow G$ given an action of $\pi_1(M)$ on $G$, which in turn gives an action of $\pi_1(M)$ on $\tilde{M}\times G$. Thus, trivial principal bundle $\tilde{M}\times G\rightarrow \tilde{M}$ induce $(\tilde{M}\times G)/\pi_1(M)\rightarrow M$. Thus, we get a principal $G$-bundle over $M$, which we denote by $\tilde{M}\times_{\rho}G\rightarrow M$. It turns out that, there exists a $\mathfrak{g}$-valued $1$-form on $\tilde{M}\times_{\rho}G\rightarrow M$ whose pullback to $\tilde{M}\times G\rightarrow \tilde{M}$ is the canonical connection on the trivial bundle. It turns out that this $\mathfrak{g}$-valued $1$-form on $\tilde{M}\times_{\rho}G\rightarrow M$ is a flat connection on the principal bundle $\tilde{M}\times_{\rho}G\rightarrow M$. Thus, we get a principal bundle $(P_\rho,A_{\rho})$. Take its equivalence class to get an element in $\mathcal{F}_{M,G}$.

It is not clear how does this answer the question:

Given a principal bundle $P\rightarrow M$, how does one know for what $\rho:\pi_1(M)\rightarrow G$, do we get that that there is an isomorphism of principal bundle $P\cong \tilde{M}\times_{\rho}G$?

I am also interested in only "different" connections in the sense if two connections on $(P,M)$ are related by an isomorphism, in the sense of pullbacks, then I am calling these to be same.

  • $\begingroup$ If anyone wants to see more details, who do not have the book, I can give more details... $\endgroup$ May 8, 2020 at 19:06
  • $\begingroup$ It sounds like the missing piece for you is this image. As I said in my comment: $\rho$ is in the image of $\mathcal F_P$ iff the $G$-bundle $(\widetilde M \times_\rho G))$ is isomorphic to $P$ as a G-bundle, where the subscript $\rho$ indicates we identify $(x, g) \sim (\gamma x, \rho(\gamma) g)$ for all $\gamma, x, g$, thinking of $\gamma \in \pi_1 M$ as a deck transformation. This is as explicit as you will ever get in full generality. If you have a specific bundle over a specific manifold (or your group G is not complicated) you may be able to get better answers by ad hoc techniques. $\endgroup$
    – mme
    May 8, 2020 at 19:18
  • $\begingroup$ @MikeMiller I have clarified that part.. Do you have a procedure to write this down explicitly if I have a principal bundle over a manifold when the group G is reasonably good.. Any reference is welcome.. I enjoyed reading chapter 13 of Taubes book.. $\endgroup$ May 9, 2020 at 5:43

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