# Holonomy bundle is a covering space

I have trouble understanding following content, this is from Foundations of Differential geometry by Kobayashi and Nomizu in section on Flat connections.

Let $\Gamma$ be a flat connection in principal bundle $P(M,G)$, where $M$ is connected and paracompact. Let $u\in P$ and $M^*=P(u)$, the holonomy bundle through $u$; $M^*$ is a principal bundle over $M$ whose structure group is the holonomy group $\Phi(u)$. Since $\Phi(u)$ is discrete by Ambrose Singer theorem and since $M^*$ is connected, $M^*$ is a covering space of $M$.

As $\Gamma$ is a flat connection, it’s curvature is zero. So, Lie algebra generated by $\Omega(p) (v,w)$ where $v,w$ are horizontal vectors at $p\in P(u)$ is zero Lie algebra. This Lie algebra, by Ambrose Singer theorem, is same as the Lie algebra of the Lie group $\Phi(u)$. I do not understand why this would mean Lie group is discrete group. I do not have strong background of Lie groups so could not understand. I think

Only Lie groups whose Lie algebras are zero Lie algebras are Discrete Lie groups.

I would like to know if there is any such result. I tried to google Lie group with trivial Lie algebra and ended up with no positive result.

Assuming the Lie group is discrete, I am almost sure that this map $M^*\rightarrow M$ is a covering map. I do not really need that this is a principal bundle. I think any fiber bundle whose fiber is discrete space can be seen as a covering space.

What I would like to understand is, Is this the reason why $M^*$ is a covering space of $M$ or is there any other reason. Any reference related to Lie group being discrete is most welcome.

Let $N$ be a normal subgroup of $\Phi(u)$ and set $M’= M^*/N$ ( this is the quotient space of $M^*$ over $N$). Then $M’$ is a principal bundle over $M$ with structure group $\Phi(u)/N$ . In particular $M’$ is a covering space of $M$.

I fail to see why $M’$ is a principal fiber bundle. Do we have a more general result that says

structure group of a principal bundle $P(M,G)$ with discrete structure group $G$ can be reduced to $G/N$ for any (normal) subgroup $N$ of $G$?

Assuming $M’$ is a principal bundle over $M$ with structure group $\Phi(u)/N$ then I think for the same reason as above $M’$ is a covering space of $M$. I did not check but I am more or less sure that quotient group of discrete group is a discrete group.

Any reference for these questions is most welcome.

## 1 Answer

Let $G$ be a Lie group with Lie algebra ${\cal G}$, the connected component of the identity of $G$ is generated by $exp(X), X\in {\cal G}$, so if ${\cal G}=0$, $G_0=\{Id\}$ and $G$ is discrete.

https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Properties

• So, you are saying if connected component consists of just identity element, then the group is discrete. – user312648 Jan 13 '18 at 15:46
• yes, if the connected component of the identity is a singleton, the group is discrete. – Tsemo Aristide Jan 13 '18 at 15:58
• Can you please give some reference for the same. – user312648 Jan 13 '18 at 16:48