# The holonomy group is a Lie group

I'm reading about holonomy groups and the book says:

Theorem. Let $P(M,G)$ be a principal fibre bundle whose base manifold $M$ is connected and paracompact. Let $\Phi(u)$ and $\Phi^0(u)$, $u \in P$, be the holonomy group and the restricted holonomy group of a connection $\Gamma$ with reference point $u$. Then $\Phi^0(u)$ is a connected Lie subgroup of $G$, $\Phi^0(u) \lhd \Phi(u)$ and $\Phi(u)/\Phi^0(u)$ is countable.
By virtue of this theorem, $\Phi(u)$ is a Lie subgroup of $G$ whose identity component is $\Phi^0(u)$.

I understand the proof, but my question is: Why is $\Phi(u)$ a Lie group?

• What is the book? – Armando j18eos Nov 16 '16 at 23:44
• Foundations of Differential Geometry Volume I, by Kobayashi and Nomizu – Di Ma Nov 17 '16 at 0:06

## 1 Answer

Each component of $\Phi(u)$ is homeomorphic to the identity component (via a translation), so you can just translate the smooth structure to the whole group. Since there are only countably many components, this is indeed a (second-countable) manifold.