In Kobayashi and Nomizu(Foundations of differential geometry Volume 1) (pg:69) Proposition 3.1 (existence and uniqueness of horizontal lift of path) it is mentioned that given a pt $u$ in $P$ and a path $x_t$ in $M$ the local triviality of the principle $G$- bundle $\pi:P \rightarrow M$ is sufficient to produce at least one lift $v_t$( not necessarily horizontal) in $P$ of the path $x_t$ lying in the base manifold $M$ such that $v_t$ begins at $u$.

I approached in the following way:

Let ($\phi_i,U_i)$ $i\in I$ (where $I$ is an index set) are local trivializations. So we have corresponding local sections $\lbrace\ s_i\rbrace $ . Now define $v_t$= $s_io x_t$ if $x_t \in U_i$. But if $x_t \in U_i\cap U_j$ then we have another choice of local section $s_j$ .

But then how will it be well defined?

I know that $s_i(x)= \phi_i o \phi_j^{-1}o s_j(x)$ when $x \in U_i\cap U_j$

I am feeling somehow I have to use the above compatibility condition . But not able to proceed much.

Also I have to make sure that the lifted path $v_t$ should be smooth.

Thanks in advance.


1 Answer 1


Let $\psi_i$ be a local trivialization of a principle bundle $P$ on the open set $U_i$ of base manifold $M$, namely,

$$\psi_i:\pi^{-1}(U_i)\to U_i \times G;u\ \mapsto (\pi(u), \phi_i(u)).$$

We arbitrarily take a local trivialization $\{s_i\}$ of $P$ with respect to the open covering $\{U_i\}$ of the base manifold $M$.

In order to get the compatible lift on each $U_i$, we consider the following: First, let $U_0$ contains the initial point of the curve $x_t$ of $M$ , that is $x_0 \in U_0$. Then we can construct the lift of $x_t$ on the each open cover $U_i$ by $v_i(t):=s_i(x_t)$.

Take a sequence of point $$t_1 <t_2<,...,<t_n$$ so that $x_{t_i} \in U_i \cap U_{i+1}$.

If $v_1(t_1)=s_1(x_{t_1})\neq v_2(t_1)=s_2(x_{t_1})$, then by the action of $G$, we can take a base $s_2'$ so that $s_1(x_{t_1}) = s_2'(x_{t_1})$, and using this procedure successively, we can obtain a lift of the path in the base manifold so that it is compatible in each open covers.

Since each local trivialization $s_i$ of $P$ and action of $G$ to $P$ are smooth, the lifted path is also smooth.

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  • $\begingroup$ Now if $U_2$ contains more than 1 points of the curve i.e let $x_t$ and $x_t{'}$ $\in U_2$ then according to your solution $s_2(t)$= $s_1(t).a_t$ and $s_2(t')$=$s_1(t{'})$.$a_t{'}$ for some $a_t$ and $a_t{'}$ $\in G$. Then how are you ensuring that $s_2(t)$ is smooth? $\endgroup$ Commented Sep 20, 2019 at 13:24
  • $\begingroup$ Since the curve is continuous, $U_2$ always contains one more points. So, what we need is to select a discrete points of curve $x_t$ such that each point is chosen from a one intersection of two open covers as an above figure. $\endgroup$ Commented Sep 20, 2019 at 14:09
  • $\begingroup$ I could not get how your argument in the comment solve "my doubt" in comment. Can you please elaborate? $\endgroup$ Commented Sep 20, 2019 at 14:21
  • $\begingroup$ If $t<t'$ and we define $s_2(t):=s_1(t)a_t$, then the second condition holds automatically, namely, $a'_{t'}=1_G$. $\endgroup$ Commented Sep 20, 2019 at 14:30
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    $\begingroup$ Can you please write down what you are assuming clearly. It looks like I understand but I have same question as the one asked by user above... Please consider editing your answer to include all these explanation $\endgroup$ Commented Sep 20, 2019 at 15:09

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