# Path lifting property induced by local triviality of a principle G bundle.

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.

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 $$t_1 so that $$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.

• 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? Commented Sep 20, 2019 at 13:24
• 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. Commented Sep 20, 2019 at 14:09
• I could not get how your argument in the comment solve "my doubt" in comment. Can you please elaborate? Commented Sep 20, 2019 at 14:21
• If $t<t'$ and we define $s_2(t):=s_1(t)a_t$, then the second condition holds automatically, namely, $a'_{t'}=1_G$. Commented Sep 20, 2019 at 14:30
• 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 Commented Sep 20, 2019 at 15:09