I wanted to show that for any smooth principal $G$-bundle $E\xrightarrow\pi B$ any smooth curve $\gamma\colon I\to B$ has a unique horizontal lift from a fixed starting point $u_0\in\pi^{-1}\left(\left\{\gamma\left(0\right)\right\}\right)$. There are two proofs I could think of/I could find. (Convention: $I=\left[0,1\right]$) **This is a long question, so even if you have an answer to any subpart, please do comment/answer.** I tried posting this on MathSE, but didn't get any answers. I saw the book by Michor Topics in
Differential Geometry
and he has a proof in 19.6. But I didn't quite understand it.

## Proof 1

Suppose we have an Ehresmann connection $u\mapsto H_u$ on the bundle. Then as we have the map $\mathrm d\pi\vert_u\colon T_uE\to T_{\pi\left(u\right)}B$ we have $T_uE/\mathrm{ker}\,\mathrm d\pi\vert_u\simeq\mathrm{im}\,\mathrm d\pi\vert_u$. But $\mathrm{ker}\,\mathrm d\pi\vert_u=V_u$ (the vertical subspace) and $T_uE=V_u\oplus H_u$ and $\mathrm{im}\,\mathrm d\pi\vert_u=T_{\pi\left(u\right)}B$ ($\mathrm d\pi\vert_u$ is surjective as the bundle is locally trivialisable). So we have $H_u\simeq T_{\pi\left(u\right)}B$.

Now for smooth $\gamma\colon I\to B$, we have the vector field $X_t=\dot\gamma\left(t\right)\equiv_\text{def}\mathrm d\gamma\vert_t\left(1\right)$. Using the above isomorphism, we have a horizontal vector at every point u such that $\pi\left(u\right)\in\mathrm{im}\,\gamma$. Choosing a $u_0\in\pi^{-1}\left(\left\{\gamma\left(0\right)\right\}\right)$ We can then find the integral curve of this vector field starting at $u_0$ and this will give a horizontal lift.

**Questions about this proof**

- How do we guarantee that the integral curve is lift (is it because it has to stay on the submanifold $\pi^{-1}\left(\mathrm{im}\,\gamma\right)$, is this even a submanifold?)?
- Do we have an existence theorem for integral curves of vector fields along submanifolds? Or an existence theorem for integral curves of vector fields defined on closed subsets? Any special case relevant here is fine too.
- Even if we have such an existence theorem, then won't it only give us a solution in some $\left[0,\epsilon\right]$ and not on all of $I$? Is there a theorem that will give me the existence of the curve on all of $I$?
- Now suppose what we constructed gives us a horizontal lift $\gamma^\mathrm{h}\colon I\to E$, then how do we get uniqueness. What I thought is, now that we have a lift, we can apply the next proof to show uniqueness. But this uses the principal $G$-bundle structure. Can we get uniqueness of horizontal lifts on general bundles with connection?

## Proof 2

We have a connection form $\omega\colon TE\to\mathrm{Lie}\, G$ ($\mathrm{Lie}\,G$ is the lie algebra corresponding to the Lie group $G$). Let us assume that we have a lift $\tilde\gamma\colon I\to E$ starting at $u_0$ of $\gamma$. Then any lift can be written as $\tilde\gamma_1\left(t\right)=\tilde\gamma\left(t\right)\cdot g\left(t\right)$ fpr some $g\colon I\to G$ (this follows from freeness and transitivity of the $G$-action on each fibre). So in particular if a horizontal lift $\gamma^\mathrm{h}$ exists, then it too can we written as $\gamma^\mathrm{h}\left(t\right)=\tilde\gamma\left(t\right)\cdot g\left(t\right)$. We have \begin{equation} \omega_{\gamma^\mathrm{h}\left(t\right)}\left(\dot\gamma^\mathrm{h}\left(t\right)\right)=0\Longleftrightarrow\omega_{\gamma^{h}\left(t\right)}\circ\mathrm d\gamma^\mathrm{h}\vert_t=0\Longleftrightarrow\left(\gamma^\mathrm{h}\right)^*\omega=0 \end{equation} Let $a\colon E\times G\to E,a_g\colon E\to E,a_u\colon G\to E$ be defined by $a\left(u,g\right)=a_g\left(u\right)=a_u\left(g\right)=u\cdot g$. Then \begin{equation} \mathrm d\gamma^\mathrm{h}\vert_t=\mathrm da_{\tilde\gamma\left(t\right)}\vert_{g\left(t\right)}\circ\mathrm dg\vert_t+\mathrm da_{g\left(t\right)}\vert_{\tilde\gamma\left(t\right)}\circ\mathrm d\tilde\gamma\vert_t \end{equation} Now, $a_{\tilde\gamma\left(t\right)}=a_{\gamma^\mathrm{h}\left(t\right)\cdot g\left(t\right)^{-1}}=a_{\gamma^\mathrm{h}\left(t\right)}\circ L_{g\left(t\right)^{-1}}$. So, \begin{equation} \omega_{\gamma^\mathrm{h}\left(t\right)}\circ\mathrm d\gamma^\mathrm{h}\vert_t=\omega_{\gamma^\mathrm{h}\left(t\right)}\circ\mathrm da_{\gamma^\mathrm{h}\left(t\right)}\vert_e\circ\mathrm dL_{g\left(t\right)^{-1}}\vert_{g\left(t\right)}\circ\mathrm dg\vert_t+\omega_{\gamma^\mathrm{h}\left(t\right)}\circ\mathrm da_{g\left(t\right)}\vert_{\tilde\gamma\left(t\right)}\circ\mathrm d\tilde\gamma\vert_t\\ =\mathrm dL_{g\left(t\right)^{-1}}\vert_{g\left(t\right)}\circ\mathrm dg\vert_t+\mathrm{Ad}_{g\left(t\right)^{-1}}\circ\omega_{\tilde\gamma\left(t\right)}\circ\mathrm d\tilde\gamma\vert_t\\ =\mathrm dL_{g\left(t\right)^{-1}}\vert_{g\left(t\right)}\circ\left(\mathrm dg\vert_t+\mathrm dR_{g\left(t\right)}\vert_e\circ\left(\tilde\gamma^*\omega\right)_t\right)=0 \end{equation} Since $L_g$ is an isomorphism, its derivative is invertible, so we have \begin{equation} \mathrm dg\vert_t=-\mathrm dR_{g\left(t\right)}\vert_e\circ\left(\tilde\gamma^*\omega\right)_t\Longleftrightarrow \dot g\left(t\right)=-\mathrm dR_{g\left(t\right)}\vert_e\left(\omega_{\tilde\gamma\left(t\right)}\left(\dot{\tilde\gamma}\left(t\right)\right)\right) \end{equation} Since $\tilde\gamma\left(0\right)=\gamma^\mathrm{h}\left(0\right)=u_0$, we have $g\left(0\right)=e$. So we have an ODE for $g$, and we have a unique solution for $t\in\left[0,\epsilon\right]$ for some $\epsilon$. We have uniqueness of lift as any lift could have written as $\tilde\gamma\left(t\right)\cdot g\left(t\right)$.

**Questions about this proof**

- We have the same extendibility problem as above. Given the final ODE for $g\left(t\right)$, how do we know it can be found on all of $I$ and not just till some $\epsilon$. (I know that for matrix groups we can do it using the path-ordered exponential.)
- How do we know that there is some lift? I tried to prove this as follows. First assume that the image of the curve lies in some open $U$ for which we have a trivialisation $\psi_U\colon \pi^{-1}\left(U\right)\xrightarrow\sim U\times G$. Then we have the obvious lift starting at $u_0$ given by $\tilde\gamma\left(t\right)=\psi^{-1}\left(\gamma\left(t\right),g_0\right)$ where $\psi\left(u_0\right)=\left(\gamma\left(0\right),g_0\right)$. Now suppose that the there $\mathrm{im}\,\gamma$ does not lie in a single trivialisation. Let $\left(U_p,\psi_p\colon\pi^{-1}\left(U_p\right)\xrightarrow\sim U_p\times G\right)$ be the local trivialisation of the bundle around $p\in E$. Then we have a cover for $\mathrm{im}\,\gamma$ given by $\left\{U_p\right\}_{p\in\mathrm{im}\,\gamma}$. Since $I$ is compact, and $\gamma$ is continuous, $\mathrm{im}\,\gamma$ is compact. So we have a finite subcover $\left\{U_{p_k}\right\}_{k=1}^n$. Now I tried to inductively create a lift. First take the open set which contains $\gamma\left(0\right)$, let is be $U_{p_{k_1}}$. Take the connected component containing $0$ in $\gamma^{-1}\left(U_{p_{k_1}}\right)$. Let the supremum of this $t_1$. We can lift $\gamma\vert_{\left[0,t_1\right)}$, call it $\tilde\gamma$. Now because $\mathrm{im}\,\gamma$ is connencted, $\gamma\left(t_1\right)$ It will belong to some other open set, let that be $U_{p_{k_2}}$. Repeat the process and lift this new strand with initial point at $\lim\limits_{t\rightarrow t_1}\tilde\gamma\left(t\right)$ (exists by compactness). repeat this process untill the enitre curve is lifted.
**The problem here is that I cannot show that this induction will end.**It will be solved if the curve enters each open set only finitely many times (I think this will follow from some smoothness criterion.)