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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.)
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  • $\begingroup$ Break this into two separate questions.. I don't know if this would be recieved positively here.. $\endgroup$ – Praphulla Koushik Oct 8 '19 at 8:30
  • $\begingroup$ Pull back the bundle via the map from the interval, so that it becomes a bundle over an interval. Then the curvature, which is a semibasic 2-form, vanishes as there is only one semibasic dimension. The local existence of a lift is, as you say, just elementary ODE. The global existence can be proven many ways; one is to glue local solutions together by gauge group action, as there is a lift with any initial choice of point of the bundle over the curve. $\endgroup$ – Ben McKay Oct 8 '19 at 8:47
  • $\begingroup$ @BenMcKay No but, how do I ensure that the gluing is a finite operation? I have a local lift of the curve along. But suppose that the first point takes me till $\frac\epsilon4$ and then the next adds an $\frac\epsilon8$ and so on. How do i reach the end? I need some sort of finiteness right? $\endgroup$ – Chetan Vuppulury Oct 8 '19 at 9:05
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There is a local lifting near each point of $I$. Any two glue together, after some gauge group action. So there is no maximal interval over which a lift exists, unless that interval is $I$. Choose a point $p_0$ of $E$ you want to lift some $x_0 \in I$ to. The intervals over which lifts taking $x_0$ to $p_0$ exist are closed under union, by uniqueness of the lift. That union is a maximal interval of $I$ on which a lift exists, so equals $I$.

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  • $\begingroup$ What is the situation on general bundles with an Ehresmann connection? $\endgroup$ – Chetan Vuppulury Oct 8 '19 at 9:38
  • $\begingroup$ Every Ehresmann connection induces a connection 1-form, and vice versa, so it is the same situation: en.wikipedia.org/wiki/Ehresmann_connection $\endgroup$ – Ben McKay Oct 8 '19 at 9:48
  • $\begingroup$ No but, then we don.t have the gauge group to glue the different pieces together. $\endgroup$ – Chetan Vuppulury Oct 8 '19 at 9:49
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    $\begingroup$ I was using the term Ehresmann connection differently; following Michor's book, as he points out on p. 209, in general there is no global lift for a fiber bundle connection. There is if all fibers are compact, or if the bundle is principal and the connection is equivariant. If every curve in the base lifts, Michor then calls the connection an Ehresmann connection, so it is a matter of convention rather than a theorem. I would have called any equivariant fiber bundle connection on a principal bundle an Ehresmann connection, as opposed to a Cartan connection. $\endgroup$ – Ben McKay Oct 8 '19 at 9:59

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