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Let's take the following definitions.

Definition 1. An Ehresmann connection is a smooth assignment of the complements of the vertical subspaces of the tangent spaces of the total space of the bundle. These complement subspaces are called vertical subspaces.

Definition 2. The horizontal lift of a differentiable curve $C$ in the base space is a curve $D$ in the total space having the following properties

  • $\pi(D)=C$
  • The tangent vectors of $D$ are horizontal at every point.

Definition 3. An Ehresmann connection on a smooth fiber bundle is curvature free if any horizontal lift of any closed differentiable curve in the base space is also closed. The connections which are not curvature free arre called curved.

Let $M$ be a simply connected 1-dimensional smooth manifold, and $\pi:E\to M$ a smooth fiber bundle.

In the special case when $E=\mathbb R^2$ , $M=\mathbb R$, and $\pi:(x,y)\mapsto (x,0)$, it is fairly obvious, that the horizontal lifts of every closed curve in $M$ are closed. But is this true also, when the fibers are not simply connected, or have more than 1-dimensions?

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  • $\begingroup$ Can you clarify your question? As written it seems different to the title. Is it the base or the fiber which is meant to be (or not be?) simply connected/1-dimensional? $\endgroup$ – Anthony Carapetis Nov 27 '17 at 9:47
  • $\begingroup$ @AnthonyCarapetisThe base space is always simply connected (as the title shows) and the fiber can be anything. One possibility that it is also simply connected and 1-dimensional (this is the trivial case). $\endgroup$ – mma Nov 27 '17 at 9:51
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The topology of the fiber plays essentially no role here - the fact that the base is $\mathbb R$ immediately rules out anything interesting happening.

First, the fact that the base is one-dimensional tells us that the connection is flat, meaning that the lift of any null-homotopic closed curve is closed. The intuition here is pretty simple: such a curve in the base has to backtrack over itself perfectly in order to close, and thus the "vertical motion" of the lift is perfectly reversed. If you know some standard theorems, the neat proof is probably to note that the curvature form $R(X,Y) = [X_H, Y_H]_V$ must vanish since the horiozontal bundle is one-dimensional.

Secondly, the fact the base is simply-connected (in addition to being flat) tells us that the connection has zero holonomy, and thus is "curvature-free" in your parlance. This is super easy: simply-connected means exactly that every curve is null-homotopic.

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  • $\begingroup$ "and thus the "vertical motion" of the lift is perfectly reversed" - I don't see the reason of this. $\endgroup$ – mma Nov 28 '17 at 7:42
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    $\begingroup$ @mma: in a local trivialization, performing parallel transport/lifting a curve amounts to solving a first-order linear ODE; so if you transport back along the same curve then you must recover the initial datum due to the uniqueness of solutions to ODEs. $\endgroup$ – Anthony Carapetis Nov 28 '17 at 8:20

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