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This is probably a rather basic question, but I can't figure out if this is a triviality (or maybe just rubbish...) ! Say one has a Lie group $G$ acting on some smooth manifold $P$, with a free, transitive and properly discontinous action. So the quotient map $p:M \to M/G$ on the orbit space is a normal covering, which hence defines a principal fibre bundle $M(M/G, G)$ with structure group $G = \pi_1(M/G)/p_*(\pi_1(M))$. Now, we can lift each curve in $M/G$ to a curve $\tilde{\gamma}$ in $M$ through the covering map $p$. My question is now the following: can one see the lifted curve as the horizontal lift of the curve with respect to some connection in the bundle? And if so, what would the corresponding connection $1$-form be?

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You probably didn't mean to say transitive or else $M/G$ is a point. Anyway, for this action to be properly discontinuous $G$ will have to be a discrete group. Hence the vertical subspace of the principal bundle at each point is $0$. Thus there's only one possible connection to choose here (the horizontal subspace is everything; the connection $1$-form is identically $0$). And then you're right that horizontal lifts are just lifts, since any curve is horizontal.

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  • $\begingroup$ Thank you for the answer! But I still can't see something which is probably rather trivial... How can one see that the vertical subspace is trivial provided the group G is discontinuous? I have in mind the example of the associated bundle to a Moebius strip where each fiber contains two points. Or maybe this cannot be seen as a covering map? $\endgroup$ – user374452 Oct 3 '16 at 15:58

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