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Jan 7, 2022 at 12:59 comment added Benoît Kloeckner Contractibility is not sufficient, the exponential map is not one-to-one on $\mathbb{R}^n$ whenever $n\ge 2$. Simply consider the vector field of rotations aruond a codimension-$2$ subspace: it generates a group isomorphic to $\mathbb{S}^1$.
Jan 7, 2022 at 10:51 comment added ABIM Yes this follows from Thursten's theorem
Jan 7, 2022 at 10:29 comment added Saúl RM The map is never surjective, see this question
Jan 7, 2022 at 10:04 comment added Saúl RM @ThiKu but the time derivative of the flow by $C^k$-diffeomorphisms will probably depend on time
Jan 7, 2022 at 9:57 comment added ThiKu The image of Exp should consists of the path-connected component of the identity in the diffeomorphism group: on the one hand the vector field X gives you a flow connecting the identity to Exp(X), on the other hand (up to checking details) a path from Exp(X) to the identity should be realised by a $C^k$-path in the diffeomorphism group (one has to check that this is still true in this infinite-dimensional manifold), which should correspond to a flow by $C^k$-diffeomorphisms, whose time derivative should give a $C^k$-vector field.
Jan 7, 2022 at 9:37 comment added ABIM Yes, I imagined it only has to do with $M$'s topological properties. I expected if $M$ is contractable then it should be okay, but I'm not certain if this is true nor if its too strong of a condition..
Jan 7, 2022 at 9:36 comment added Ben McKay The Riemannian metric is surely irrelevant, as it has no role in the definition of compact sets, vector fields, or the exponential map of a vector field. Every manifold admits a complete Riemannian metric, so this is not a special property of the manifold.
Jan 7, 2022 at 9:23 history asked ABIM CC BY-SA 4.0