6

Let $M$ be a connected and complete Riemannian manifold of positive dimension, $k$ be a positive integer, and let $\mathfrak{X}^k_c$ be the set of class $C^k$-vector fields on $M$ of compact support. It is well known that $Diff_c^k(M)$, the set of class $C^k$ compactly supported homeomorphisms on $M$ is an infinite-dimensional Lie group with Lie algebra $\mathfrak{X}_c^k$; with exponential map $$ Exp:\mathfrak{X}_c^k\ni V \mapsto \operatorname{Solution}\left[\frac{d}{dt} \phi_t(x) = V(\phi_t(p)),\, \phi_0=1_M\right] \in Diff_c^k(M) $$ and that this map is continuous but not $1-1$ nor onto. My question is, are the conditions on $M$'s geometry which ensure that $M$ is $1-1$?

ABIM
  • 5,039
  • 2
    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. – Ben McKay Jan 07 '22 at 09:36
  • 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.. – ABIM Jan 07 '22 at 09:37
  • 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.

     – ThiKu Jan 07 '22 at 09:57
  • @ThiKu but the time derivative of the flow by $C^k$-diffeomorphisms will probably depend on time – Saúl RM Jan 07 '22 at 10:04
  • 3
    The map is never surjective, see this question – Saúl RM Jan 07 '22 at 10:29
  • Yes this follows from Thursten's theorem – ABIM Jan 07 '22 at 10:51
  • 1
    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 $\mathbb{S}^1$. – Benoît Kloeckner Jan 07 '22 at 12:59

0 Answers0