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Two Questions: (1) Under what conditions(if any) can the logarithm map from a point on a Riemannian manifold, $q_1\in Q$, to the Tangent Space $T_{q_0}Q$, locally, be a contraction mapping?

Or more generally, in terms of a Lipschitz constant,

(2) Given a flow on TQ and the canonical projection $\pi_Q$, if $\pi_Q\circ\Phi_h(q_0,v_0)=q_1$ and $\pi_Q\circ\Phi_h(q_0,\tilde{v}_0)=\tilde{q}_1$, then $exp(v_0)=q_0$ and $exp(\tilde{v}_0)=\tilde{q}_1$. Does there exist a constant depending on h, $C_h$, such that

$\|v_0-\tilde{v}_0\|=\|log(q_1)-log(\tilde{q}_1)\|\leq C_h\|q_1-\tilde{q}_1\|$

if we restrict ourselves to a local neighbourhood of $q_0$, so that the exponential map is injective?

If the exponential map is defined locally and the domain is shrunk it becomes arbitrarily close to an isometry. Is there a way to measure how close it is to an isometry given the domain has been shrunk to a particular size?

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    $\begingroup$ What are $TQ$ and $Q$? It seems to me that $Q$ is a Riemannian manifold and $TQ$ its tangent bundle. In this case your first example is wrong, since the exponential map on a manifold homeomorphic to $\mathbb R$ is always isometric. The logarithm is (typically) only locally defined and arbitrarily close to being an isometry when the domain is shrinked. If your logarithm means something else, these remarks don't apply, but you could make your question clearer. $\endgroup$
    – Joonas Ilmavirta
    Commented Oct 4, 2014 at 19:40
  • $\begingroup$ Thanks for the feedback. Q is a Riemannian manifold and TQ it's tangent bundle. In particular, we are looking at the logarithm map in a neighborhood around $q_0$ where it is injective. I'll edit my post. $\endgroup$
    – the42nd
    Commented Oct 4, 2014 at 22:00
  • $\begingroup$ Correction: "..we are looking at the logarithm map in a neighbourhood around $q_0$, where the exponential map is injective." $\endgroup$
    – the42nd
    Commented Oct 4, 2014 at 22:13
  • $\begingroup$ You still haven't corrected the wrong example noticed by Joonas. On $\mathbb{R}$ the logarithm is not contracting, you seem to be confusing between the "usual" logarithm and local inverses of the exponential map for Riemannian manifolds, which, in the case of $\mathbb{R}$, is not the logarithm map. $\endgroup$
    – YCor
    Commented Oct 4, 2014 at 22:29
  • $\begingroup$ Yes, I was hoping to gain some insight from what appears to be a bad analogy. I've deleted that part. $\endgroup$
    – the42nd
    Commented Oct 4, 2014 at 23:06

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I interpret the question to be intuitively

the map of a small piece of a manifold back to its tangent distorts distances

expanding for positive curvature and contracting for negative curvature

and neither for zero curvature

so the answer intuitively sought is : for negatively curved spaces the inverse of the exp map is contracting

dennis sullivan

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  • $\begingroup$ Thank you for the insight. That definitely helps intuition wise, and seems reasonable. So perhaps I can get a measure on how close it is to an isometry by looking at the curvature near $q_0$ and then taking into account the distance from $q_0$ to $q_1$ $\endgroup$
    – the42nd
    Commented Oct 5, 2014 at 18:55

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