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Let $(M,g)$ be a Riemannian manifold, geodesically complete, and assume logarithms are well defined and smooth.

Let $c: I\to M $ be a smooth path in $M$, and $x\in M$. Can we say something about $$\Vert\nabla_{\dot{c}(t)}\log_{c(t)}(x)\Vert_{c(t)} ?$$

I can easily prove that $$\Vert\nabla_{\dot{c}(t)}\log_{x}(c(t))\Vert_{x}=\Vert \dot{c}(t)\Vert_{c(t)},$$ by differentiating the equality $$\exp_{x}(\log_x(c(t)))=c(t),$$ and using the Gauss lemma to justify that $d(\exp_x)_{\log(c(t))}:T_x M\to T_{c(t)}M$ is a radial isometry.

Is it also true that $$\Vert\nabla_{\dot{c}(t)}\log_{c(t)}(x)\Vert_{c(t)}=\Vert \dot{c}(t)\Vert_{c(t)}?$$

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I believe the answer to your question is "no". I am not familiar with the term "Riemannian logarithmic map," but I imagine you mean the inverse of the Riemannian exponential map. Your condition is thus that the exponential map from any point of $M$ is a diffeomorphism (in particular this implies that $M$ is diffeomorphic to $\mathbb{R}^n$).

Since the exponential map from $x$ is a diffeomorphism, there exist geodesic normal co-ordinates about $x$ for the whole manifold, and so the corresponding distance function $r=d(x,\cdot)$ is smooth on the whole manifold (except at $x$).

At each point $y$ in $M\setminus\{x\}$, one has that $\operatorname{exp}_y(-r\nabla r) =x$. So, in your terminology, $\operatorname{log}_y(x)=-r\nabla r$.

You ask essentially whether, at each point $c$ in $M$, for each vector $v$ in $T_cM$, $$\lVert\nabla_{v}(-r\nabla r)\rVert_c=\lVert v\rVert_c;$$ equivalently whether the map from $T_cM$ to $T_cM$, $$v\mapsto \nabla_v(-r\nabla r),$$ is always an isometry (of vector spaces). In general it is not. For example, in the hyperbolic plane, if I calculate correctly, $$\nabla_{\frac{\partial}{\partial\theta}}(r\nabla r)=r\coth(r)\frac{\partial}{\partial\theta}.$$

This is closely related to the famous question of the "Hessian of the square of the distance function" (since $\nabla (-rdr)=-\tfrac{1}{2}\operatorname{Hess}(r^2)$), which is equal to the identity only in flat manifolds.

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  • $\begingroup$ Thanks a lot for your help! Ok, not an isometry... but can we have $\Vert\nabla_v(-r\nabla r)\Vert_c \leq \alpha \Vert v\Vert_c$ or $\Vert\nabla_v(-r\nabla r)\Vert_c \geq \alpha \Vert v\Vert_c$, for some $\alpha>0$, (ideally $\leq$ with $\alpha\leq 1$) ? Feel free to make assumptions on the curvature, like a lower or upper bound $\endgroup$
    – tisydi
    Dec 13, 2018 at 22:04
  • $\begingroup$ Yes, certainly. Have a look at Petersen's book on Riemannian geometry, the subsection entitled "Basic comparison estimates" (section 6.5 in my edition). Letting $A:T_cM\to T_cM$ be the operator $A(v):=\nabla_v(-r\nabla r)$, one has, eg, $||Av||\leq ||v||$ if the curvature is nonnegative (which BTW will often conflict with having $\exp$ invertible); $||Av||\geq ||v||$ if the curvature is nonpositive; $||Av||\leq r \coth(r)||v||$ if the curvature is everywhere at least $-1$. $\endgroup$
    – macbeth
    Dec 14, 2018 at 18:21

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