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I asked before this question on MSE but I was not able to work out the details on my own.

Suppose $M$ is a smooth compact Riemannian manifold, take $p \in M$ and consider the map $$ T_pM \ni v \mapsto \exp_p(tv)\in M $$ where $t \in (0, \text{inj}(M))$ is a fixed parameter and $\text{inj}(M)$ is the (positive) injectivity radius of $M$.

Is is true that this map in Lipschitz uniformly in $p$? More precisely, is it possibile to prove that there exist $\delta>0$ and $C>0$ s.t. $$d(\exp_p(tv_1), \exp_p(tv_2)) \le Ct \|v_1-v_2\|_p $$

for every $v_1, v_2 \in T_pM$, $p \in M$ and $t \in (0, \delta)$? Clearly with $d$ I mean the Riemannian distance on $M$ and with $\| \cdot \|_p$ the norm in $T_pM$ induced by the Riemannian metric.

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    $\begingroup$ Yes, the exponential map is smooth and therefore is Lipschitz on any compact domain. $\endgroup$ – Anton Petrunin Jul 10 '19 at 20:02
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    $\begingroup$ The parameter $t$ is irrelevant --- your question is equivalent to the following inequality $$d(\exp_pv_1,\exp_pv_2)\le C\cdot|v_1-v_2|.$$ If $v_1$ and $v_2$ lie in a compact domain, then, as I said, the inequality follow from smoothness of $\exp_p$. BUT if $v_1$ and $v_2$ are arbitrary tangent vectors, then there is no such constant --- an example is a cone with slightly smoothed vertex and the point $p$ slightly aside. (Sorry I did not read the question carefully.) $\endgroup$ – Anton Petrunin Jul 10 '19 at 20:10
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    $\begingroup$ The constant $C$ can not be fixed --- consider Lobachevsky plane with constant curvature near $-\infty$. If you have a lower bound on curvature, say $\ge -1$ then yes --- you can assume that $C_p=\sinh r$ in $B_r(0)\subset \mathrm{T}_p$ if $r<\textrm{inj}M$. The latter follows from Toponogov comparison. $\endgroup$ – Anton Petrunin Jul 10 '19 at 21:13
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    $\begingroup$ No, for fixed manifold and radius there is a Lipschitz constant, but there is no way to make this constant fixed for all manifolds without additional assumption. $\endgroup$ – Anton Petrunin Jul 10 '19 at 21:33
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    $\begingroup$ $\mathrm{T}$ has a metric induced by $g$. For the previous question: actually it depends on $p$, but if the manifold is compact then you can choose one $C$ for all points. Another thing $\exp_p$ is smooth even behind the injectivity radius so no need to use it unless you want to apply Toponogov. $\endgroup$ – Anton Petrunin Jul 11 '19 at 20:38
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I believe that the answer is positive. Let me consider an autonomous differential equation $\dot x=f(x)$ and let us assume that $f$ is Lipschitz-continuous. The flow $\phi(t,y)$ is defined by $$ \dot \phi(t, y) = f\bigl(\phi(t,y)\bigr), \quad \phi(0, y)=y. $$ As a result, we have $ \phi(t, y_2)-\phi(t, y_1)=y_2-y_1+\int_0^t\left( f(\phi(s,y_2))-f(\phi(s,y_2)) \right) ds, $ so that, at least in a coordinate chart, $$ \rho(t)=\Vert\phi(t, y_2)-\phi(t, y_1)\Vert\le \Vert y_2-y_1\Vert+ C_{\text{Lip}}\int_0^t\Vert\phi(s, y_2)-\phi(s, y_1)\Vert ds=R(t), $$ and then $ \dot R=C_{\text{Lip}} \rho\le C_{\text{Lip}} R. $ Gronwall's inequality implies $$ \Vert\phi(t, y_2)-\phi(t, y_1)\Vert\le \underbrace{R\le R(0) e^{C_{\text{Lip}} t}}_{ \text{follows from Gronwall}}=\Vert y_2-y_1\Vert e^{C_{\text{Lip}} t}, $$ proving that the flow is Lipschitz-continuous with an estimate of the Lipschitz constant of the flow by the Lipschitz constant of the flux $f$ and time. Checking a linear scalar ODE proves that this estimate is essentially optimal. There are variants of this argument in the non-autonomous case.

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  • $\begingroup$ Thank you for your answer. I don't think I completely understand it. How is it connected to my question? $\endgroup$ – Bremen000 Jul 13 '19 at 7:15
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    $\begingroup$ @Bremen000 $exp_p(tv)$ is the solution of a second-order differential equation with initial data $p,v$. Since a second-order ODE becomes first-order by increasing the dimension, your question is "How do the solution of my ODE depend on the initial data?" I gave above a way to get Lipschitz-continuity wrt to the initial data from Lipschitz continuity of the flux. $\endgroup$ – Bazin Jul 13 '19 at 7:59
  • $\begingroup$ Thank you, now I see. The last part should be to express this in terms of the Riemannian distance on $M$. How can I do that? $\endgroup$ – Bremen000 Jul 27 '19 at 16:55
  • $\begingroup$ I mean, now I know that there exists a constant $C=C(M)>0$ s.t. $$ \| \varphi(\exp_p(v_1)) - \varphi(\exp_p(v_2)) \| \le C \|v_1 - v_2\| $$ for every $p \in M$ and for every $v_1, v_2 \in T_p M$ sufficiently small. I used the notation $\varphi$ to indicate a coordinate chart around $p$. Now, how can I express the LHS w.r.t. the Riemannian distance between $\exp_p(v_1)$ and $\exp_p(v_2)$? $\endgroup$ – Bremen000 Jul 28 '19 at 14:15
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$\newcommand{\norm}[1]{\lVert#1\rVert}$ $\newcommand{\lin}{\mathsf{L}}$ $\newcommand{\et}[1]{\mathsf{T}_{#1}}$ $\newcommand{\ft}{\mathsf{T}}$ $\newcommand{\bxr}[2]{\mathbb{U}(#1,#2)}$ $\newcommand{\Ck}[1]{\mathsf{C^{#1}}}$ By compacity, we may take $\delta>0$ such that the exponential is defined on the compact subset of $\ft M$ given by $K:=\{v\in \ft M| \,\lVert v\rVert\leq\delta\}$. The map defined on the domain of the exponential by $v_p\mapsto \norm{(\exp_p)_{\ast v_p}}_{\lin(\et{p}M,\et{\exp_p v_p}M)}$ (i.e. the operator norm of the tangent map to $\exp_p$ at $v_p$) is continuous, hence its $\sup$ on $K$, say $C$, is finite. Then, for each $p\in M$ and for each $v_p,w_p$ in the open ball $\bxr{0_p}{\delta}\subset\et{p}M$, we have $$d(\exp_p v_p,\exp_p w_p)\leq C\norm{v_p - w_p}_p.$$

To see this, take any sectionally $\Ck{1}$ curve $\gamma$ on $\bxr{0_p}{\delta}\subset\et{p}M$ joining $v_p$ to $w_p$. Then $\Gamma:=\exp_p\circ\gamma$ is a sectionally $\Ck{1}$ curve on $M$ joining $\exp_p v_p$ to $\exp_p w_p$, whose length $\ell(\Gamma)$ is $\leq C\ell(\gamma)$. Thus, $d(\exp_p v_p,\exp_p w_p)\leq\ell(\Gamma)\leq C\ell(\gamma)$. Taking the infimum over the set of sectionally $\Ck{1}$ curves on $\bxr{0_p}{\delta}$ joining $v_p$ to $w_p$ yields the asserted inequality.

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