# Lipschitz constant of exponential map

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.

• Yes, the exponential map is smooth and therefore is Lipschitz on any compact domain. Jul 10, 2019 at 20:02
• 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.) Jul 10, 2019 at 20:10
• 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. Jul 10, 2019 at 21:13
• 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. Jul 10, 2019 at 21:33
• $\mathrm{T}$ 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. Jul 11, 2019 at 20:38

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.
• @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. Jul 13, 2019 at 7:59
• 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? Jul 27, 2019 at 16:55
• 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 $\exp_p(v_2)$? Jul 28, 2019 at 14:15
$$\newcommand{\norm}{\lVert#1\rVert}$$ $$\newcommand{\lin}{\mathsf{L}}$$ $$\newcommand{\et}{\mathsf{T}_{#1}}$$ $$\newcommand{\ft}{\mathsf{T}}$$ $$\newcommand{\bxr}{\mathbb{U}(#1,#2)}$$ $$\newcommand{\Ck}{\mathsf{C^{#1}}}$$ 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.